# NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism

Chapter 4 Physics Class 12 Moving Charges and Magnetism explains concepts like a magnetic field, magnets, behaviour of charges, magnetic force and many others. This chapter will help students understand the science behind magnets and their functioning. The Class 12 Physics Chapter 4 NCERT Solutions is exclusively available for students appearing for CBSE exams and competitive exams like JEE or NEET. Every solution set consists of stepwise explanations in simple language that students can refer to for an easy read and thorough understanding.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 4

Here’s a detailed breakup of topics under NCERT Solutions Class 12 Physics Chapter 4

 Ex 4.1 Introduction Ex 4.2 Magnetic Force Ex 4.2.1 Sources and Fields Ex 4.2.2 Magnetic Field, Lorentz Force Ex 4.2.3 Magnetic force on a current-carrying conductor Ex 4.3 Motion in a magnetic field Ex 4.4 Motion in Combined Electric and Magnetic fields Ex 4.4.1 Velocity Selector Ex 4.4.2 Cyclotron Ex 4.5 Magnetic field due to a current element, Biot-savart law Ex 4.6 Magnetic Field on the Axis Of A Circular Current Loop Ex 4.7 Ampere’s Circuital Law Ex 4.8 The Solenoid and the Toroid Ex 4.8.1 The solenoid Ex 4.8.2 The Toroid Ex 4.9 Force Between Two Parallel Currents, The Ampere Ex 4.10 Torque on Current Loop, Magnetic Dipole Ex 4.10.1 Torque on a rectangular current loop in a uniform magnetic field Ex 4.10.2 Circular current loop as a magnetic dipole Ex 4.10.3 The magnetic dipole moment of a revolving electron Ex 4.11 The moving coil galvanometer

4.1 Introduction

The first topic under NCERT Solutions Class 12 Physics Chapter 4 is an introduction to the magnetic field. The magnetic field imposes a force on moving charges. Moving charges are perpendicular to the plane.

4.2 Magnetic Force

The magnetic force is an effect of electromagnetic force and is one of the four fundamental forces of nature. This force results in the motion of charges. Any two objects containing a charge with the same direction of motion have a magnetic attraction force among them. Also, objects with charges moving in opposite directions have a repulsive force among them.

4.2.1 Sources and Fields

In this subunit of NCERT solutions class 12 Physics chapter 4.2, students will learn about the cause of magnetism. The motion of the electrical charge causes magnetism. Substances consist of atoms, which have protons, electrons and neutrons. The magnetic field describes the interaction between moving electric charges and magnetic materials. A charge moving in any magnetic field experiences a perpendicular force.

4.2.2 Lorentz Force

Lorentz forces the force exerted on a charged particle q moving with velocity v through an electric field E and magnetic field B. The electromagnetic force F on the charged particle is called the Lorentz force (after the Dutch physicist Hendrik A. Lorentz) and is given by F = QE + qv × B.

4.2.3 Magnetic forces on a current-carrying conductor

The NCERT solutions class 12 Physics chapter 4.2.3 explain magnetic forces. An electric current flowing via a conductor produces a magnetic field. The field exerts a force on a magnet located within the conductor’s area. The magnet additionally exerts an equal and opposite force at the current-carrying conductor.

4.3 Effect of Motion in a magnetic field

NCERT solutions class 12 Physics Chapter 4.3 describes the effects of motion in a magnetic field. The velocity component, perpendicular to the magnetic field, creates circular movement. In contrast, the velocity component, which is parallel to the field, moves the particle alongside a straight line. The pitch is the horizontal distance between consecutive circles. The ensuing motion is a spiral.

4.4 Effect of Motion in Combined Electric and Magnetic fields

When a moving charge comes under the effect of magnetic and electric fields, it experiences Lorentz force. Lorentz force is calculated as the vector sum of forces created by magnetic and electric fields. F = F(electric) + F(Magnetic) = q (E = v x B)

Students may refer to NCERT solutions class 12 Physics chapter 4.4 to know more.

4.4.1 Velocity Selector

A Wien filter, also called a velocity selector, is a tool such as perpendicular electric or magnetic fields, used as a velocity filter for charged particles, as an example in electron microscopes and spectrometers. It is utilized in accelerator mass spectrometry to select particles based on their speed.

4.4.2 Cyclotron

NCERT solutions class 12 Physics chapter 4.4.2 describes cyclotron and its effects on the magnetic field. A cyclotron is a compact particle accelerator that produces radioactive isotopes used for imaging procedures. Stable, non-radioactive isotopes are positioned into the cyclotron, which accelerates charged particles (protons) to excessive energy in a magnetic field.

4.5 Effect of Magnetic Field due to a current element, Biot-savant law

A small current-carrying conductor of length dl carrying a current I is an elementary source of a magnetic field. The force on some other comparable conductor can be expressed quite simply in phrases of magnetic field dB because of the first. The dependence of the magnetic field was first identified using Biot and Savart.

4.6 The Effect of Magnetic Field on the Axis Of A Circular Current Loop

In this section of NCERT solutions class 12 Physics chapter 4, students learn about the effect of the magnetic field on the axis of a circular current loop. Magnetic field dB at point P due to current element idl, making a right angle to the line joining point P and current element, will be given by Biot-Savart law as:

dB = (μo/4π)idl sin(90°)/r2 = (μo/4π)idl/r2

4.7 Ampere’s Circuital Law

NCERT solutions class 12 physics chapter 4.7 describes Ampere’s Law and Ampere’s Circuital Law.

Ampere’s law states that “The magnetic field produced by an electric current is proportional to the size of that electric current, given  that a constant of proportionality equal to the permeability of free space is defined.”

Ampere’s Circuital Law states that the line integral of the magnetic field surrounding the closed loop equals the number of times the algebraic sum of currents passing through the loop.

H.dL=Ienc

Suppose a conductor carries a current I, then the current flow generates a magnetic field surrounding the wire. The equation’s left side illustrates an imaginary path encircling the wire, and the magnetic field is added at every point.

Students may refer to NCERT solutions class 12 Physics Chapter 4.7 for more information on Ampere’s law.

4.8 The Solenoid and the Toroid

4.8.1 The solenoid

A coil of wire designed to generate a strong magnetic field inside the coil is referred to as a solenoid. Wrapping the same wire around a cylinder creates a strong magnetic field whilst an electric current is passed. The number of turns of a solenoid is denoted by N.

B=0NIl

Here, B=solenoid magnetic flux density

0=magnetic constant

N=number of turns

I=current

l=length of the solenoid

4.8.2 The Toroid

A toroid is a hollow circular ring in the shape of a doughnut, with many turns of enamelled wire, wound so close that there is little space between the two turns.

4.9 Two Parallel Currents & their forces

Suppose one Ampere of current passes through each of two parallel conductors of infinite length, separated by one meter in space free of other magnetic fields. It causes a force of exactly two × 10−7 N/m on each conductor. To know more, students may refer to NCERT Solutions Class 12 Physics Chapter 4.

4.10 Effect of Torque on Current Loop, Magnetic Dipole

A twisting force that causes the rotation of objects is called torque. The point where an object rotates is the axis of rotation. Students will learn about the effect of torque on current loop and magnetic dipole under NCERT solutions class 12 Physics chapter 4.10.

A magnetic dipole is generally a tiny magnet of microscopic to subatomic dimensions, equivalent to a flow of electric charge around a loop. Electrons are magnetic dipoles that circulate atomic nuclei; electrons spin on their axes and rotate positively charged atomic nuclei.

4.10.1 Torque on a rectangle-shaped current loop in a uniform magnetic field

qzIf a rectangular loop carrying a steady current is located in a uniform magnetic field, it will experience a torque. The total force on the loop will be zero. Students may refer to NCERT solutions class 12 Physics chapter 4.10.1 to know more.

4.10.2 Effect of Circular current loop as a magnetic dipole

The field lines resemble that of a magnet. Hence, it behaves as a magnetic dipole. Its poles can be found from its field lines (Field lines go away from the magnet from the north pole and enter from the south pole).

4.10.3 Effect of the magnetic dipole moment of a revolving electron

The value of eh/4πm is referred to as the Bohr magneton. In addition to the magnetic moment, the electron possesses a magnetic moment because of its spin. The resultant magnetic moment can be stated as the vector sum of the orbital and spin magnetic moment.

4.11 The moving coil galvanometer

It is a device used to measure electric currents. It is a sensitive electromagnetic device that may measure low currents even of the order of some microamperes.

Students may click on the respective chapter names to access a more detailed NCERT solutions class 12 Physics chapter 4.

### List of NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism

Students may click on the below links to view NCERT Solutions Class 12 Physics Chapter 4:

• Chapter 4: Exercise 4.1 Solutions
• Chapter 4: Exercise 4.2.1 Solutions
• Chapter 4: Exercise 4.2.2 Solutions
• Chapter 4: Exercise 4.2.3 Solutions
• Chapter 4: Exercise 4.3 Solutions
• Chapter 4: Exercise 4.4.1 Solutions
• Chapter 4: Exercise 4.4.2 Solutions
• Chapter 4: Exercise 4.5 Solutions
• Chapter 4: Exercise 4.6 Solutions
• Chapter 4: Exercise 4.7 Solutions
• Chapter 4: Exercise 4.8.1 Solutions
• Chapter 4: Exercise 4.8.2 Solutions
• Chapter 4: Exercise 4.9 Solutions
• Chapter 4: Exercise 4.10.1 Solutions
• Chapter 4: Exercise 4.10.2 Solutions
• Chapter 4: Exercise 4.10.3 Solutions
• Chapter 4: Exercise 4.11 Solutions

Students may click on the respective exercise of NCERT solutions class 12 Physics chapter 4 to access the solution provided by Extramarks. In addition, they can also access NCERT Solutions for different classes in the links below.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

### Key Features of NCERT Solutions Class 12 Physics Chapter 4

Physics Chapter 4 Class 12 covers information about the magnetic field, magnetic force, and charges’ behaviours. In exams, numerical problems may be asked from this section. All the NCERT solutions class 12 Physics chapter 4 comprise a detailed explanation with diagrams, graphs and derivations for formulae. Class 12 Physics Chapter 4 NCERT Solutions are prepared by Subject Experts at Extramarks. All solutions are in accordance with  CBSE guidelines, and every question is answered according to the weightage of marks.

Q.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Ans.

$\begin{array}{l}\text{Distance of the point from the wire},\text{r}=\text{2}0\text{cm}=0.\text{2 m}\\ \text{Magnitude of magnetic field at this point is given by}\text{\hspace{0.17em}}\\ \text{the}\text{\hspace{0.17em}}\text{relation}\\ \text{B=}\frac{{\mu }_{0}}{4\pi }\frac{2I}{r}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =\text{4}\pi ×\text{1}{0}^{–\text{7}}{\text{Tm A}}^{–\text{1}}\\ B=\frac{4\pi ×\text{1}{0}^{–\text{7}}×2×35}{4\pi ×0.2}=3.5×\text{1}{0}^{–5}\text{\hspace{0.17em}}T\\ \therefore \text{The magnitude of the magnetic field at a point}\\ \text{2}0\text{cm from the wire is 3}.\text{5}×\text{1}{0}^{–\text{5}}\text{\hspace{0.17em}}\text{T}.\end{array}$

Q.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Ans.

$\begin{array}{l}\text{Distance of the point from the wire},\text{r}=\text{2}0\text{cm}=0.\text{2 m}\\ \text{Magnitude of magnetic field at this point is given by}\text{\hspace{0.17em}}\\ \text{the}\text{\hspace{0.17em}}\text{relation}\\ \text{B=}\frac{{\mu }_{0}}{4\pi }\frac{2I}{r}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =\text{4}\pi ×\text{1}{0}^{–\text{7}}{\text{Tm A}}^{–\text{1}}\\ B=\frac{4\pi ×\text{1}{0}^{–\text{7}}×2×35}{4\pi ×0.2}=3.5×\text{1}{0}^{–5}\text{\hspace{0.17em}}T\\ \therefore \text{The magnitude of the magnetic field at a point}\\ \text{2}0\text{cm from the wire is 3}.\text{5}×\text{1}{0}^{–\text{5}}\text{\hspace{0.17em}}\text{T}.\end{array}$

Q.3

$\begin{array}{l}\text{A long straight wire in the horizontal plane carries a current of 50 A in north to south}\\ \text{direction. Give the magnitude and direction of}\stackrel{\to }{\mathrm{B}}\text{at a point 2.5 m east of the wire.}\end{array}$

Ans.

$\begin{array}{l}\text{Here, \hspace{0.17em}current flowing\hspace{0.17em}through the wire, I = 50 A}\\ \text{Distance of the point from the wire, r = 2.5 m\hspace{0.17em} east\hspace{0.17em}of\hspace{0.17em}wire}\\ \text{Magnitude of magnetic field at that point is given as:}\\ \text{B =}\frac{{\text{μ}}_{\text{0}}\text{2I}}{\text{4}\mathrm{\pi }\text{r}}\\ {\text{Here,\hspace{0.17em}μ}}_{\text{0}}\text{= 4}\mathrm{\pi }{\text{×10}}^{\text{–7}}{\text{TmA}}^{\text{–1}}\\ \text{B =}\frac{\text{4}\mathrm{\pi }{\text{×10}}^{\text{–7}}{\text{TmA}}^{\text{-1}}\text{×2×50 A}}{\text{4}\mathrm{\pi }\text{×2.5 m}}{\text{= 4×10}}^{\text{–6}}\text{\hspace{0.17em}T}\\ \text{The point is located normal to the plane\hspace{0.17em}of\hspace{0.17em}wire at a distance of 2.5 m.}\\ \text{The current flowing\hspace{0.17em}through the wire is\hspace{0.17em}directed vertically downwards.}\\ \therefore \text{According to the Maxwell’s\hspace{0.17em}right hand thumb rule, the direction of the magnetic field}\\ \text{at the given point is vertically upward.}\end{array}$

Q.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Ans.

$\begin{array}{l}\text{Current carried}\text{\hspace{0.17em}}\text{by the power line},\text{I}=\text{9}0\text{A}\\ \text{Point lies below the power line at distance},\text{r}=\text{1}.\text{5 m}\\ \therefore \text{Magnetic field at that point is given by the relation},\\ B=\frac{{\mu }_{0}2I}{4\pi r}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =\text{4}\pi ×\text{1}{0}^{–\text{7}}{\text{TmA}}^{–\text{1}}\\ B=\frac{\text{4}\pi ×\text{1}{0}^{–\text{7}}×2×90}{\text{4}\pi ×1.5}=1.2×\text{1}{0}^{–5}\text{\hspace{0.17em}}T\\ \text{Since}\text{\hspace{0.17em}}\text{the current is flowing from East to West}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}t\text{he}\\ \text{point is below the power line,}\\ \therefore \text{According to Maxwell}’\text{s right hand thumb rule},\text{}\\ \text{the}\text{\hspace{0.17em}}\text{direction}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{magnetic}\text{\hspace{0.17em}}\text{field}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{towards}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{south}\text{.}\end{array}$

Q.5 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west.
A wire carrying current of 7.0 A in the north to south direction passes through this region.
What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}strength\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{m}\text{agnetic field},\text{B}=\text{1}.\text{5 T}\\ \text{Radius of cylindrical region},\text{r}=\text{1}0\text{cm}=\text{}0.\text{1 m}\\ \text{Current flowing\hspace{0.17em}through the wire passing through the}\\ \text{cylindrical region},\text{I}=\text{7 A}\\ \mathrm{}\left(\text{a}\right)\text{If the wire intersects the axis},\text{then the length of}\\ \text{the wire is the diameter of the cylindrical region}.\\ \text{Thus},\text{l}=\text{2r}=\text{}0.\text{2 m}\\ \text{Angle between current and magnetic field},\text{}\mathrm{\theta }\text{}=\text{9}0\mathrm{°}\\ \text{Magnetic force acting on the wire is given as:}\\ \text{F}=\text{BIl sin}\mathrm{\theta }=\text{1}.\text{5}×\text{7}×\text{}0.\text{2}×\text{sin 9}0\mathrm{°}=\text{2}.\text{1 N}\\ \therefore \text{A force of 2}.\text{1 N acts on the wire vertically}\\ \text{downwards}.\\ \left(\text{b}\right)\text{New length of the wire after turning it to the}\\ \text{North\hspace{0.17em}east}-\text{North\hspace{0.17em}west direction\hspace{0.17em}is\hspace{0.17em}given as}:\\ {\text{l}}_{\text{1}}=\frac{\mathrm{l}}{\text{sin}\mathrm{\theta }}\\ :\text{Angle between magnetic field and current},\text{}\mathrm{\theta }=\text{45}\mathrm{°}\\ \text{Force acting\hspace{0.17em}on the wire\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by},\\ \text{F}={\text{BIl}}_{\text{1}}\text{sin}\mathrm{\theta }\\ =\text{BIl}\\ =1.5×7×0.2=2.1\text{\hspace{0.17em}}\mathrm{N}\\ \therefore \text{A force of 2}.\text{1 N acts on the wire\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}vertically}\\ \text{downward direction\hspace{0.17em}}.\text{}\\ \left(\text{c}\right)\text{The wire in\hspace{0.17em}the\hspace{0.17em}N-S\hspace{0.17em}direction\hspace{0.17em}is lowered from the}\\ \text{axis by distance},\text{d}=\text{6}.0\text{cm}\\ \mathrm{Suppose}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{wire}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{passing}\text{\hspace{0.17em}}\mathrm{normally}\text{\hspace{0.17em}}\mathrm{to}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{axis}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\\ \mathrm{cylinderical}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{field},\text{\hspace{0.17em}}\mathrm{then}\text{\hspace{0.17em}}\mathrm{lowering}\text{\hspace{0.17em}}6\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{means}\\ \mathrm{displacing}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{wire}\text{\hspace{0.17em}}6\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{from}\text{\hspace{0.17em}}\mathrm{its}\text{\hspace{0.17em}}\mathrm{initial}\text{\hspace{0.17em}}\mathrm{position}\text{\hspace{0.17em}}\mathrm{towards}\\ \mathrm{to}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{end}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{area}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{cross}-\mathrm{section}.\end{array}$ $\begin{array}{l}\mathrm{From}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{figure},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ \text{\hspace{0.17em}}\mathrm{x}=\sqrt{{10}^{2}-{6}^{2}}=8\text{\hspace{0.17em}}\mathrm{cm}\\ \therefore \mathrm{Length}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{wire}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{field},\text{\hspace{0.17em}}{\mathrm{l}}_{2}=2\mathrm{x}=2×8\\ =16\text{\hspace{0.17em}}\mathrm{cm}\text{= 0.16 m}\\ \mathrm{Force}\text{\hspace{0.17em}}\mathrm{on}\text{\hspace{0.17em}}\mathrm{wire},\text{\hspace{0.17em}}\mathrm{F}={\mathrm{il}}_{2}\mathrm{Bsin}{90}^{0}\\ \mathrm{F}=7×0.16×1.5=1.68\text{\hspace{0.17em}}\mathrm{N}\\ \mathrm{According}\text{\hspace{0.17em}}\mathrm{to}\text{\hspace{0.17em}}\mathrm{screw}\text{\hspace{0.17em}}\mathrm{rule},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{direction}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{this}\text{\hspace{0.17em}}\mathrm{force}\text{\hspace{0.17em}}\mathrm{is}\\ \mathrm{vertically}\text{\hspace{0.17em}}\mathrm{downwards}.\\ \therefore \text{A force of 1}.\text{68 N acts\hspace{0.17em}on the wire in the\hspace{0.17em}vertically}\\ \text{downward direction}.\end{array}$

Q.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}l\text{ength of wire},\text{l}=\text{3 cm}=\text{}0.0\text{3 m}\\ \text{Current flowing through the wire},\text{I}=\text{1}0\text{A}\\ \text{Magnitude}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{}0.\text{27 T}\\ \text{Angle between current and magnetic field},\text{}\theta =\text{9}0°\\ \text{Magnetic force exerted on the wire is given by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\\ \text{relation,}\\ \text{F}=\text{BIlsin}\theta =\text{}0.\text{27}×\text{1}0×0.0\text{3 sin9}0°\\ =\text{8}.\text{1}×\text{1}{0}^{–\text{2}}\text{N}\\ \therefore \text{The magnetic force on the wire is 8}.\text{1}×\text{1}{0}^{–\text{2}}\text{N}.\text{}\end{array}$

Q.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Ans.

$\begin{array}{l}\text{Current flowing through wire A},{\text{I}}_{\text{A}}\text{}=\text{8}.0\text{A}\\ \text{Current flowing through wire B},{\text{I}}_{\text{B}}\text{}=\text{5}.0\text{A}\\ \text{Distance between the wires},\text{r}=\text{4}.0\text{cm}=0.0\text{4 m}\\ \text{Length of the section of wire A},\text{l}=\text{1}0\text{cm}=\text{}0.\text{1 m}\\ \text{Force exerted on length l due to the magnetic field is}\\ \text{given by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation,}\\ \text{B=}\frac{{\mu }_{0}2{I}_{A}{I}_{B}l}{4\pi r}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =\text{4}\pi ×\text{1}{0}^{–\text{7}}{\text{TmA}}^{–\text{1}}\\ \therefore \text{B=}\frac{\text{4}\pi ×\text{1}{0}^{–\text{7}}×2×8×5×0.1}{\text{4}\pi ×0.04}\\ =\text{2}×\text{1}{0}^{–\text{5}}\text{N}\\ \text{This is an attractive force normal to A towards B}\text{.This}\text{\hspace{0.17em}}\\ \text{is}\text{\hspace{0.17em}}\text{because, the direction of current in the both}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}the\\ \text{wires is same}.\end{array}$

Q.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}l\text{ength of solenoid},\text{l}=\text{8}0\text{cm}=\text{}0.\text{8 m}\\ \text{Since}\text{\hspace{0.17em}}t\text{here are}\text{\hspace{0.17em}}\text{5 layers of windings of 4}00\text{turns}\\ \text{each on the solenoid,}\\ \therefore \text{Total number of turns}\text{\hspace{0.17em}}\text{on}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}\text{solenoid},\text{N}=\text{5}×\text{4}00\text{}\\ =\text{2}000\\ \text{Diameter of solenoid},\text{D}=\text{1}.\text{8 cm}=\text{}0.0\text{18 m}\\ \text{Current flowing}\text{\hspace{0.17em}}\text{through the solenoid},\text{I}=\text{8}.0\text{A}\\ \text{Magnitude of magnetic field inside the solenoid near}\\ \text{its centre is given as:}\\ \text{B=}\frac{{\mu }_{0}NI}{l}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =\text{4}\pi ×\text{1}{0}^{–\text{7}}{\text{TmA}}^{–\text{1}}\\ \text{B=}\frac{\text{4}\pi ×\text{1}{0}^{–\text{7}}×2000×8}{0.8}=8\pi ×\text{1}{0}^{–3}\\ =2.512×\text{1}{0}^{–2}\text{\hspace{0.17em}}T\\ \therefore \text{The magnitude of the magnetic field inside the}\\ \text{solenoid near its centre is 2}.\text{512}×\text{1}{0}^{–\text{2}}\text{T}.\end{array}$

Q.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30o with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}l\text{ength of one side of the square coil},\text{l}=\text{1}0\text{cm}\\ =\text{}0.\text{1 m}\\ \text{Current flowing through the coil},\text{I}=\text{12 A}\\ \text{Number of turns on the coil},\text{n}=\text{2}0\\ \text{Angle made by the plane of coil with uniform}\text{\hspace{0.17em}}\text{magnetic}\\ \text{field},\text{}\theta =\text{3}0°\\ \text{Magnitude of magnetic field},\text{B}=\text{}0.\text{8}0\text{T}\\ \text{Magnitude of the magnetic torque experienced by the}\\ \text{coil in the magnetic field is given}\text{\hspace{0.17em}}as:\\ \tau \text{}=\text{nBIAsin}\theta \\ \text{Area of square coil,}\text{\hspace{0.17em}}\text{A=l}×\text{l}=\text{}0.\text{1}×0.\text{1}=\text{}0.0{\text{1 m}}^{\text{2}}\\ \therefore \tau \text{}=\text{2}0\text{}×\text{}0.\text{8}×\text{12}×\text{}0.0\text{1}×\text{sin3}{0}^{o}=\text{}0.\text{96 Nm}\\ \therefore T\text{he magnitude of torque experienced by the coil}\text{\hspace{0.17em}}\text{is}\\ 0.\text{96 Nm}.\end{array}$

Q.10 Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30,
A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42,
A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

Ans.

$\begin{array}{l}\\ \text{In}\text{\hspace{0.17em}}\text{case}\text{\hspace{0.17em}}{\text{of coil meter M}}_{\text{1}}:\\ \text{Resistance},{\text{R}}_{\text{1}}=\text{1}0\text{}\Omega \text{\hspace{0.17em}};\text{\hspace{0.17em}}\text{Number of turns},{\text{N}}_{\text{1}}=\text{3}0\\ C\text{ross}-\text{sectional}\text{\hspace{0.17em}}\text{area},{\text{A}}_{\text{1}}\text{}=\text{3}.\text{6}×\text{1}{0}^{–\text{3}}{\text{m}}^{\text{2}}\\ \text{Strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},{\text{B}}_{\text{1}}\text{}=\text{}0.\text{25 T}\\ {\text{Spring constant K}}_{\text{1}}\text{}=\text{K}\\ \text{In}\text{\hspace{0.17em}}\text{case}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}coil{\text{meter M}}_{\text{2}}:\\ \text{Resistance},{\text{R}}_{\text{2}}\text{}=\text{14}\Omega \\ \text{Number of turns},{\text{N}}_{\text{2}}\text{}=\text{42}\\ \text{Cross}-\text{sectional}\text{\hspace{0.17em}}\text{area},{\text{A}}_{\text{2}}\text{}=\text{1}.\text{8}×\text{1}{0}^{–\text{3}}{\text{m}}^{\text{2}}\\ \text{Strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},{\text{B}}_{\text{2}}\text{}=\text{}0.\text{5}0\text{T}\\ \text{Spring constant},{\text{K}}_{\text{2}}\text{}=\text{K}\\ \left(\text{a}\right)\text{Current sensitivity of coil meter}\text{\hspace{0.17em}}{\text{M}}_{\text{1}}\text{is given as}:\\ {I}_{s1}=\frac{{\text{N}}_{\text{1}}{\text{B}}_{\text{1}}{\text{A}}_{\text{1}}}{{\text{K}}_{\text{1}}}\\ \text{Current sensitivity of coil meter}\text{\hspace{0.17em}}{\text{M}}_{\text{2}}\text{is given as}:\\ {I}_{s2}=\frac{{\text{N}}_{2}{\text{B}}_{2}{\text{A}}_{2}}{{\text{K}}_{2}}\\ \therefore \frac{{I}_{s2}}{{I}_{s1}}=\frac{{\text{N}}_{2}{\text{B}}_{2}{\text{A}}_{2}{\text{K}}_{\text{1}}}{{\text{N}}_{\text{1}}{\text{B}}_{\text{1}}{\text{A}}_{\text{1}}{\text{K}}_{2}}=\frac{\text{42}×0.\text{5}0×\text{1}.\text{8}×\text{1}{0}^{–\text{3}}×K}{K×30×0.25×3.6×{10}^{-3}}=1.4\\ \therefore {\text{The ratio of current sensitivity of M}}_{\text{2}}{\text{to M}}_{\text{1}}\text{is 1}.\text{4}.\\ \left(\text{b}\right)\text{Voltage sensitivity for meter}\text{\hspace{0.17em}}{\text{M}}_{\text{2}}\text{is given as}:\\ {V}_{s2}=\frac{{\text{N}}_{2}{\text{B}}_{2}{\text{A}}_{2}}{{\text{K}}_{2}{R}_{2}}\\ \text{Voltage sensitivity for}\text{\hspace{0.17em}}{\text{meter M}}_{\text{1}}\text{is given as}:\\ {V}_{s1}=\frac{{\text{N}}_{1}{\text{B}}_{1}{\text{A}}_{1}}{{\text{K}}_{1}{R}_{1}}\\ \therefore \frac{{V}_{s2}}{{V}_{s1}}=\frac{{\text{N}}_{2}{\text{B}}_{2}{\text{A}}_{2}{\text{K}}_{1}{R}_{1}}{{\text{N}}_{1}{\text{B}}_{1}{\text{A}}_{1}{\text{K}}_{2}{R}_{2}}=\frac{\text{42}×0.\text{5}0×\text{1}.\text{8}×\text{1}{0}^{–\text{3}}×10×K}{K×14×30×0.25×3.6×{10}^{-3}}=1\\ \therefore {\text{The ratio of voltage sensitivity of M}}_{\text{2}}{\text{to M}}_{\text{1}}\text{is 1}.\end{array}$

Q.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 106 m s–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e =1.6 × 10–19 C, me= 9.1×10–31 kg)

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{6}.\text{5 G}\\ =\text{6}.\text{5}×\text{1}{0}^{–\text{4}}\text{T}\\ \text{Speed of electron},\text{v}=\text{4}.\text{8}×\text{1}{0}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{Charge on electron},\text{e}=\text{1}.\text{6}×\text{1}{0}^{–\text{19}}\text{C}\\ \text{Mass of electron},{\text{m}}_{\text{e}}\text{}=\text{9}.\text{1}×\text{1}{0}^{–\text{31}}\text{kg}\\ \text{Angle between the shot electron and magnetic field},\text{}\theta \\ =\text{9}0°\\ \text{Magnetic force exerted on the electron in the magnetic}\\ \text{field is given by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation,}\\ \text{F}=\text{evB sin}\theta \\ \text{This force provides centripetal force to the moving}\\ \text{electron}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{the electron starts}\text{\hspace{0.17em}}\text{moving in a circular}\\ \text{path of radius r}.\\ \therefore \text{Centripetal force exerted on the electron}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by},\\ {\text{F}}_{\text{c}}\text{=}\frac{m{v}^{2}}{r}\\ \text{At equilibrium},\text{the centripetal force exerted on the}\\ \text{electron is equal to the magnetic}\text{\hspace{0.17em}}\text{force i}.\text{e}.,\\ {\text{F}}_{\text{c}}\text{=}F\\ \frac{m{v}^{2}}{r}=\text{evB sin}\theta \\ r=\frac{mv}{\text{Besin}\theta }\\ =\frac{\text{9}.\text{1}×\text{1}{0}^{–\text{31}}×\text{4}.\text{8}×\text{1}{0}^{\text{6}}}{\text{6}.\text{5}×\text{1}{0}^{–\text{4}}×\text{1}.\text{6}×\text{1}{0}^{–\text{19}}×\text{sin}{90}^{0}}\\ =4.2\text{}x\text{}{10}^{-2}\text{}m\\ \therefore \text{The radius of the circular orbit of the electron is}\\ \text{4}.\text{2 cm}.\end{array}$

Q.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{6}.\text{5}×\text{1}{0}^{-\text{4}}\text{T}\\ \text{Charge on}\text{\hspace{0.17em}}\text{electron},\text{e}=\text{1}.\text{6}×\text{1}{0}^{-\text{19}}\text{C}\\ \text{Mass of electron},{\text{m}}_{\text{e}}\text{}=\text{9}.\text{1}×\text{1}{0}^{-\text{31}}\text{kg}\\ \text{Velocity of electron},\text{v}=\text{4}.\text{8}×\text{1}{0}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{Radius of orbit},\text{r}=\text{4}.\text{2 cm}=\text{}0.0\text{42 m}\\ \text{Frequency of revolution of electron}=\text{}\nu \\ \text{Angular frequency of electron}=\text{}\omega \text{}=\text{2}\pi \nu \\ \text{The}\text{\hspace{0.17em}}\text{relation}\text{\hspace{0.17em}}\text{connecting}\text{\hspace{0.17em}}v\text{elocity of electron to angular}\text{\hspace{0.17em}}\\ \text{frequency}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}as:\\ \text{v}=\text{r}\omega \\ \text{In the circular orbit},\text{the magnetic force on an}\text{\hspace{0.17em}}\text{electron}\\ \text{is balanced by the centripetal}\text{\hspace{0.17em}}\text{force}.\text{}\\ \therefore evB=\frac{m{v}^{2}}{r}\\ eB=\frac{m}{r}\left(\text{r}\omega \right)=\frac{m}{r}\left(\text{r2}\pi \nu \right)\\ \therefore \nu =\frac{Be}{2\pi m}=\frac{\text{6}.\text{5}×\text{1}{0}^{-\text{4}}×\text{1}.\text{6}×\text{1}{0}^{-\text{19}}}{2×3.14×\text{9}.\text{1}×\text{1}{0}^{-\text{31}}}=18.2×\text{1}{0}^{\text{6}}\text{\hspace{0.17em}}Hz\\ \therefore \text{The frequency of the electron is around 18 MHz}\\ \text{and is independent of the speed}\text{\hspace{0.17em}}\text{of the electron}.\end{array}$

Q.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60o with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{\hspace{0.17em}}n\text{umber of turns on the circular coil},\text{n}=\text{3}0\\ \text{Radius of coil},\text{r}=\text{8}.0\text{cm}=\text{}0.0\text{8 m}\\ \text{Area of coil=}\pi {\text{r}}^{\text{2}}=\pi {\left(0.0\text{8}\right)}^{2}=0.0201\text{\hspace{0.17em}}{m}^{2}\\ \text{Current flowing through the coil},\text{I}=\text{6}.0\text{A}\\ \text{Strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{1 T}\\ \text{Angle between the magnetic}\text{\hspace{0.17em}}\text{field lines and normal}\\ \text{with the coil surface},\theta \text{}=\text{6}0°\\ \text{The coil experiences a torque in the magnetic field}\text{\hspace{0.17em}}\\ \text{and it turns}.\text{The counter torque exerted to prevent}\\ \text{the coil from turning is given as:}\\ \tau \text{}=\text{n IBA sin}\theta \to \text{}\left(\text{i}\right)\\ =\text{3}0\text{}×\text{6}×\text{1}×\text{}0.0\text{2}0\text{1}×\text{sin6}0°=\text{3}.\text{133 Nm}\\ \left(\text{b}\right)\text{From relation (i),}\text{\hspace{0.17em}}\text{it can be concluded that the}\\ \text{magnitude of the applied torque is not dependent on}\\ \text{the shape of the coil}.\text{However,}\text{\hspace{0.17em}}i\text{t depends on the area}\\ \text{of the coil}.\\ \therefore \text{The}\text{\hspace{0.17em}}torque\text{\hspace{0.17em}}will\text{\hspace{0.17em}}remain\text{unchanged if the circular coil}\\ \text{in the above case is replaced by a planar coil of some}\\ \text{irregular shape that encloses the same area}.\end{array}$

Q.14 Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Ans.

$\begin{array}{l}For\text{\hspace{0.17em}}coil\text{\hspace{0.17em}}X:\\ \text{Radius},{\text{r}}_{\text{1}}\text{}=\text{16 cm}=\text{}0.\text{16 m}\\ \text{Number of turns,}\text{\hspace{0.17em}}{\text{n}}_{\text{1}}\text{}=\text{2}0\\ \text{Current},{\text{I}}_{\text{1}}\text{}=\text{16 A}\\ For\text{\hspace{0.17em}}coil\text{\hspace{0.17em}}Y:\\ \text{Radius},{\text{r}}_{\text{2}}\text{}=\text{1}0\text{cm}=\text{}0.\text{1 m}\\ \text{Number of turns},{\text{n}}_{\text{2}}\text{}=\text{25}\\ \text{Current},{\text{I}}_{\text{2}}\text{}=\text{18 A}\\ \text{Magnetic field due to coil X at their centre is given as:}\\ {\text{B}}_{\text{1}}\text{=}\frac{{\mu }_{0}{\text{n}}_{\text{1}}{\text{I}}_{\text{1}}}{2{\text{r}}_{\text{1}}}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}\text{\hspace{0.17em}}=\text{Permeability of free space}\\ =4\pi ×{10}^{-7}\text{\hspace{0.17em}}Tm{A}^{-1}\\ \therefore {\text{B}}_{\text{1}}\text{=}\frac{4\pi ×{10}^{-7}×20×16}{2×0.16}=4\pi ×{10}^{-4}\text{\hspace{0.17em}}T\text{\hspace{0.17em}}\left(towards\text{\hspace{0.17em}}East\right)\\ \text{Magnetic field due to coil Y at their centre is given by},\\ {\text{B}}_{2}\text{=}\frac{{\mu }_{0}{\text{n}}_{2}{\text{I}}_{2}}{2{\text{r}}_{2}}=\frac{4\pi ×{10}^{-7}×25×18}{2×0.10}\\ =9\pi ×{10}^{-4}\text{\hspace{0.17em}}T\text{\hspace{0.17em}}\left(towards\text{\hspace{0.17em}}West\right)\\ \therefore \text{Net magnetic field,}\text{\hspace{0.17em}}B={B}_{2}-{B}_{1}\\ =9\pi ×{10}^{-4}-4\pi ×{10}^{-4}\\ =5\pi ×{10}^{-4}\text{\hspace{0.17em}}T\\ =1.57×{10}^{-3}\text{\hspace{0.17em}}T\text{\hspace{0.17em}}\left(towards\text{\hspace{0.17em}}West\right)\end{array}$

Q.15 A magnetic field of 100 G (1 G = 10−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10−3 m2. The maximum current carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{1}00\text{G}\\ =\text{1}00×\text{1}{0}^{-\text{4}}\text{\hspace{0.17em}}\text{T}\\ \text{Number of turns per unit length},\text{n}=\text{1}000\text{turns}\text{\hspace{0.17em}}{\text{m}}^{-\text{1}}\\ \text{Current flowing through the coil},\text{I}=\text{15 A}\\ \text{Permeability of free space},\text{}{\mu }_{\text{0}}=4\pi ×{10}^{-7}\text{\hspace{0.17em}}Tm{A}^{-1}\\ \text{Magnetic field is given as:}\\ B={\mu }_{\text{0}}nI\\ \therefore nI=\frac{B}{{\mu }_{\text{0}}}=\frac{\text{1}00×\text{1}{0}^{-\text{4}}}{4\pi ×{10}^{-7}}=7957.74\\ \approx 8000\text{\hspace{0.17em}}A{m}^{-1}\\ \text{If we}\text{\hspace{0.17em}}\text{take}\text{\hspace{0.17em}}\text{the length of the coil as 5}0\text{cm},\text{radius 4 cm},\\ \text{number of turns 4}00,\text{and current 1}0\text{A},\text{then these}\\ \text{values may give the magnetic field of same order}\text{.}\\ \text{So, these values are not unique and we may have other}\\ \text{suitable parameters too}\text{.}\end{array}$

Q.16 For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distancex from its centre is given by, B= μ 0 IR 2 N 2 ( x 2 +R 2 ) 3/2 ( a ) Show that this reduces to the familiar result for field at the centre of the coil. ( b ) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction,and separated by a distance R. Show thatthe field on the axis around the mid-point between the coils is uniform over a distancethat is small as compared to R, and is given by,B = 0.72 μ 0 NI R , approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Ans.

$Here, radius of the circular coil = R Number of turns on the coil = N Current flowing through the coil = I Magnetic field at a point on its axis at distance x is given as: B= μ 0 IR 2 N 2 ( x 2 +R 2 ) 3/2 Here, μ 0 = Permeability of free space ( a ) For magnetic field at the centre of the coil, x=0 ∴B = μ 0 IR 2 N 2R 3 = μ 0 IN 2R This is the familiar result for field at the centre of the coil. ( b ) Radii of two parallel co-axial circular coils = R Number of turns on each coil = N Current flowing through both coils = I Separation between both the coils = R Consider point Q at distance d from the centre. Then, one coil is at a distance of R 2 +d from point Q. ∴Magnetic field at Q is given as: B 1 = μ 0 NIR 2 2 [ ( R 2 +d ) 2 +R 2 ] 3/2 The other coil is at a distance of R 2 -d from point Q. ∴Magnetic field due to this coil is given as: B 2 = μ 0 NIR 2 2 [ ( R 2 -d ) 2 +R 2 ] 3/2 Total magnetic field, B = B 1 + B 2 = μ 0 IR 2 2 [ { ( R 2 -d ) 2 +R 2 } −3/2 + { ( R 2 +d ) 2 +R 2 } −3/2 ]$ $= μ 0 IR 2 2 [ ( 5R 2 4 +d 2 -Rd ) −3/2 + ( 5R 2 4 +d 2 +Rd ) −3/2 ] = μ 0 IR 2 2 × ( 5R 2 4 ) −3/2 [ ( 1+ 4 5 d 2 R 2 – 4 5 d R ) −3/2 + ( 1+ 4 5 d 2 R 2 + 4 5 d R ) −3/2 ] For d = R, neglecting d 2 R 2 , we have, ≈ μ 0 IR 2 2 × ( 5R 2 4 ) −3/2 ×[ ( 1- 4 5 d R ) −3/2 + ( 1+ 4 5 d R ) −3/2 ] ≈ μ 0 IR 2 N 2R 3 × ( 4 5 ) 3/2 [ 1- 6d 5R +1+ 6d 5R ] B = ( 4 5 ) 3/2 μ 0 IN R = 0.72( μ 0 IN R ) ∴It is proved that the magnetic field on the axis around the mid-point between the coils is uniform.$

Q. 17 A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(a) outside the toroid,
(b) inside the core of the toroid, and
(c) in the empty space surrounded by the toroid.

Ans.

Inner radius of toroid, r1 = 25 cm = 0.25 m

$\begin{array}{l}\text{Outer radius of toroid}\text{\hspace{0.17em}}\text{core},{\text{r}}_{\text{2}}\text{}=\text{26 cm}=\text{}0.\text{26 m}\\ \text{Number of turns on the coil},\text{N}=\text{35}00\\ \text{Current flowing}\text{\hspace{0.17em}}\text{through the coil},\text{I}=\text{11 A}\\ \left(\text{a}\right)\text{Magnetic field outside the toroid is zero}.\text{}\\ \left(\text{b}\right)\text{Magnetic field inside the core of a toroid is given}\\ \text{by},\text{\hspace{0.17em}}\text{B}=\frac{{\mu }_{0}NI}{l}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =4\pi ×{10}^{-7}\text{\hspace{0.17em}}Tm{A}^{-1}\\ \text{Length of toroid,}\text{\hspace{0.17em}}\text{l}=2\pi \left[\frac{{r}_{1}+{r}_{2}}{2}\right]=\pi \left(0.25+0.26\right)\\ =0.51\text{\hspace{0.17em}}\pi \\ \therefore B=\frac{4\pi ×{10}^{-7}×3500×11}{0.51\text{\hspace{0.17em}}\pi }\approx 3.0×{10}^{-2}\text{\hspace{0.17em}}T\\ \left(\text{c}\right)\text{The}\text{\hspace{0.17em}}m\text{agnetic field in the empty space surrounded}\\ \text{by the toroid is zero}.\end{array}$

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Ans.

(a) Since the initial velocity of the particle is either parallel or anti-parallel to the magnetic field, therefore, it travels along a straight path without suffering any deflection in the field.

(b) The final speed of the charged particle will be equal to its initial speed. This is because the magnetic force acting on the charged particle which is perpendicular to direction of its velocity can change the direction of velocity, but not its magnitude.

(c) When an electron travelling from West to East will enter a chamber having a uniform electrostatic field in the North-South direction, it will be deflected by the electric field towards north. The moving electron can remain undeflected only if the electric force acting on it is equal and opposite of the magnetic field.
Since the magnetic force is directed towards the South. Therefore, according to Fleming’s left hand rule, the magnetic field should be applied in a vertically downward direction to nullify the deflection.

Q.19 An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30o with the initial velocity.

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{strength}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}m\text{agnetic field},\text{}\\ \text{B}=\text{}0.\text{15 T}\\ \text{Charge on electron},\\ \text{e}=\text{1}.\text{6}×\text{1}{0}^{-\text{19}}\text{C}\\ \text{Mass of electron},\\ \text{m}=\text{9}.\text{1}×\text{1}{0}^{-\text{31}}\text{kg}\\ \text{Potential difference},\\ \text{V}=\text{2}.0\text{kV}\\ =\text{2}×\text{1}{0}^{\text{3}}\text{V}\\ \therefore \text{K}\text{.E}\text{. of the electron}=\text{eV}\\ ⇒\text{eV}=\frac{1}{2}m{v}^{2}\\ \\ v=\sqrt{\frac{2eV}{m}}\to \left(i\right)\\ \text{Here},\text{v}=\text{velocity of electron}\\ \left(\text{a}\right)\text{Magnetic force on the electron provides the}\\ \text{required centripetal force to the electron}.\\ \therefore \text{The electron traces a circular path of radius r}.\\ \text{Magnetic force on the electron=Bev}\\ \text{Centripetal force=}\frac{m{v}^{2}}{r}\\ \therefore \text{Bev}=\frac{m{v}^{2}}{r}\\ r=\frac{mv}{\text{Be}}\to \left(ii\right)\\ \text{From equations}\left(i\right)\text{and}\left(ii\right),\text{we have}\\ r=\frac{m}{\text{Be}}{\left[\frac{2eV}{m}\right]}^{\frac{1}{2}}\\ =\frac{\text{9}.\text{1}×\text{1}{0}^{-\text{31}}}{0.\text{15}×\text{1}.\text{6}×\text{1}{0}^{-\text{19}}}×{\left(\frac{2×\text{1}.\text{6}×\text{1}{0}^{-\text{19}}×2×{10}^{3}}{\text{9}.\text{1}×\text{1}{0}^{-\text{31}}}\right)}^{\frac{1}{2}}\\ \\ =100.55×\text{1}{0}^{-5}\\ =1.01×\text{1}{0}^{-3}\text{\hspace{0.17em}}m=1\text{\hspace{0.17em}}mm\\ \therefore \text{The electron has a circular trajectory of radius 1}.0\text{}\\ \text{mm normal to the direction}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{magnetic}\text{\hspace{0.17em}}\text{field}.\\ \left(\text{b}\right)\text{When the magnetic}\text{\hspace{0.17em}}\text{field makes an angle}\left(\theta \right)\text{of 3}0°\text{}\\ \text{with initial velocity},\text{\hspace{0.17em}}\text{the initial velocity}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{\hspace{0.17em}}{v}_{1}=v\mathrm{sin}\theta \\ \text{From equation}\left(\text{2}\right),\text{\hspace{0.17em}}\text{the expression for new radius is}\text{\hspace{0.17em}}\\ \text{given}\text{\hspace{0.17em}}\text{as}:\\ {r}_{1}=\frac{m{v}_{1}}{Be}=\frac{mv\mathrm{sin}\theta }{Be}\\ =\frac{\text{9}.\text{1}×\text{1}{0}^{-\text{31}}}{0.15×\text{1}.\text{6}×\text{1}{0}^{-\text{19}}}×{\left[\frac{2×\text{1}.\text{6}×\text{1}{0}^{-\text{19}}×2×{10}^{3}}{\text{9}.\text{1}×\text{1}{0}^{-\text{31}}}\right]}^{\frac{1}{2}}×\mathrm{sin}\text{3}0°\\ =0.5×{10}^{-3}\text{\hspace{0.17em}}m=0.5\text{\hspace{0.17em}}mm\\ \therefore \text{The electron has a helical trajectory of radius}\\ 0.\text{5 mm along the direction}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{magnetic field}\text{\hspace{0.17em}}.\end{array}$

Q.20 A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10−5 V m−1, make a simple guess as to what the beam contains. Why is the answer not unique?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}s\text{trength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{}0.\text{75 T}\\ \text{Accelerating voltage},\text{V}=\text{15 kV}=\text{15}×\text{1}{0}^{\text{3}}\text{V}\\ \text{Electrostatic field},\text{E}=\text{9}×\text{1}{0}^{\text{5}}{\text{V m}}^{-\text{1}}\\ \text{Mass of electron}=\text{m}\\ \text{Charge of electron}=\text{e}\\ \text{Velocity of electron}=\text{v}\\ \text{Kinetic energy of electron}=\text{eV}\\ ⇒\frac{1}{2}m{v}^{2}=\text{eV}\\ \therefore \frac{e}{m}=\frac{{v}^{2}}{2V}\to \left(i\right)\\ \text{As the particle remains undeflected by electric and}\\ \text{magnetic fields},\\ \therefore \text{The electric field is balancing the magnetic field}.\\ \therefore eE=evB\\ v=\frac{E}{B}\to \left(ii\right)\\ \text{Substituting equation}\left(ii\right)\text{in equation}\left(i\right),\text{we get}\\ \\ \frac{e}{m}=\frac{1}{2}\frac{{\left(\frac{E}{B}\right)}^{2}}{V}=\frac{{E}^{2}}{2V{B}^{2}}=\frac{{\left(\text{9}×\text{1}{0}^{\text{5}}\right)}^{2}}{2×15000×{\left(0.75\right)}^{2}}\\ =4.8×\text{1}{0}^{7}\text{\hspace{0.17em}}Ck{g}^{-1}\\ \text{This value of specific charge}\frac{e}{m}\text{corresponds}\text{\hspace{0.17em}}\text{to}\text{\hspace{0.17em}}\\ \text{deuteron or deuterium ions}.\\ \text{However,}\text{\hspace{0.17em}}t\text{his is not a unique answer}.\text{The}\text{\hspace{0.17em}}o\text{ther}\\ {\text{possible answers are He}}^{++},{\text{Li}}^{+++},\text{etc}.\end{array}$

Q. 21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 ms−2.

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}l\text{ength of rod},\text{l}=\text{}0.\text{45 m}\\ \text{Mass suspended}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{vertical}\text{\hspace{0.17em}}\text{wires},\text{m}=\text{6}0\text{g}\\ \text{}=\text{6}0×\text{1}{0}^{-\text{3}}\text{kg}\\ \text{Acceleration due to gravity},\text{g}=\text{9}.{\text{8 ms}}^{\text{-2}}\\ \text{Current in the rod passing through the wire},\text{I}=\text{5 A}\\ \left(\text{a}\right)\text{Magnetic field strength}\text{\hspace{0.17em}}\text{is equal and opposite to the}\\ \text{weight}\text{\hspace{0.17em}}\text{of the wire i}.\text{e}.,\\ BIl=mg\\ \therefore B=\frac{mg}{Il}=\frac{\text{6}0×\text{1}{0}^{-\text{3}}×\text{9}.\text{8}}{5×0.\text{45}}=0.26\text{\hspace{0.17em}}T\\ \text{A horizontal magnetic field of}0.\text{26 T perpendicular to}\\ \text{the length of the conductor should be set}\text{\hspace{0.17em}}\text{up in order}\\ \text{to get zero tension in the wire}.\text{}\\ \text{The magnetic field must be such that}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}\text{Fleming}’\text{s left}\\ \text{hand rule gives an upward magnetic force}.\\ \left(\text{b}\right)\text{If}\text{\hspace{0.17em}}\text{we reverse}\text{\hspace{0.17em}}\text{the direction of current},\text{then the}\\ \text{force due to magnetic field and}\text{\hspace{0.17em}}\text{the weight of the wire}\\ \text{act in the vertically downward direction}.\\ \therefore \text{Total tension in the wire}=\text{BIl}+\text{mg}\\ \text{=}0.\text{26}×5×0.\text{45}+\left(\text{6}0×\text{1}{0}^{-\text{3}}\right)×9.8\\ =1.176\text{\hspace{0.17em}}N\end{array}$

Q.22 The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}c\text{urrent flowing}\text{\hspace{0.17em}}\text{through both wires},\text{I}=\text{3}00\text{A}\\ \text{Separation between the wires},\text{r}=\text{1}.\text{5 cm}=0.0\text{15 m}\\ \text{Length of each wire},\text{l}=\text{7}0\text{cm}=\text{}0.\text{7 m}\\ \text{Force between the two wires is given as:}\\ \text{F=}\frac{{\mu }_{0}{I}^{2}}{2\pi r}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =4\pi ×{10}^{-7}\text{\hspace{0.17em}}Tm{A}^{-1}\\ \therefore F=\frac{4\pi ×{10}^{-7}×{\left(300\right)}^{2}}{2\pi ×0.0\text{15}}=1.2\text{\hspace{0.17em}}N{m}^{-1}\\ \text{As the direction of current in the wires is opposite},\\ \text{a repulsive force exists}\text{\hspace{0.17em}}\text{between them}.\end{array}$

Q.23 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west.
A wire carrying current of 7.0 A in the north to south direction passes through this region.
What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}strength\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{m}\text{agnetic field},\text{B}=\text{1}.\text{5 T}\\ \text{Radius of cylindrical region},\text{r}=\text{1}0\text{cm}=\text{}0.\text{1 m}\\ \text{Current flowing\hspace{0.17em}through the wire passing through the}\\ \text{cylindrical region},\text{I}=\text{7 A}\\ \mathrm{}\left(\text{a}\right)\text{If the wire intersects the axis},\text{then the length of}\\ \text{the wire is the diameter of the cylindrical region}.\\ \text{Thus},\text{l}=\text{2r}=\text{}0.\text{2 m}\\ \text{Angle between current and magnetic field},\text{}\mathrm{\theta }\text{}=\text{9}0\mathrm{°}\\ \text{Magnetic force acting on the wire is given as:}\\ \text{F}=\text{BIl sin}\mathrm{\theta }=\text{1}.\text{5}×\text{7}×\text{}0.\text{2}×\text{sin 9}0\mathrm{°}=\text{2}.\text{1 N}\\ \therefore \text{A force of 2}.\text{1 N acts on the wire vertically}\\ \text{downwards}.\\ \left(\text{b}\right)\text{New length of the wire after turning it to the}\\ \text{North\hspace{0.17em}east}-\text{North\hspace{0.17em}west direction\hspace{0.17em}is\hspace{0.17em}given as}:\\ {\text{l}}_{\text{1}}=\frac{\mathrm{l}}{\text{sin}\mathrm{\theta }}\\ :\text{Angle between magnetic field and current},\text{}\mathrm{\theta }=\text{45}\mathrm{°}\\ \text{Force acting\hspace{0.17em}on the wire\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by},\\ \text{F}={\text{BIl}}_{\text{1}}\text{sin}\mathrm{\theta }\\ =\text{BIl}\\ =1.5×7×0.2=2.1\text{\hspace{0.17em}}\mathrm{N}\\ \therefore \text{A force of 2}.\text{1 N acts on the wire\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}vertically}\\ \text{downward direction\hspace{0.17em}}.\text{}\\ \left(\text{c}\right)\text{The wire in\hspace{0.17em}the\hspace{0.17em}N-S\hspace{0.17em}direction\hspace{0.17em}is lowered from the}\\ \text{axis by distance},\text{d}=\text{6}.0\text{cm}\\ \mathrm{Suppose}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{wire}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{passing}\text{\hspace{0.17em}}\mathrm{normally}\text{\hspace{0.17em}}\mathrm{to}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{axis}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\\ \mathrm{cylinderical}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{field},\text{\hspace{0.17em}}\mathrm{then}\text{\hspace{0.17em}}\mathrm{lowering}\text{\hspace{0.17em}}6\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{means}\\ \mathrm{displacing}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{wire}\text{\hspace{0.17em}}6\text{\hspace{0.17em}}\mathrm{cm}\text{\hspace{0.17em}}\mathrm{from}\text{\hspace{0.17em}}\mathrm{its}\text{\hspace{0.17em}}\mathrm{initial}\text{\hspace{0.17em}}\mathrm{position}\text{\hspace{0.17em}}\mathrm{towards}\\ \mathrm{to}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{end}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{area}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{cross}-\mathrm{section}.\end{array}$ $\begin{array}{l}\mathrm{From}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{figure},\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{have}\\ \text{\hspace{0.17em}}\mathrm{x}=\sqrt{{10}^{2}-{6}^{2}}=8\text{\hspace{0.17em}}\mathrm{cm}\\ \therefore \mathrm{Length}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{wire}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{field},\text{\hspace{0.17em}}{\mathrm{l}}_{2}=2\mathrm{x}=2×8\\ =16\text{\hspace{0.17em}}\mathrm{cm}\text{= 0.16 m}\\ \mathrm{Force}\text{\hspace{0.17em}}\mathrm{on}\text{\hspace{0.17em}}\mathrm{wire},\text{\hspace{0.17em}}\mathrm{F}={\mathrm{il}}_{2}\mathrm{Bsin}{90}^{0}\\ \mathrm{F}=7×0.16×1.5=1.68\text{\hspace{0.17em}}\mathrm{N}\\ \mathrm{According}\text{\hspace{0.17em}}\mathrm{to}\text{\hspace{0.17em}}\mathrm{screw}\text{\hspace{0.17em}}\mathrm{rule},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{direction}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{this}\text{\hspace{0.17em}}\mathrm{force}\text{\hspace{0.17em}}\mathrm{is}\\ \mathrm{vertically}\text{\hspace{0.17em}}\mathrm{downwards}.\\ \therefore \text{A force of 1}.\text{68 N acts\hspace{0.17em}on the wire in the\hspace{0.17em}vertically}\\ \text{downward direction}.\end{array}$

Q.24 A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium? Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{strength}\text{\hspace{0.17em}}of\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{3}000\text{G}\\ =\text{3}000×\text{1}{0}^{-\text{4}}\text{T}=\text{}0.\text{3 T}\\ \text{Length of rectangular loop},\text{l}=\text{1}0\text{cm}\\ \text{Width of rectangular loop},\text{b}=\text{5 cm}\\ \therefore \text{Area of the}\text{\hspace{0.17em}}\text{loop},\text{\hspace{0.17em}}\text{A}=\text{l}×\text{b}=\text{1}0×\text{5}=\text{5}0{\text{cm}}^{\text{2}}\\ =\text{5}0×\text{1}{0}^{-\text{4}}{\text{m}}^{\text{2}}\\ \text{Current flowing}\text{\hspace{0.17em}}\text{through the loop},\text{I}=\text{12 A}\\ \text{Taking the anti}-\text{clockwise direction of current as}\\ \text{positive and vice}-\text{versa}:\\ \left(\text{a}\right)\text{Torque}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation},\\ \stackrel{\to }{\tau }=I\stackrel{\to }{A}×\stackrel{\to }{B}\\ \text{From the given figure},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}can\text{\hspace{0.17em}}observe\text{that A is normal}\\ \text{to the y}-\text{z plane and B is}\text{\hspace{0.17em}}\text{directed along the z}-\text{axis}.\\ \therefore \tau =12×\left(50×{10}^{-4}\right)\stackrel{\wedge }{i}×0.3\stackrel{\wedge }{k}\\ =-1.8×{10}^{-2}\text{\hspace{0.17em}}\stackrel{\wedge }{j}\text{\hspace{0.17em}}Nm\\ \text{The torque is N m along the}\text{\hspace{0.17em}}\text{negative y}-\text{direction}.\text{The}\\ \text{force on the loop is zero}\text{\hspace{0.17em}}\text{because the angle between A}\\ \text{and B is zero}.\\ \left(\text{b}\right)\text{This}\text{\hspace{0.17em}}case\text{\hspace{0.17em}}is\text{\hspace{0.17em}}similar\text{\hspace{0.17em}}to\text{\hspace{0.17em}}case\text{\hspace{0.17em}}\left(a\right),\text{\hspace{0.17em}}therefore,\text{\hspace{0.17em}}the\text{\hspace{0.17em}}\\ answer\text{\hspace{0.17em}}is\text{\hspace{0.17em}}same\text{\hspace{0.17em}}as\text{\hspace{0.17em}}\left(a\right).\\ \left(\text{c}\right)\text{Torque}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \stackrel{\to }{\tau }=I\stackrel{\to }{A}×\stackrel{\to }{B}\\ \text{From the given figure},\text{we observe that A is}\\ \text{perpendicular to the x}-\text{z plane and B is}\text{\hspace{0.17em}}\text{directed along}\\ \text{the z}-\text{axis}.\\ \therefore \tau =-12×\left(50×{10}^{-4}\right)\stackrel{\wedge }{j}×0.3\stackrel{\wedge }{k}\\ =-1.8×{10}^{-2}\text{\hspace{0.17em}}\stackrel{\wedge }{i}\text{\hspace{0.17em}}Nm\\ \therefore \text{The torque is}1.8×{10}^{-2}\text{\hspace{0.17em}}\text{Nm along the negative x}\\ \text{direction}.\text{\hspace{0.17em}}The\text{\hspace{0.17em}}force\text{\hspace{0.17em}}is\text{\hspace{0.17em}}zero.\\ \left(\text{d}\right)\text{Magnitude of torque,}\text{\hspace{0.17em}}|\tau |=IAB\\ =12×50×{10}^{-4}×0.3=1.8×{10}^{-2}\text{\hspace{0.17em}}Nm\\ \text{Torque is}1.8×{10}^{-2}\text{\hspace{0.17em}}\text{Nm at an angle of 24}0°\text{with}\\ \text{positive x direction}.\text{In}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{case,}\text{\hspace{0.17em}}t\text{he force is zero}.\\ \left(\text{e}\right)\text{Torque}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \stackrel{\to }{\tau }=I\stackrel{\to }{A}×\stackrel{\to }{B}=\left(50×{10}^{-4}×12\right)\stackrel{\wedge }{k}×0.3\stackrel{\wedge }{k}=0\\ \therefore \text{Torque is zero}.\text{The}\text{\hspace{0.17em}}f\text{orce is also zero}\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}\text{this}\text{\hspace{0.17em}}\text{case}.\\ \left(\text{f}\right)\text{Torque}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \stackrel{\to }{\tau }=I\stackrel{\to }{A}×\stackrel{\to }{B}=\left(50×{10}^{-4}×12\right)\stackrel{\wedge }{k}×0.3\stackrel{\wedge }{k}=0\\ \therefore T\text{orque is zero}.\text{Force is also zero}.\\ \text{In case}\left(\text{e}\right),\text{}I\stackrel{\to }{A}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\stackrel{\to }{B}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}\text{along}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{same direction}\\ \text{and the angle between them is zero}.\text{\hspace{0.17em}}\text{If disturbed},\text{they}\\ \text{come back to an equilibrium}.\text{}\\ \therefore \text{Its equilibrium is stable}.\\ \text{In case}\left(\text{f}\right),\text{}I\stackrel{\to }{A}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\stackrel{\to }{B}\text{\hspace{0.17em}}\text{are}\text{\hspace{0.17em}}in\text{\hspace{0.17em}}opposite\text{directions and}\\ \text{the angle between them is18}0°.\text{If disturbed},\text{\hspace{0.17em}}it\text{\hspace{0.17em}}will\text{\hspace{0.17em}}not\text{}\\ \text{restore}\text{\hspace{0.17em}}\text{its original position}.\text{}\\ \therefore I\text{ts equilibrium is}\text{\hspace{0.17em}}\text{unstable}.\end{array}$

Q.25 A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10−5 m2, and the free electron density in copper is given to be about 1029 m−3.)

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}n\text{umber of turns on the circular coil},\text{n}=\text{2}0\\ \text{Radius of coil},\text{r}=\text{1}0\text{cm}=\text{}0.\text{1 m}\\ \text{Strength}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}m\text{agnetic field},\text{B}=\text{}0.\text{1}0\text{T}\\ \text{Current flowing}\text{\hspace{0.17em}}\text{through the coil},\text{I}=\text{5}.0\text{A}\\ \left(\text{a}\right)\text{As}\text{\hspace{0.17em}}\text{the field is uniform,}\text{\hspace{0.17em}}\text{therefore,}\text{\hspace{0.17em}}t\text{he total torque}\\ \text{acting}\text{\hspace{0.17em}}\text{on the coil is zero}.\\ \left(\text{b}\right)\text{\hspace{0.17em}}\text{As}\text{\hspace{0.17em}}\text{the field}\text{\hspace{0.17em}}\text{is uniform,}\text{\hspace{0.17em}}\text{the total force on the coil is zero.}\\ \left(\text{c}\right)\text{Area}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}c\text{ross}-\text{section of copper coil},\text{A}=\text{1}{0}^{-\text{5}}{\text{m}}^{\text{2}}\\ \text{Number of free electrons per cubic meter in copper},\text{N=}\\ \text{1}{0}^{\text{29}}{\text{m}}^{\text{-3}}\\ \text{Charge on electron},\text{e}=\text{1}.\text{6}×\text{1}{0}^{-\text{19}}\text{C}\\ \text{Magnetic force}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{F}={\text{Bev}}_{\text{d}}\\ \text{Here},{\text{v}}_{\text{d}}\text{}=\text{Drift velocity}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{electrons}\text{\hspace{0.17em}}\text{=}\frac{I}{NeA}\\ \therefore F=\frac{BeI}{NeA}=\frac{0.10×5.0}{\text{1}{0}^{\text{29}}×{10}^{-5}}=5×{10}^{-25}\text{\hspace{0.17em}}N\\ \therefore A\text{verage force on each electron is}\text{\hspace{0.17em}}5×{10}^{-25}\text{\hspace{0.17em}}N.\end{array}$

Q.26 A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g= 9.8 m s−2

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}l\text{ength of solenoid},\text{L}=\text{6}0\text{cm}=\text{}0.\text{6 m}\\ \text{Radius of solenoid},\text{r}=\text{4}.0\text{cm}=\text{}0.0\text{4 m}\\ \text{As there are 3 layers of windings of 3}00\text{turns each,}\\ \therefore \text{Total number of turns},\text{n}=\text{3}×\text{3}00\text{}=\text{9}00\\ \text{Length of wire},\text{l}=\text{2 cm}=\text{}0.0\text{2 m}\\ \text{Mass of wire},\text{m}=\text{2}.\text{5 g}=\text{2}.\text{5}×\text{1}{0}^{-\text{3}}\text{kg}\\ \text{Current passing through the wire},\text{i}=\text{6 A}\\ \text{Acceleration due to gravity},\text{g}=\text{9}.{\text{8 ms}}^{\text{-2}}\\ \text{Magnetic field produced inside the solenoid}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as:}\\ \text{B=}\frac{{\mu }_{0}nI}{L}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{0}=\text{Permeability of free space}\\ =4\pi ×{10}^{-7}\text{\hspace{0.17em}}Tm{A}^{-1}\\ \text{I}=\text{Current passing through the windings of the}\\ \text{solenoid}\\ \text{Magnetic force is given as:}\\ \text{F=Bil=}\frac{{\mu }_{0}nI}{L}×\text{il=}\frac{{\mu }_{0}nI\text{il}}{L}\\ \text{The force acting}\text{\hspace{0.17em}}\text{on the wire is equal to the weight of}\\ \text{the wire}.\\ \therefore mg=\frac{{\mu }_{0}nI\text{il}}{L}\\ I=\frac{mgL}{{\mu }_{0}n\text{il}}=\frac{2.5×{10}^{-3}×9.8×0.6}{4\pi ×{10}^{-7}×900×0.02×6}=108\text{\hspace{0.17em}}A\\ \therefore T\text{he current flowing through the solenoid is 1}0\text{8 A}.\end{array}$

Q.27 A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}r\text{esistance of galvanometer coil},\text{G}=\text{12}\Omega \\ \text{Current for which there is the}\text{\hspace{0.17em}}\text{full scale deflection},\text{\hspace{0.17em}}{I}_{g}\\ =\text{3 mA}=\text{3}×\text{1}{0}^{-\text{3}}\text{A}\\ \text{As}\text{\hspace{0.17em}}the\text{\hspace{0.17em}}r\text{ange of the voltmeter is}0\text{\hspace{0.17em}}to\text{\hspace{0.17em}}\text{18 V,}\\ \therefore \text{V}=\text{18 V}\\ \text{Let a resistor of resistance R be connected in series}\\ \text{with the galvanometer in}\text{\hspace{0.17em}}\text{order}\text{\hspace{0.17em}}\text{to convert it}\text{\hspace{0.17em}}\text{into a}\\ \text{voltmeter}.\text{}\\ \text{This resistance is given by,}\\ \text{R=}\frac{V}{{I}_{g}}-G=\frac{18}{\text{3}×\text{1}{0}^{-\text{3}}}-12=6000-12=5988\text{\hspace{0.17em}}\Omega \\ \therefore \text{A resistor of resistance}\text{\hspace{0.17em}}5988\text{\hspace{0.17em}}\Omega \text{is to be connected in}\\ \text{series with the}\text{\hspace{0.17em}}\text{galvanometer}.\end{array}$

Q.28 A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

Ans.

Here, resistance of the galvanometer coil, G = 15 Ω

$\begin{array}{l}\text{Current at which the galvanometer shows full scale}\\ \text{deflection},\text{\hspace{0.17em}}{I}_{g}=\text{4 mA}=\text{4}×\text{1}{0}^{-\text{3}}\text{A}\\ \text{As}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}r\text{ange of the ammeter is}0\text{to 6 A,}\\ \therefore \text{Current},\text{I}=\text{6 A}\\ \text{A shunt resistor of resistance S should}\text{\hspace{0.17em}}\text{be connected in}\\ \text{parallel with the galvanometer to}\text{\hspace{0.17em}}\text{convert it into an}\\ \text{ammeter}.\text{}\\ \text{The value of S is given by}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{relation,}\\ \text{S=}\frac{{I}_{g}G}{I-{I}_{g}}=\frac{\text{4}×\text{1}{0}^{-\text{3}}×15}{6-4×\text{1}{0}^{-\text{3}}}=\frac{6×\text{1}{0}^{-2}}{5.996}\approx 0.01\text{\hspace{0.17em}}\Omega =10\text{\hspace{0.17em}}m\Omega \\ \therefore \text{A shunt resistor of}\text{\hspace{0.17em}}\text{resistance}\text{\hspace{0.17em}}10\text{\hspace{0.17em}}m\Omega \text{\hspace{0.17em}}should\text{be}\\ \text{connected in parallel with the galvanometer}.\end{array}$