# NCERT Solutions Class 12 Physics Chapter 5

Class 12 Physics Chapter 5 NCERT Solutions gives students a brief understanding of the important questions and the best ways to answer them. They help students in understanding the concepts detailed in the chapter and prepare them for any upcoming examination. Using NCERT Solutions for Class 12 Physics Chapter 5, students can easily understand the briefly explained concepts and attempt all questions in exams, including numerical problems.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 5

NCERT Class 12 Physics Chapter 5 details Magnetism and matter. Some of the important questions from this topic are based on bar magnet, Gauss’ law, magnetic properties of different materials, permanent magnets, etc. The NCERT Solutions Class 12 Physics Chapter 5 Magnets is equally divided into all sub-sections of the chapter and is briefly explained below.

5.1 Introduction

Magnetism is a force produced by moving electric charges. The phenomenon is associated with magnetic fields, and this motion can take many forms. Depending on the object and associated magnetic field, it can be an attractive or repulsive force.

5.2 The Bar Magnet

A bar magnet is a rectangular piece of an object made of iron, steel, or other ferromagnetic substance or composite, indicating everlasting magnetic properties.

5.2.1 The magnetic field lines

These are visible tools used to symbolise magnetic fields. They describe the path of the magnetic force on a north monopole at any given position. The density of the lines suggests the magnitude of the field. The magnetic field is more potent and crowded close to the poles of a magnet.

5.2.2 Bar Magnet as an equivalent solenoid

• A solenoid is an extended coil of circular loops of insulated copper wire. Magnetic field lines are produced across the solenoid while a current is authorised to flow via it. The magnetic field produced with the aid of using it’s far much like the magnetic field of a bar magnet.
• The field lines produced in a current-carrying solenoid are proven withinside the above figure. When the north pole of a bar magnet is introduced close to the end connected to the battery’s negative terminal, the solenoid repels the bar magnet.
• Since poles repel each other, the end connected to the battery’s negative terminal behaves like the North Pole of the solenoid, and the alternative end behaves like the South Pole. Hence, one end of the solenoid behaves because of the North Pole, and the alternative end behaves like the South Pole.

5.2.3 The dipole in a uniform magnetic field

When a magnetic rod (which may be taken as a magnetic dipole) is saved in a uniform magnetic field, the North Pole senses a force equal to the multiplication of the magnetic field intensity and the pole strength withinside the magnetic field direction.

5.2.4 The electrostatic analogue

Let’s evaluate the equation derived above with the equations of the electric dipole in an electric field. We find that the magnetic field because of a bar magnet at a massive distance is similar to that of an electric dipole in an electric powered field.

5.3 Magnetism and Gauss’ law

Gauss’ law for Magnetism states that the magnetic flux through any closed surface is zero; this law is consistent with the commentary that remote magnetic poles (monopoles) no longer exist.

5.4 The earth’s magnetism

Earth’s Magnetism is generated via convection currents of molten iron and nickel within the earth’s core. These currents bring streams of charged particles and generate magnetic fields.

5.4.1 Magnetic declination and dip

It is the angle between the path of the overall magnetic field of the earth and a horizontal line in the magnetic meridian.

5.5 Magnetisation and magnetic intensity

The magnetic intensity defines the forces that the poles of a magnet report in a magnetic field, while the intensity of magnetisation explains the alternate withinside the magnetic moment of a magnet per unit volume.

5.6 Magnetic properties of materials

Property 1: Intensity of magnetisation (I): The electrons circulating around the nucleus have a magnetic moment. When the material isn’t magnetised, the magnetic dipole moment sums as much as 0. When the material is stored in an outside magnetic field, the magnetic moments are aligned in a specific direction, and the material receives a net non-zero dipole moment. The net dipole moment in keeping with unit volume is described as magnetisation or intensity of magnetisation.

Property 2: Magnetic Field (H) or Magnetic intensity: The magnetic field produced simplest with the aid of using the electrical current flowing in a solenoid is known as the magnetic intensity. It is the external magnetic field that induces magnetic property in a material.

Property 3: Magnetic susceptibility: When a material is positioned in an external magnetic field, the material receives magnetisation. For a small magnetising field, the intensity of magnetisation (I) received with the aid of using the material is without delay proportional to the magnetic field (H).

I ∝ H

I = χmH, χm is the susceptibility of the material.

Property 4: Retentivity: The capacity of a material to hold or withstand magnetisation is known as retentivity.

Property 5: Coercivity: The coercivity of a material is the capacity to face up to the outside magnetic field without becoming demagnetised.

5.6.1 Diamagnetism

Diamagnetism is a completely susceptible form of Magnetism that is brought about with the aid of using an alternate withinside the orbital movement of electrons because of an implemented magnetic field. This Magnetism is nonpermanent and persists only withinside the presence of an external field.

5.6.2 Paramagnetism

Paramagnetism is a form of Magnetism wherein a few substances are weakly attracted with the aid of using an externally implemented magnetic field and form internal, brought about magnetic fields withinside the direction of the implemented magnetic field.

5.6.3 Ferromagnetism

Ferromagnetism is a type of Magnetism that is related to iron, cobalt, nickel, and a few alloys or compounds containing one or extra of those elements. It additionally happens in gadolinium and some different rare-earth elements.

5.7 Permanent Magnets and electromagnets

Similarities:

Both the magnets own imaginary magnetic field lines. The magnets have the north and south poles, whose conduct relies upon the geographic north pole and south pole of the earth. Both the magnets showcase the properties of Magnetism.

Differences between Permanent Magnet and Electromagnet:

 Permanent (Bar) Magnet Electromagnet They are completely magnetised. These are briefly magnetised. These are typically products of tough substances. They are typically products of tender substances. The strength of the magnetic field line is consistent, i.e. it can’t be numerous. The strength of the magnetic field lines may be numerous in step with our needs. The poles of a Permanent magnet can’t be changed. The poles of an electromagnet may be altered. An example of an everlasting magnet is a Bar Magnet . An example of a temporary magnet is a solenoid wounded throughout a nail and connected to a battery.

### NCERT Solutions Class 12 Physics Chapter 5 Current Electricity – Article Link

This Solution set for Chapter 5 Physics Class 12 is prepared by subject matter experts at Extramarks. The Solution also contains numerical problems and important theoretical questions from previous year’s question papers, NCERT textbooks and study guides. For scoring good marks, students may refer to the study materials given below.

Click on the below links to view NCERT Solutions Class 12 Physics Chapter 5:

• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.1
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.2
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.2.1
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.2.2
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.2.3
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.2.4
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.3
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.4
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.4.1
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.5
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.6
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.6.1
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.6.2
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.6.3
• NCERT Solutions Class 12 Physics Chapter 5: Exercise 5.7

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### Key Features of NCERT Solutions Class 12 Physics Chapter 5

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Q.1 Answer the following questions regarding earth’s magnetism:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18o. Would you expect a greater or smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or South Pole?
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T−1 located at its centre. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

Ans.

(a) The three independent quantities used for specifying earth’s magnetic field are: – Magnetic declination (θ), Angle of dip (δ), and Horizontal component of earth’s magnetic field (H).

(b)The angle of dip at a point depends on the location of the point w.r.t. the North Pole or the South Pole. We expect greater angle of dip in Britain (it is about 70°) than in southern India because the location of Britain is closer to the magnetic North Pole.

(c)As Melbourne is situated in southern hemisphere, which behaves as the North pole of earth’s magnet, therefore, field lines due to the earth’s magnetism would seem to come out of the ground at Melbourne.

(d)At the poles, the angle of dip is 90o; therefore, earth’s magnetic field at the poles is exactly in vertical direction. If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to rotate in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In this situation, the compass can point in any direction.

$\begin{array}{l}\left(\text{e}\right)\text{Here,\hspace{0.17em}}\mathrm{m}\text{agnetic moment},\text{M}=\text{8}×\text{1}{0}^{\text{22}}{\text{\hspace{0.17em}JT}}^{-\text{1}}\\ \text{Radius of the\hspace{0.17em}}\mathrm{E}\text{arth},\text{r}=\text{6}.\text{4}×\text{1}{0}^{\text{6}}\text{m}\\ \text{Magnetic field strength\hspace{0.17em}of\hspace{0.17em}Earth},\text{\hspace{0.17em}}\mathrm{B}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{M}}{4{\mathrm{\pi r}}^{3}}\\ \text{Here},\\ {\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}\\ =4\mathrm{\pi }×\text{1}{0}^{\text{-7}}{\text{\hspace{0.17em}TmA}}^{\text{-1}}\\ \therefore \mathrm{B}=\frac{4\mathrm{\pi }×\text{1}{0}^{\text{-7}}×\text{8}×\text{1}{0}^{\text{22}}}{4\mathrm{\pi }×{\left(\text{6}.\text{4}×\text{1}{0}^{\text{6}}\right)}^{3}}\\ =0.3\text{\hspace{0.17em}}\mathrm{G}\\ \text{This value is of the order of magnitude of the}\\ \text{observed values\hspace{0.17em}of\hspace{0.17em}field on Earth}.\end{array}$

(f) There are several local poles on earth’s surface oriented in different directions. This is because of the presence of deposits of magnetized materials. The magnetised mineral deposit is an example of a local N-S pole.

(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f) Interstellar space has an extremely weak magnetic field of the order of 10−12 T. Can such a weak field be of any significant consequence? Explain.
[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]

Ans.

(a) The earth’s magnetic field changes with time. Earth takes a few hundred years to change by an appreciable amount. The change in earth’s magnetic field with the time cannot be neglected.
(b) Earth’s core contains iron in the molten form which is is not ferromagnetic. Therefore, it cannot be considered as a source of earth’s magnetism.
(c) The radioactivity in the interior of Earth is the source of energy that sustains the currents in the outer conducting regions of the core of Earth. These charged currents are considered to be responsible for earth’s magnetism, but it is not certain.
(d) Earth reversed the direction of its magnetic field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in certain rocks during their solidification. The analysis of this rock magnetism can reveal the geomagnetic history of Earth.
(e) Earth’s magnetic field departs from its dipole shape substantially at large distances (greater than about 30,000 km) due to the presence of the ionosphere. In this region, earth’s field gets modified due to the magnetic field produced by the motion of single ions.
(f) An extremely weak magnetic field can deflect charged particles moving in a circular path. This may be less noticeable for a large radius path. Therefore, over the gigantic interstellar space, the deflection can affect the passage of charged particles.

Q.3 A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Ans.

$\begin{array}{l}\text{The}\text{\hspace{0.17em}}\text{Earth}’\text{s magnetic field at the given place},\text{H}\\ =0.\text{36 G}\\ As\text{\hspace{0.17em}}the\text{\hspace{0.17em}}null\text{\hspace{0.17em}}po\mathrm{int}s\text{\hspace{0.17em}}lie\text{\hspace{0.17em}}on\text{\hspace{0.17em}}the\text{\hspace{0.17em}}axis\text{\hspace{0.17em}}of\text{\hspace{0.17em}}the\text{\hspace{0.17em}}magnet,\\ \therefore \text{Magnetic field at a distance d},\text{on the axis of the}\\ \text{magnet}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ {\text{B}}_{\text{1}}=\frac{{\mu }_{o}}{4\pi }\frac{2M}{{d}^{3}}\\ =H\to \left(i\right)\\ \text{Here},=\text{Permeability of free space}\\ \text{M}=\text{Magnetic moment}\\ \text{Magnetic field at the same distance d},\text{on the}\\ \text{equatorial line of the magnet}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ \text{\hspace{0.17em}}{\text{B}}_{\text{1}}=\frac{{\mu }_{o}}{4\pi }\frac{M}{{d}^{3}}\\ =\frac{H}{2}\to \left(ii\right)\\ \text{Total magnetic field},B={B}_{1}+{B}_{2}\\ =H+\frac{H}{2}\\ =0.\text{36}+\frac{0.\text{36}}{2}\\ =0.54\text{\hspace{0.17em}}G\\ \therefore \text{The total}\text{\hspace{0.17em}}\text{magnetic field =}0.\text{54 G}\\ \text{It}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{along the direction of earth}’\text{s magnetic field}.\end{array}$

Q.4 A short bar magnet of magnetic moment m = 0.32 J T−1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its
(a) stable, and
(b) unstable equilibrium?
What is the potential energy of the magnet in each case?

Ans.

\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{magnetic}\text{\hspace{0.17em}}m\text{oment of bar magnet},\text{M}=0.{\text{32 JT}}^{-\text{1}}\\ \text{External magnetic field},\text{B}=0.\text{15 T}\\ \left(\text{a}\right)\text{In stable equilibrium,}\text{\hspace{0.17em}}t\text{he bar magnet is aligned}\\ \text{along the direction}\text{\hspace{0.17em}}\text{magnetic field}\text{\hspace{0.17em}}\\ \therefore \text{Angle between}\text{\hspace{0.17em}}\text{the bar magnet and the}\text{\hspace{0.17em}}\text{magnetic}\\ \text{field,}\theta \text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}0°.\\ \text{Potential energy=-MBcos}\theta \\ =-0.32×0.\text{15}×{\text{cos0}}^{\text{0}}\\ =-4.8×{10}^{-2}\text{\hspace{0.17em}}J\\ \left(\text{b}\right)\text{In}\text{\hspace{0.17em}}\text{unstable}\text{\hspace{0.17em}}\text{equilibrium,}\text{\hspace{0.17em}}t\text{he bar magnet is oriented}\\ \text{at}\text{\hspace{0.17em}}\text{18}0°\text{to the magnetic field}.\text{}\\ \therefore \theta =\text{18}0°\\ \text{Potential energy}=-\text{MB cos}\theta \\ =-0.32×0.\text{15}×{\text{cos180}}^{\text{0}}\\ =-0.32×0.\text{15}×-1\\ =4.8×{10}^{-2}\text{\hspace{0.17em}}J\end{array}

Q.5 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10−4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10−2 T is set up at an angle of 30o with the axis of the solenoid?

Ans.

Here, number of turns on the solenoid, n = 2000
Cross-sectional area of the solenoid, A = 1.6 × 10−4 m2
Current flowing in the solenoid, I = 4 A
(a)Magnetic moment associated with solenoid is given by,
M = nAI = 2000 × 1.6 × 10−4 × 4 = 1.28 Am2
(b)Here, magnetic field, B = 7.5 × 10−2 T
As the magnetic field is uniform,
Force on the solenoid = 0

$\begin{array}{l}\text{Angle between the magnetic field and the axis of the}\\ \text{solenoid},\text{}\theta \text{}=\text{3}0°\\ \therefore \text{Torque},\tau =MB\mathrm{sin}\theta \\ =\text{1}.\text{28}×\text{7}.\text{5}×\text{1}{0}^{-\text{2}}×\mathrm{sin}\text{3}0°\\ =\text{1}.\text{28}×\text{7}.\text{5}×\text{1}{0}^{-\text{2}}×\frac{1}{2}\\ =4.8×\text{1}{0}^{-\text{2}}\text{\hspace{0.17em}}Nm\\ \therefore F\text{orce on the solenoid =0}\\ \text{Torque on the solenoid =}4.8×\text{1}{0}^{-\text{2}}\text{\hspace{0.17em}}Nm\end{array}$

Q.6 If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

Ans.

Here, applied magnetic field, B = 0.25 T
Magnetic moment, M = 0.6 T−1
Angle between the axis of the solenoid and the direction of the applied field, θ=30°

$\begin{array}{l}\text{Torque acting on the solenoid\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}}\mathrm{as},\\ \mathrm{\tau }=\mathrm{MBsin\theta }\\ \therefore \mathrm{\tau }=0.6×0.25\mathrm{sin}{30}^{\mathrm{o}}\\ =0.075\text{\hspace{0.17em}}\mathrm{J}.\end{array}$

Q.7 A bar magnet of magnetic moment 1.5 JT−1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b) What is the torque on the magnet in cases (i) and (ii)?

Ans.

(a) Here, magnetic moment, M = 1.5 J T−1
Magnetic field strength, B = 0.22 T

\begin{array}{l}\left(\text{i}\right)\text{Initial angle between the axis\hspace{0.17em}of\hspace{0.17em}bar\hspace{0.17em}magnet and the}\\ \text{magnetic field},\text{}{\mathrm{\theta }}_{\text{1}}=0\mathrm{°}\\ \text{Final angle between the axis of\hspace{0.17em}bar\hspace{0.17em}magnet\hspace{0.17em}and the}\\ \text{magnetic field},\text{}{\mathrm{\theta }}_{\text{2}}=\text{9}0\mathrm{°}\\ \text{Work required to align the magnetic moment of\hspace{0.17em}the\hspace{0.17em}}\\ \mathrm{bar}\text{\hspace{0.17em}}\mathrm{magnet}\text{\hspace{0.17em}}\mathrm{normal}\text{\hspace{0.17em}}\mathrm{to}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{direction}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{field},\\ \mathrm{W}=-\mathrm{MB}\left({\mathrm{cos\theta }}_{2}-{\mathrm{cos\theta }}_{1}\right)\\ =-1.5×0.22\left(\mathrm{cos}\text{9}0\mathrm{°}-\mathrm{cos}0\mathrm{°}\right)\\ =-0.33\left(0-1\right)=0.33\text{\hspace{0.17em}}\mathrm{J}\\ \left(\text{ii}\right)\text{Initial angle between the axis of\hspace{0.17em}bar\hspace{0.17em}magnet\hspace{0.17em}and}\\ \text{the magnetic field},\text{}{\mathrm{\theta }}_{\text{1}}=\text{}0\mathrm{°}\\ \text{Final angle between the axis\hspace{0.17em}of\hspace{0.17em}bar\hspace{0.17em}magnet and the}\\ \text{magnetic field},\text{}{\mathrm{\theta }}_{\text{2}}=\text{18}0\mathrm{°}\\ \text{Work required to align the magnetic moment of\hspace{0.17em}the\hspace{0.17em}}\\ \mathrm{bar}\text{\hspace{0.17em}}\mathrm{magnet}\text{\hspace{0.17em}}\mathrm{opposite}\text{\hspace{0.17em}}\mathrm{to}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{direction}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{field},\\ \mathrm{W}=-\mathrm{MB}\left({\mathrm{cos\theta }}_{2}-{\mathrm{cos\theta }}_{1}\right)\\ =-1.5×0.22\left(\mathrm{cos}180\mathrm{°}-\mathrm{cos}0\mathrm{°}\right)\\ =-0.33\left(-1-1\right)=0.66\text{\hspace{0.17em}}\mathrm{J}\\ \left(\text{b}\right)\mathrm{Torque}\text{\hspace{0.17em}}\mathrm{on}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{magnet}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{by}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{relation},\text{\hspace{0.17em}}\\ \mathrm{\tau }=\mathrm{MBsin\theta }\\ \text{For case}\left(\text{i}\right):\\ \mathrm{\theta }={90}^{\mathrm{o}}\\ \therefore \text{Torque},\mathrm{\tau }=1.5×0.22\mathrm{sin}{90}^{\mathrm{o}}\\ =0.33\text{\hspace{0.17em}}\mathrm{J}\\ \text{For case}\left(\text{ii}\right):\\ \mathrm{\theta }={180}^{\mathrm{o}}\\ \therefore \text{Torque},\mathrm{\tau }=1.5×0.22\mathrm{sin}{180}^{\mathrm{o}}\\ =0\text{\hspace{0.17em}}\mathrm{J}\end{array}

Q.8 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10−4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10−2 T is set up at an angle of 30o with the axis of the solenoid?

Ans.

Here, number of turns on the solenoid, n = 2000
Cross-sectional area of the solenoid, A = 1.6 × 10−4 m2
Current flowing in the solenoid, I = 4 A
(a)Magnetic moment associated with solenoid is given by,
M = nAI = 2000 × 1.6 × 10−4 × 4 = 1.28 Am2
(b)Here, magnetic field, B = 7.5 × 10−2 T
As the magnetic field is uniform,
Force on the solenoid = 0

$\begin{array}{l}\text{Angle between the magnetic field and the axis of the}\\ \text{solenoid},\text{}\theta \text{}=\text{3}0°\\ \therefore \text{Torque},\tau =MB\mathrm{sin}\theta \\ =\text{1}.\text{28}×\text{7}.\text{5}×\text{1}{0}^{-\text{2}}×\mathrm{sin}\text{3}0°\\ =\text{1}.\text{28}×\text{7}.\text{5}×\text{1}{0}^{-\text{2}}×\frac{1}{2}\\ =4.8×\text{1}{0}^{-\text{2}}\text{\hspace{0.17em}}Nm\\ \therefore F\text{orce on the solenoid =0}\\ \text{Torque on the solenoid =}4.8×\text{1}{0}^{-\text{2}}\text{\hspace{0.17em}}Nm\end{array}$

Q.9 A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10−2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s−1. What is the moment of inertia of the coil about its axis of rotation?

Ans.

Here, number of turns in the circular coil, N = 16
Radius of coil, r = 10 cm
= 0.1 m
Cross-sectional area of the coil, A = πr2
= π × (0.1)2 m2
Current in the circular coil, I = 0.75 A
Magnetic field strength, B = 5.0 × 10−2 T

$\begin{array}{l}\text{Frequency of oscillations of the coil},\text{v}=\text{2}.0{\text{s}}^{-\text{1}}\\ \therefore \text{Magnetic moment},\text{M}=\text{NIA}\\ \text{=NI}{\mathrm{\pi r}}^{2}\\ =\text{16}×0.\text{75}×\mathrm{\pi }×{\left(0.\text{1}\right)}^{\text{2}}\\ =0.{\text{377 JT}}^{-\text{1}}\\ \mathrm{Frequency}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{by},\text{\hspace{0.17em}}\mathrm{\nu }=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}\\ \mathrm{Here},\text{\hspace{0.17em}}\mathrm{I}=\mathrm{M}.\mathrm{I}.\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{circular}\text{\hspace{0.17em}}\mathrm{coil}\\ \therefore \mathrm{I}=\frac{\mathrm{MB}}{4{\mathrm{\pi }}^{2}{\mathrm{\nu }}^{2}}\\ =\frac{0.\text{377}×5×{10}^{-2}}{4{\mathrm{\pi }}^{2}{\left(2\right)}^{2}}\\ =1.19×{10}^{-4}\text{\hspace{0.17em}}{\mathrm{kgm}}^{2}\\ \therefore \mathrm{M}\text{oment of inertia of the coil about its axis}\\ \text{of rotation is\hspace{0.17em}}1.19×{10}^{-4}\text{\hspace{0.17em}}{\mathrm{kgm}}^{2}.\end{array}$

Q.10 A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22o with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{h}\text{orizontal component of thearth}’\text{s magnetic field},\text{}\\ {\text{B}}_{\text{H}}=0.\text{35 G}\\ \text{Angle subtended by the needle with the horizontal}\\ \text{plane}=\text{Angle of dip}=\mathrm{\delta }\\ ={22}^{\mathrm{o}}\\ \text{Let\hspace{0.17em}strength\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}Earth}’\text{s magnetic field}=\text{B}\\ {\text{The\hspace{0.17em}relation\hspace{0.17em}between\hspace{0.17em}B and B}}_{\text{H}}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}as}:\\ {\text{B}}_{\text{H}}=\text{B}\mathrm{cos\theta }\\ \therefore \text{B}=\frac{{\text{B}}_{\text{H}}}{\mathrm{cos\delta }}\\ =\frac{0.\text{35}}{\mathrm{cos}{22}^{\mathrm{o}}}\\ =0.377\text{\hspace{0.17em}}\mathrm{G}\\ \therefore \text{The strength of the\hspace{0.17em}}\mathrm{E}\text{arth}’\text{s magnetic field at the}\\ \text{given location =\hspace{0.17em}}0.\text{377 G}.\end{array}$

Q.11 At a certain location in Africa, a compass points 12o west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60o above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}a\text{ngle of declination},\theta =\text{12}°\\ \text{Angle of dip},\text{\hspace{0.17em}}\delta ={60}^{o}\\ \text{Horizontal component of the}\text{\hspace{0.17em}}E\text{arth}’\text{s magnetic field},{\text{B}}_{\text{H}}\text{}\\ =0.\text{16 G}\\ \text{Let}\text{\hspace{0.17em}}\text{Earth}’\text{s magnetic field at the given location}=\text{B}\\ \text{The}\text{\hspace{0.17em}}\text{relation}\text{\hspace{0.17em}}{\text{between B and B}}_{\text{H}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as}:\\ {\text{B}}_{\text{H}}=B\mathrm{cos}\delta \\ \therefore B=\frac{{\text{B}}_{\text{H}}}{\mathrm{cos}\delta }\\ =\frac{0.\text{16}}{\mathrm{cos}{60}^{o}}\\ =0.32\text{\hspace{0.17em}}G\\ \\ \text{The}\text{\hspace{0.17em}}\text{Earth}’\text{s magnetic field lies in the vertical plane},\text{12}°\\ \text{west of geographic meridian}\text{\hspace{0.17em}}\text{at an angle of 6}0°\text{}\\ \left(\text{upward}\right)\text{with the horizontal direction}.\text{}\\ \text{The magnitude of}\text{\hspace{0.17em}}\text{Earth}’\text{s magnetic field}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}0.\text{32 G}.\end{array}$

Q.12 A short bar magnet has a magnetic moment of 0.48 J T−1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Ans.

Here, magnetic moment of bar magnet, M = 0.48 JT−1

Distance, d = 10 cm

= 0.1 m

$\begin{array}{l}\text{(a)}\text{\hspace{0.17em}}\text{Magnetic field on the axis of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{magnet}\text{\hspace{0.17em}}\text{is given by}\\ \text{\hspace{0.17em}}\text{the relation}:\\ B=\frac{{\mu }_{o}}{4\pi }\frac{2M}{{d}^{3}}\\ \text{Here},\text{\hspace{0.17em}}{\mu }_{o}=\text{Permeability of free space}=4\pi ×{10}^{-7}\text{\hspace{0.17em}}Tm{A}^{-1}\\ \therefore B=\frac{4\pi ×{10}^{-7}×2×0.48}{4\pi ×{\left(0.1\right)}^{3}}\\ =0.96×{10}^{-4}\text{\hspace{0.17em}}T\\ =0.96\text{\hspace{0.17em}}G\\ \text{It}\text{\hspace{0.17em}}\text{is along the S}-\text{N direction}.\\ \left(\text{b}\right)\text{The magnetic field on the equatorial line of}\text{\hspace{0.17em}}\text{the}\\ \text{magnet}\text{\hspace{0.17em}}\text{is given as}:\\ B=\frac{{\mu }_{o}}{4\pi }\frac{M}{{d}^{3}}={10}^{-7}×\frac{0.48}{{\left(0.1\right)}^{3}}\\ =0.48×{10}^{-4}\text{\hspace{0.17em}}T\text{\hspace{0.17em}}\\ \text{It}\text{\hspace{0.17em}}\text{is along the N}-\text{S direction}.\end{array}$

Q.13. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null−point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)

Ans.

$\begin{array}{l}\text{The}\text{\hspace{0.17em}}\text{Earth}’\text{s magnetic field at the given place},\text{H}\\ =0.\text{36 G}\\ As\text{\hspace{0.17em}}the\text{\hspace{0.17em}}null\text{\hspace{0.17em}}po\mathrm{int}s\text{\hspace{0.17em}}lie\text{\hspace{0.17em}}on\text{\hspace{0.17em}}the\text{\hspace{0.17em}}axis\text{\hspace{0.17em}}of\text{\hspace{0.17em}}the\text{\hspace{0.17em}}magnet,\\ \therefore \text{Magnetic field at a distance d},\text{on the axis of the}\\ \text{magnet}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ {\text{B}}_{\text{1}}=\frac{{\mu }_{o}}{4\pi }\frac{2M}{{d}^{3}}\\ =H\to \left(i\right)\\ \text{Here},=\text{Permeability of free space}\\ \text{M}=\text{Magnetic moment}\\ \text{Magnetic field at the same distance d},\text{on the}\\ \text{equatorial line of the magnet}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ \text{\hspace{0.17em}}{\text{B}}_{\text{1}}=\frac{{\mu }_{o}}{4\pi }\frac{M}{{d}^{3}}\\ =\frac{H}{2}\to \left(ii\right)\\ \text{Total magnetic field},B={B}_{1}+{B}_{2}\\ =H+\frac{H}{2}\\ =0.\text{36}+\frac{0.\text{36}}{2}\\ =0.54\text{\hspace{0.17em}}G\\ \therefore \text{The total}\text{\hspace{0.17em}}\text{magnetic field =}0.\text{54 G}\\ \text{It}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{along the direction of earth}’\text{s magnetic field}.\end{array}$

Q.14 If the bar magnet in exercise 5.13 is turned around by 180o, where will the new null points be located?

Ans.

$\begin{array}{l}\text{Magnetic field on the axis of the magnet at a distance}\\ {\text{d}}_{\text{1}}=\text{14 cm},\text{can be written}\text{\hspace{0.17em}}\text{as}:\\ {B}_{1}=\frac{{\mu }_{o}}{4\pi }\frac{2M}{{\left({d}_{1}\right)}^{3}}\\ =H\\ \text{Here},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{M}=\text{Magnetic moment}\\ {\mu }_{o}=\text{Permeability of free space}\\ \text{H}={\text{Horizontal component of magnetic field at d}}_{\text{1}}\\ \text{If bar magnet is turned around}\text{\hspace{0.17em}}\text{by 18}0°,\text{then the}\\ \text{neutral point will lie on the equatorial}\text{\hspace{0.17em}}\text{line}.\\ \therefore \text{Magnetic field on the equatorial line}\text{\hspace{0.17em}}{\text{at a distance d}}_{\text{2}}\\ \text{of the magnet can be}\text{\hspace{0.17em}}\text{written as}:\\ {B}_{2}=\frac{{\mu }_{o}}{4\pi }\frac{M}{{\left({d}_{2}\right)}^{3}}\\ =H\to \left(ii\right)\\ \text{Equating equation}\left(i\right)\text{and equation}\text{\hspace{0.17em}}\left(ii\right),\text{we obtain}\\ \frac{2}{{\left({d}_{1}\right)}^{3}}=\frac{1}{{\left({d}_{2}\right)}^{3}}\\ \therefore {\left(\frac{{d}_{2}}{{d}_{1}}\right)}^{3}=\frac{1}{2}\\ \therefore {d}_{2}={d}_{1}×{\left(\frac{1}{2}\right)}^{\frac{1}{3}}\\ =14×0.794\\ =11.1\text{\hspace{0.17em}}cm\\ \therefore \text{New null points will be located 11}.\text{1 cm on the normal}\\ \text{bisector}.\end{array}$

Q.15 A short bar magnet of magnetic moment 5.25 × 10−2 J T−1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the Magnet, the resultant field is inclined at 45o with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{m}\text{agnetic moment of the bar magnet},\text{}\\ \text{M}=\text{5}.\text{25}×\text{1}{0}^{-\text{2}}{\text{J T}}^{-\text{1}}\\ \text{Magnitude of earth}’\text{s magnetic field at the\hspace{0.17em}given place},\text{H}\\ =0.\text{42 G}=0.\text{42}×\text{1}{0}^{-\text{4}}\text{T}\\ \left(\text{a}\right)\text{Magnetic field at a distance R from the centre}\\ \text{of the magnet on the normal bisector is given by,}\\ \text{B=}\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{R}}^{3}}\\ \text{Here},\\ {\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}=\text{4}\mathrm{\pi }×\text{1}{0}^{-\text{7}}{\text{Tm A}}^{-\text{1}}\\ \text{When the resultant field is inclined at 45}\mathrm{°}\text{with earth}’\text{s}\\ \text{field},\text{\hspace{0.17em}}\mathrm{then}\text{B}=\text{H}\mathrm{}\\ \therefore \frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{\mathrm{M}}{{\mathrm{R}}^{3}}=\mathrm{H}\\ =0.42×{10}^{-4}\\ {\mathrm{R}}^{3}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{M}}{0.42×{10}^{-4}×4\mathrm{\pi }}\\ =\frac{4\mathrm{\pi }×\text{1}{0}^{-\text{7}}×\text{5}.\text{25}×\text{1}{0}^{-\text{2}}}{4\mathrm{\pi }×0.42×{10}^{-4}}\\ =12.5×\text{1}{0}^{-5}\\ \therefore \mathrm{R}=0.05\text{\hspace{0.17em}}\mathrm{m}\\ =5\text{\hspace{0.17em}}\mathrm{cm}\\ \left(\text{b}\right)\text{Magnetic field at a distance\hspace{0.17em}}\mathrm{R}‘\text{from the centre}\\ \text{of the magnet on its axis is\hspace{0.17em}given by}:\\ \text{B’}=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\left(\mathrm{R}‘\right)}^{3}}\\ \mathrm{Since}\text{\hspace{0.17em}}\mathrm{t}\text{he resultant field is inclined at 45}\mathrm{°}\text{with earth}’\text{s}\\ \text{field,}\\ \therefore \text{B’}=\mathrm{H}\\ \frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{{\left(\mathrm{R}‘\right)}^{3}}=\mathrm{H}\\ \therefore {\left(\mathrm{R}‘\right)}^{3}=\frac{{\mathrm{\mu }}_{\mathrm{o}}}{4\mathrm{\pi }}\frac{2\mathrm{M}}{\mathrm{H}}\\ =\frac{4\mathrm{\pi }×\text{1}{0}^{-\text{7}}×2×\text{5}.\text{25}×\text{1}{0}^{-\text{2}}}{4\mathrm{\pi }×0.42×{10}^{-4}}\\ =25×{10}^{-5}\\ \therefore \mathrm{R}‘=0.063\text{\hspace{0.17em}}\mathrm{m}\\ =6.3\text{\hspace{0.17em}}\mathrm{cm}\end{array}$

(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?

Ans.

(a)Due to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced due to the reduced random thermal motion. That is why, a paramagnetic sample displays greater magnetisation when cooled.
(b) In a diamagnetic substance, the induced dipole moment is always opposite to the magnetising field. Therefore, the random motion of the atoms (which is related to temperature) does not affect the diamagnetism of a material.
(c)As bismuth is a diamagnetic substance, therefore, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
(d)The permeability of a ferromagnetic material is not independent of the applied magnetic field. From the hysteresis curve it is clear that, μ is greater for a lower field.
(e)The permeability of a ferromagnetic material is always greater than one. That is why the magnetic field lines are always nearly normal to the surface of such materials at every point.
(f)Yes, maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. However, this requires very high magnetising fields for saturation, which are hard to achieve.

(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?
(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

Ans.

The hysteresis curve (B-H curve) of a ferromagnetic material is shown in the given figure.
(a) From the given curve, it can be observed that magnetisation persists even when the external field is removed. This indicates the irreversibility in the magnetization curve of a ferromagnet.
(b) The energy dissipated per cycle is directly proportional to the area of the hysteresis loop. Since the carbon steel piece has a greater hysteresis curve area, therefore, it dissipates greater heat energy.
(c) The value of magnetisation of a ferromagnet depends not only on the magnetising field, but also on the history of magnetisation (i.e., how many cycles of magnetisation it has undergone). Hence, value of magnetisation of a specimen is the record of memory of the cycles of magnetization it has undergone. The system displaying such a hysteresis loop can be used for storing information.
(d) Ferrites are the ferromagnetic materials used for coating magnetic tapes in cassette players and for building memory stores in modern computers. Some common ferrites are FeFe2O4, CoFe2O4, and MnFe2O4 etc.
(e) A certain region of space can be shielded from magnetic fields by surrounding it by soft iron ring. In such arrangement, the magnetic field lines are drawn into the ring and the enclosed region will become free of magnetic field.

Q.18 A long straight horizontal cable carries a current of 2.5 A in the direction 10o south of west to 10° north of east. The magnetic meridian of the place happens to be 10o west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable).
(At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}}\mathrm{c}\text{urrent in the wire},\text{I}=\text{2}.\text{5 A}\\ \text{Angle of dip at the given location},\text{\hspace{0.17em}}\mathrm{\delta }={0}^{\mathrm{o}}\\ \text{Earth}’\text{s magnetic field\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}given\hspace{0.17em}location},\text{B}\\ =0.\text{33 G}\\ =0.\text{33}×\text{1}{0}^{-\text{4}}\text{T}\\ \text{Horizontal component of earth}’\text{s magnetic field is}\\ \text{given by,}\\ {\text{B}}_{\text{H}}=\text{B cos}\mathrm{\delta }\text{\hspace{0.17em}}\\ \text{=\hspace{0.17em}}0.\text{33}×\text{1}{0}^{-\text{4}}×\text{cos}{0}^{\mathrm{o}}\\ =0.\text{33}×\text{1}{0}^{-\text{4}}\text{\hspace{0.17em}}\mathrm{T}\\ \text{Magnetic field at the neutral point at a distance R}\\ \text{from the cable is given by,}\\ {\text{B}}_{\text{H}}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi R}}\\ \text{Here},\text{\hspace{0.17em}}{\mathrm{\mu }}_{\mathrm{o}}\text{\hspace{0.17em}}=\text{Permeability of free space}\\ =\text{\hspace{0.17em}}4\mathrm{\pi }×\text{1}{0}^{-7}\text{\hspace{0.17em}}{\mathrm{TmA}}^{-1}\\ \therefore \mathrm{R}=\frac{{\mathrm{\mu }}_{\mathrm{o}}\mathrm{I}}{2\mathrm{\pi }{\text{B}}_{\text{H}}}\\ =\frac{4\mathrm{\pi }×\text{1}{0}^{-7}×2.5}{2\mathrm{\pi }×0.\text{33}×\text{1}{0}^{-\text{4}}}\\ =15.15×\text{1}{0}^{-3}\text{\hspace{0.17em}}\mathrm{m}\\ =1.51\text{\hspace{0.17em}}\mathrm{cm}\\ \therefore \mathrm{N}\text{eutral points parallel to and above the\hspace{0.17em}cable are}\\ \text{located at a normal\hspace{0.17em}distance of 1}.\text{51 cm}.\end{array}$

Q.19 A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35o. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?

Ans.

$\begin{array}{l}\text{Here,}\text{\hspace{0.17em}}\text{Number of horizontal wires},\text{n}=\text{4}\\ \text{Current flowing}\text{\hspace{0.17em}}\text{in each wire},\text{I}=\text{1}.0\text{A}\\ \text{Earth}’\text{s magnetic field at the}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{place},\text{R}=0.\text{39 G}\\ =0.\text{39}×\text{1}{0}^{-\text{4}}\text{T}\\ \text{Angle of dip at the given}\text{\hspace{0.17em}}place,\text{}\delta =\text{35}°\\ \text{Angle of declination}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{place},\text{}\theta \sim 0°\\ \text{At}\text{\hspace{0.17em}}\text{a point 4 cm below the cable}:\\ \text{Distance},\text{r}=\text{4 cm}\\ =0.0\text{4 m}\\ \\ \text{The horizontal component of earth}’\text{s magnetic field,}\\ {\text{H}}_{\text{h}}=\text{Rcos}\delta \text{}-\text{B}\\ \text{Here},\\ \text{B}=\text{Magnetic field at 4 cm because}\text{\hspace{0.17em}}\text{of current I in the}\\ \text{four}\text{\hspace{0.17em}}\text{wires}=\text{4}×\frac{{\mu }_{o}I}{2\pi r}\\ {\mu }_{o}=\text{Permeability of free space}=\text{4}\pi ×\text{1}{0}^{-\text{7}}\text{\hspace{0.17em}}{\text{Tm A}}^{-\text{1}}\\ \therefore B=\text{4}×\frac{\text{4}\pi \text{}×\text{1}{0}^{-\text{7}}×1}{2\pi ×0.0\text{4}}\\ =0.\text{2}×\text{1}{0}^{-\text{4}}\text{T}\\ =0.\text{2 G}\\ \therefore {H}_{h}=0.\text{39 cos 35}°-0.\text{2}\\ =0.\text{39}×0.\text{819}-0.\text{2}\approx \text{}0.\text{12 G}\\ \text{Vertical component of earth}’\text{s magnetic field,}\\ {\text{H}}_{\text{v}}=\text{Rsin}\delta \\ =0.\text{39 sin 35}°\\ =0.\text{22 G}\\ \text{The angle made by the field with its horizontal}\\ \text{component is given by,}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}{\text{tan}}^{\text{-1}}\frac{{\text{H}}_{\text{v}}}{{H}_{h}}\\ =\text{\hspace{0.17em}}{\text{tan}}^{\text{-1}}\frac{0.\text{22}}{0.\text{12}}\\ ={61.39}^{o}\\ \\ \text{Resultant magnetic}\text{\hspace{0.17em}}\text{field at the point is given by,}\\ {\text{H}}_{\text{1}}=\sqrt{{\left({\text{H}}_{\text{v}}\right)}^{2}+{\left({\text{H}}_{h}\right)}^{2}}\\ =\sqrt{{\left(0.22\right)}^{2}+{\left(0.12\right)}^{2}}\\ =0.25\text{\hspace{0.17em}}G\\ \text{At a point 4 cm above the cable}:\\ \text{Horizontal component of earth}’\text{s magnetic fieldis}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\\ \text{by,}\\ {\text{H}}_{\text{h}}=\text{Rcos}\delta +\text{B}\\ =0.\text{39 cos 35}°+0.\text{2}\\ =0.\text{52 G}\\ \text{Vertical component of earth}’\text{s magnetic field}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ {\text{H}}_{\text{v}}=\text{Rsin}\delta \\ =0.\text{39 sin 35}°\\ =0.\text{22 G}\\ \text{Angle},\text{}\theta ={\mathrm{tan}}^{-1}\frac{{\text{H}}_{\text{v}}}{{\text{H}}_{\text{h}}}\text{}\\ \text{=}{\mathrm{tan}}^{-1}\frac{0.22}{0.\text{52}}\\ \text{=22}.\text{9}°\\ \text{Resultant magnetic}\text{\hspace{0.17em}}\text{field}\text{\hspace{0.17em}}\text{at}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{point}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ {\text{H}}_{2}=\sqrt{{\left({\text{H}}_{\text{v}}\right)}^{2}+{\left({\text{H}}_{h}\right)}^{2}}\\ =\sqrt{{\left(0.22\right)}^{2}+{\left(0.52\right)}^{2}}\\ =0.56\text{\hspace{0.17em}}G\end{array}$

Q.20 A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45o with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90o in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Ans.

Here, number of turns in the circular coil, N = 30
Radius of circular coil, r = 12 cm = 0.12 m
Current in the circular coil, I = 0.35 A
Angle of dip, δ = 45°

$\begin{array}{l}\left(\text{a}\right)\text{Magnetic field due to current I},\text{at a distance r},\text{}\\ \text{is given by\hspace{0.17em}the\hspace{0.17em}relation},\\ \text{B}=\frac{{\mathrm{\mu }}_{\mathrm{o}}2\mathrm{\pi NI}}{4\mathrm{\pi r}}\\ \text{Here},\text{\hspace{0.17em}}{\mathrm{\mu }}_{\mathrm{o}}=\text{Permeability of free space}\\ =\text{4}\mathrm{\pi }×\text{1}{0}^{-\text{7}}{\text{Tm A}}^{-\text{1}}\\ \therefore \text{B}=\frac{\text{4}\mathrm{\pi }×\text{1}{0}^{-\text{7}}×2\mathrm{\pi }×30×0.35}{\text{4}\mathrm{\pi }×0.12}\\ =\text{5}.\text{49}×\text{1}{0}^{-\text{5}}\text{\hspace{0.17em}T}\\ \text{As\hspace{0.17em}}\mathrm{t}\text{he compass needle points from West to East}.\text{}\\ \therefore \text{The horizontal component of\hspace{0.17em}earth}’\text{s magnetic}\\ \text{field is given as}:\\ {\text{B}}_{\text{H}}=\text{Bsin}\mathrm{\delta }\\ =\text{5}.\text{49}×\text{1}{0}^{-\text{5}}\text{sin 45}\mathrm{°}\text{}\\ =\text{3}.\text{88}×\text{1}{0}^{-\text{5}}\text{T}\end{array}$

(b) When the current in the coil is reversed and the coil is rotated by an angle of 90o anticlockwise, the needle will reverse its original direction (i.e., the needle will point from East to West).

Q.21 A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60o, and one of the fields has a magnitude of 1.2 × 10−2 T. If the dipole comes to stable equilibrium at an angle of 15o with this field, what is the magnitude of the other field?

Ans.

Magnitude of one of the magnetic fields, B1=1.2 × 10−2 T

Magnitude of the other magnetic field = B2

Angle between the two fields, θ = 60°

In equilibrium, the angle between the dipole and field B1, θ1 = 15°

Angle between the dipole and field B2, θ2 = θ − θ1

= 60° − 15°

= 45°

$\begin{array}{l}\text{At equilibrium},\text{torques between the two fields must}\\ \text{balance each other}.\\ \therefore {\text{Torque due to field B}}_{\text{1}}={\text{Torque due to field B}}_{\text{2}}\\ {\text{MB}}_{\text{1}}\text{sin}{\theta }_{\text{1}}={\text{MB}}_{\text{2}}\text{sin}{\theta }_{\text{2}}\\ \text{Here},\text{M}=\text{Magnetic moment of the dipole}\\ \therefore {\text{B}}_{\text{2}}=\frac{{\text{B}}_{\text{1}}\text{sin}{\theta }_{\text{1}}}{\text{sin}{\theta }_{\text{2}}}\\ =\frac{\text{1}.\text{2}×\text{1}{0}^{-\text{2}}×\text{sin}\text{\hspace{0.17em}}{15}^{o}}{\text{sin}\text{\hspace{0.17em}}{45}^{o}}\\ =4.39×\text{1}{0}^{-3}\text{\hspace{0.17em}}T\\ \therefore \text{The magnitude of the other magnetic field is}\\ \text{4}.\text{39}×\text{1}{0}^{-\text{3}}\text{\hspace{0.17em}}\text{T}.\end{array}$

Q.22 A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me= 9.11 × 10−19 C). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]

Ans.

Here, energy of the electron beam, E = 18 keV

= 18 ×103 eV

Charge on electron, e = 1.6 × 10−19 C

E = 18 × 103 × 1.6 × 10−19 J

Magnetic field, B = 0.04 G

Mass of electron, me = 9.11 × 10−19 kg

Distance up to which the electron beam travels, d = 30 cm = 0.3 m

$\begin{array}{l}\text{Kinetic energy of the electron beam is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{by,}\\ E=\frac{1}{2}m{v}^{2}\\ \therefore v=\sqrt{\frac{2E}{m}}\\ =\sqrt{\frac{2×18×{10}^{3}×1.6×{10}^{-19}×{10}^{-15}}{9.11×{10}^{-31}}}\\ =0.795×{10}^{8}\text{\hspace{0.17em}}m{s}^{-1}\\ \text{The electron beam bends along a circular path of}\\ \text{radius},\text{r}.\\ \text{The force due to magnetic field balances the}\\ \text{centripetal force of the}\text{\hspace{0.17em}}\text{circular path}.\\ \text{Let the up and down deflection of the electron beam,}\text{\hspace{0.17em}}\text{x}\\ \text{=r}\left(1-\mathrm{cos}\theta \right)\\ \text{Here},\theta =\text{Angle of declination}\\ \text{sin}\theta \text{\hspace{0.17em}}\text{=}\frac{d}{r}\\ =\frac{0.3}{11.3}\\ \therefore \theta ={\text{sin}}^{-1}\left(\frac{0.3}{11.3}\right)={1.521}^{o}\\ \therefore x=11.3\left(1-\mathrm{cos}{1.521}^{o}\right)\\ =0.0039\text{\hspace{0.17em}}m\\ =3.9\text{\hspace{0.17em}}mm\\ \therefore \text{The up and down deflection of the beam is 3}.\text{9 mm}.\end{array}$

Q.23 A sample of paramagnetic salt contains 2.0 × 1024 atomic dipoles each of dipole moment 1.5 × 10−23 J T−1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)

Ans.

Here, number of atomic dipoles, n = 2.0 × 1024

Dipole moment of each dipole, M = 1.5 × 10−23 J T−1

For the magnetic field, B1 = 0.64 T

The sample is cooled to the temperature, T1 = 4.2° K

Total dipole moment of the sample, Mtot = n × M

= 2 × 1024 × 1.5 × 10−23

= 30 JT−1

As, magnetic saturation achieved = 15%.

$\begin{array}{l}\therefore \text{Effective dipole moment},\\ {M}_{1}=\frac{15}{100}×30\\ =4.5\text{\hspace{0.17em}}J{T}^{-1}\\ \text{When the magnetic field},{\text{B}}_{\text{2}}=\text{}0.\text{98 T}\\ \text{Temperature},{\text{T}}_{\text{2}}=\text{2}.\text{8}°\text{K}\\ \text{Total dipole moment}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{sample}={\text{M}}_{\text{2}}\\ \text{According to Curie}’\text{s law},\text{\hspace{0.17em}}\text{the ratio of two magnetic}\\ \text{dipoles is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as}:\\ \frac{{\text{M}}_{\text{2}}}{{M}_{1}}=\frac{{B}_{2}}{{B}_{1}}×\frac{{\text{T}}_{1}}{{\text{T}}_{\text{2}}}\\ \therefore {\text{M}}_{\text{2}}=\frac{{B}_{2}{\text{T}}_{1}{M}_{1}}{{B}_{1}{\text{T}}_{\text{2}}}\\ =\frac{0.\text{98}×4.2×4.5}{\text{2}.\text{8}×0.64}\\ =10.336\text{\hspace{0.17em}}J{T}^{-1}\\ \therefore T\text{otal dipole moment of the sample for}\text{\hspace{0.17em}}\text{a magnetic}\\ \text{field of}0.\text{98 T and a temperature of 2}.\text{8 K=10}\text{.336}\text{\hspace{0.17em}}{\text{JT}}^{\text{-1}}\end{array}$

Q.24 A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Ans.

$\begin{array}{l}\text{Here},\text{mean radius of the Rowland ring},\text{r}=\text{15 cm}\\ =0.\text{15 m}\\ \text{Number of turns on the ferromagnetic core},\text{N}=\text{35}00\\ \text{Relative permeability of the core material},\text{}{\mu }_{\text{r}}=\text{8}00\\ \text{Magnetising current},\text{I}=\text{1}.\text{2 A}\\ \text{Magnetic field is given by the relation}:\\ \text{B=}\frac{{\mu }_{\text{r}}{\mu }_{o}IN}{2\pi r}\\ \text{Here},{\mu }_{0}=\text{Permeability of free space}\\ \therefore B=\frac{\text{8}00×\text{4}\pi ×\text{1}{0}^{-\text{7}}×\text{1}.\text{2}×\text{35}00}{2\pi ×0.15}\\ \text{=4}\text{.48 T}\\ \therefore \text{The magnetic field in the core is 4}.\text{48 T}.\end{array}$

Q.25

$\begin{array}{l}{\text{The magnetic moment vectors μ}}_{\text{s}}{\text{and μ}}_{\text{l}}\text{}\\ \text{associated with the intrinsic spin angular}\\ \text{momentum}\stackrel{\to }{\text{S}}\text{and orbital angular momentum}\stackrel{\to }{\text{l}}\text{,}\\ \text{respectively, of an electron are predicted by}\\ \text{quantum theory (andverified experimentally}\\ \text{to a high accuracy) to be given by:}\\ {\text{μ}}_{\text{s}}\text{= –}\left(\text{e/m}\right)\text{}\stackrel{\to }{\text{S}}\text{,}\\ {\text{μ}}_{\text{l}}\text{= –}\left(\text{e/2m}\right)\stackrel{\to }{\text{l}}\\ \text{Which of these relations is in accordance with}\\ \text{the result expected classically?Outline the}\\ \text{derivation of the classical result.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Out}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{two}\text{\hspace{0.17em}}\mathrm{relations},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{moment}\text{\hspace{0.17em}}\\ \mathrm{associated}\text{\hspace{0.17em}}\mathrm{with}\text{\hspace{0.17em}}\mathrm{orbital}\text{\hspace{0.17em}}\mathrm{angular}\text{\hspace{0.17em}}\mathrm{momentum}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{valid}\text{\hspace{0.17em}}\mathrm{with}\\ \mathrm{classical}\text{\hspace{0.17em}}\mathrm{physics}.\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{Magnetic moment associated with orbital angular}\\ \text{momentum is given by,\hspace{0.17em}}{\mathrm{\mu }}_{\text{l}}=-\left(\frac{\mathrm{e}}{2\mathrm{m}}\right)\mathrm{l}\\ \text{For current i and area of cross}-\text{section A},\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{relation}\text{\hspace{0.17em}}\\ \mathrm{for}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{moment}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{given}\text{\hspace{0.17em}}\mathrm{by},\\ {\mathrm{\mu }}_{\text{l}}=\mathrm{iA}\\ \mathrm{From}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{definitions}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}{\mathrm{\mu }}_{\text{l}}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathrm{A},\text{\hspace{0.17em}}\mathrm{it}\text{\hspace{0.17em}}\mathrm{follows}\text{\hspace{0.17em}}\mathrm{that}\\ {\mathrm{\mu }}_{\text{l}}=\left(\frac{-\mathrm{e}}{\mathrm{T}}\right){\mathrm{\pi r}}^{2}\to \left(\mathrm{i}\right)\\ \text{Here},\text{e}=\text{Charge of the electron\hspace{0.17em}}\\ \text{r}=\text{Radius of the orbit}\\ \text{T}=\text{Time period\hspace{0.17em}of\hspace{0.17em}revolution\hspace{0.17em}of\hspace{0.17em}electron}\\ \text{Orbital\hspace{0.17em}}\mathrm{a}\text{ngular momentum},\text{l}=\text{mvr}\\ \mathrm{l}=\mathrm{m}×\frac{2\mathrm{\pi r}}{\mathrm{T}}×\mathrm{r}\to \left(\mathrm{ii}\right)\\ \\ \text{Here},\text{m}=\text{Mass of electron}\\ \text{v}=\text{Velocity of electron}\\ \text{r=\hspace{0.17em}Radius\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}orbit}\\ \text{Dividing equation}\left(\mathrm{i}\right)\text{by equation}\left(\mathrm{ii}\right),\text{we get}:\\ \frac{{\mathrm{\mu }}_{\text{l}}}{\mathrm{l}}=-\left(\frac{\mathrm{e}}{2\mathrm{m}}\right)\\ \therefore {\mathrm{\mu }}_{\text{l}}=-\left(\frac{\mathrm{e}}{2\mathrm{m}}\right)\mathrm{l}\\ \therefore \mathrm{The}\text{\hspace{0.17em}}\mathrm{magnetic}\text{\hspace{0.17em}}\mathrm{moment}\text{\hspace{0.17em}}\mathrm{associated}\text{\hspace{0.17em}}\mathrm{with}\text{\hspace{0.17em}}\mathrm{orbital}\text{\hspace{0.17em}}\mathrm{angular}\\ \mathrm{momentum}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}\mathrm{valid}\text{\hspace{0.17em}}\mathrm{with}\text{\hspace{0.17em}}\mathrm{classical}\text{\hspace{0.17em}}\mathrm{physics}.\text{\hspace{0.17em}}\end{array}$