Important Questions Class 12 Physics Chapter 4: Moving Charges and Magnetism

Important Questions Class 12 Physics Chapter 4: Moving Charges and Magnetism

Moving Charges and Magnetism explains how a moving charge or a current produces a magnetic field and how a magnetic field acts on a moving charge or a current-carrying wire. The chapter includes the Biot-Savart law, Ampere's circuital law, the motion of a charge in a magnetic field, the force between two parallel wires, torque on a current loop, and the moving coil galvanometer.

Here are 12 important questions from Moving Charges and Magnetism with detailed answers from Extramarks experts. The set has field at the centre of arcs and loops, force on charged particles, solenoid and toroid problems, galvanometer conversion and derivations that are asked often in the CBSE board exam. Read the question, try it yourself, and then compare with the given solution.

Q1.

separately for the contribution of the arc and the two straight sections.

Answer: The net magnetic field at O is the sum of the fields produced by the semicircular arc and the two semi-infinite straight sections.

The magnetic field due to each semi-infinite straight section at a perpendicular distance R is

Hence, the total field due to the two straight sections is

All three magnetic-field contributions are in the same direction. Therefore,

Thus,

The direction is perpendicular to the plane of the wire and is determined using the right-hand thumb rule. For clockwise current around the arc,

Q2.

Answer: The force per unit length between two long parallel current-carrying wires separated by a distance d is

Since the currents in P and Q are opposite, the force exerted by P on Q is repulsive:

Similarly, the force exerted by R on Q is

These forces have equal magnitudes and opposite directions. Hence,

To examine stability, suppose wire Q is displaced slightly towards wire R. Its distance from R decreases, while its distance from P increases.

Since

the repulsive force due to R becomes greater than that due to P. The resulting force acts back towards the original position.

Therefore,

Q3.

direction of the magnetic field.

Find:

(a) the radius of the helical path followed by the proton

(b) the pitch of the helical path

(c) how the radius and pitch change if the speed of the proton is doubled while the angle of entry remains unchanged.

Answer: The velocity of the proton is resolved into two components:

(a)The perpendicular component of velocity produces circular motion. The magnetic force provides the required centripetal force:

Therefore,

Substituting

we get

2qB

(b) Pitch of the helical path

The time period of the circular motion is

The pitch p is the distance travelled parallel to the magnetic field in one time period:

Therefore,

(c) Effect of doubling the speed

If the speed becomes 2v, then both velocity components are doubled:

the new radius is

Also, since

the new pitch is

Thus, both the radius and pitch of the helical path are doubled.

Q4. A solenoid of length 0.1 m and negligible diameter is wound with two layers of wire. The inner layer has 50 turns, and the outer layer has 40 turns. Currents of 3 A flow in opposite directions through the two layers. Calculate the magnetic induction at the centre of the solenoid.

Answer: The two layers of the solenoid carry currents in opposite directions, so their magnetic fields oppose each other. The net magnetic induction at the centre is therefore the difference between the fields produced by each layer. Since the solenoid is short and its diameter is negligible, the field inside is approximately uniform and depends on the number of turns per unit length and the current in each layer

Q5. In a physics lab, three particles enter a uniform magnetic field and follow different curved paths on a screen. Based on the direction of each path, can you identify which particle is positively charged, which is negatively charged, and which is neutral?

Answer: (1) Positive charge = According to Right-Hand Rule positive charge will be move in circular upward direction.] (2) Neutral ⇒ q = 0, F = 0 No magnetic force No deviation (3) Negative charge = According to Right-Hand Rule negative charge will be move in circular downward direction.

Q6. The figure shows four arrangements of circular loops of radius r or 2r, centered on vertical axes (perpendicular to the loops) and carrying identical currents in the directions indicated.

(a) Which arrangement produces the maximum magnetic field at the dot located midway between the loops on the central axis?

(b) Rank all four arrangements in order of the magnitude of the net magnetic field at the dot, starting from the greatest to the least.

Answer: Arrangement with maximum magnetic field:

Let's say the magnetic field due to smaller loop at the dot is B1 and magnetic field due to larger loop at dot is B2. Using Right hand thumb rule we can find direction of magentic field at axis

(a) Two loops of radius r, currents in the same direction → strongest field. Bnet =B1+ B1= 2B1

(c) One loop of radius rand one of radius 2r, currents in the opposite direction → Bnet= B1+ B2 – B1–B2 = O.

(b) Two loops of radius 2r, currents in the Opposite direction → Bnet= B2– B2= O.

(d) Two loops of radius r, currents in same directions →Bnet=B1 +B1 + B2 = 2B1 + B2.

Order: (d) > (a) > (b)= (c).

Q7. A circular loop of radius R = 20 cm is placed in a uniform magnetic field B = 2 T in the xy-plane as shown in the figure. The loop carries a currenti= 1.0 A in the direction indicated.

(a) What is the net force acting on the loop in the magnetic field? (b) Calculate the magnetic moment of the loop. (c) Determine the magnitude of the torque acting on the loop.

Answer:

(c) B= 2 T Magnitude of torque is given by

θ = angle between M and B= 90∘

Q8. 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

Answer: Step 1: Given values: B=0.75T,E=9.0×10−5V/m,Vacc=15kV Step 2: For zero deflection, electric and magnetic forces balance:eE=evB ⟹ v=EB Step 3: Relate kinetic energy to accelerating voltage:12mv2=eVacc ⟹ em=v22Vacc Step 4: Substitute v=EB into emem=12Vacc(EB)2=E22B2Vacc=(9.0×10−5)22(0.75)2(1.5×104)≈4.8×107C/kgThis specific charge corresponds to deuterons.The answer is not unique because any ion with the same charge-to-mass ratio (for example He++) would yield the same e/m value. Step 1:

with the same charge-to-mass ratio (for example He++) would yield the same e/m value.

Q9. 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)

Answer: Given: V=2.0×103V,B=0.15T,e=1.6×10−19C,m=9.1×10−31kg.1. Compute the electron speed:v=2eVm=2×1.6×10−19×2.0×1039.1×10−31≈2.65×107m/s.2. Case (a): Field transverse to velocity.The magnetic force provides the centripetal force, sor=mveB.r=9.1×10−31×2.65×1071.6×10−19×0.15≈1.0×10−3m=1.0mm.Thus the electron moves in a circle of radius 1.0mm.3. Case (b): Field at 30∘ to velocity.Resolve v into perpendicular and parallel components:v⊥=vsin⁡30∘=v2≈1.325×107m/s,v∥=vcos⁡30∘.The perpendicular component causes circular motion of radiusr⊥=mv⊥eB≈9.1×10−31×1.325×1071.6×10−19×0.15≈0.5×10−3m=0.5mm.Since there is also a parallel component, the overall

there is also a parallel component, the overall trajectory is a helix of radius 0.5mm.

Q10. Derive the expression for force per unit length between two parallel current-carrying wires. Explain why this force is attractive for currents in the same direction and repulsive for opposite directions.

Answer:

This field is perpendicular to wire 2.

Direction Analysis:Using the right-hand rule and Fleming's left-hand rule:

This explains why parallel conductors carrying current in the same direction attract each other, while those carrying current in opposite directions repel.

Q11. A charged particle enters a uniform magnetic field perpendicularly. Derive the expression for the radius of its circular path

Answer: When a charged particle enters a uniform magnetic field perpendicularly, the magnetic force provides the centripetal force for circular motion.

Equating these forces we have

The time period of circular motion is given by

Here, the time period is independent of both velocity and radius, depending only on the charge-to-mass ratio and magnetic field strength.

Q12. Explain the relationship between electric and magnetic fields produced by moving charges. Derive the expression for the magnetic force on a moving charge and discuss its characteristics.

Answer: Moving charges create both electric and magnetic fields, establishing a fundamental connection between electricity and magnetism. A

magnetic field given by the Biot-Savart law.

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• Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism
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