Physics · Magnetism · chapter 4 of 14

Class 12 Physics Chapter 4: Moving Charges and Magnetism MCQs with answers

10 multiple-choice questions on Moving Charges and Magnetism, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
A charged particle moves with velocity v through a region of uniform magnetic field B. The magnetic force on it is zero when
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Answer: (a) v is parallel or anti-parallel to B

The magnetic force is q(v × B), of magnitude qvB sin θ. It vanishes when θ = 0° or 180°, that is, when v is parallel or anti-parallel to B.

Question 2
A proton goes round a circle in a uniform magnetic field. The work done on it by the magnetic force during half a revolution is
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Answer: (a) zero

The magnetic force is always perpendicular to the velocity, so it does no work along any part of the path; the speed and kinetic energy stay the same.

Question 3
The tesla, the SI unit of magnetic field, is equivalent to
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Answer: (c) N A⁻¹ m⁻¹

From F = qvB, B = F/(qv), with unit N s C⁻¹ m⁻¹. Since 1 C s⁻¹ = 1 A, this is N A⁻¹ m⁻¹.

Question 4
In a horizontal tube, a beam of electrons travels due north. A uniform magnetic field in the region points vertically upward. The beam is deflected
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Answer: (a) towards the west

Take east, north and up as x̂, ŷ, ẑ. v × B is along ŷ × ẑ = +x̂, east. The electron's charge is negative, so the force q(v × B) is towards the west. It cannot be upward: the force is perpendicular to B.

Question 5
Protons are to be bent into a circle of radius 5.0 cm by a uniform magnetic field of 0.40 T normal to their velocity. The speed of the protons must be (e = 1.6 × 10⁻¹⁹ C, mass of proton = 1.67 × 10⁻²⁷ kg)
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Answer: (d) 1.9 × 10⁶ m s⁻¹

From r = mv/qB, v = qBr/m = (1.6 × 10⁻¹⁹ × 0.40 × 0.050)/(1.67 × 10⁻²⁷) = 1.9 × 10⁶ m s⁻¹.

Question 6
An electron circles in a uniform magnetic field. If its speed is doubled, keeping it normal to the field, its period of revolution
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Answer: (b) stays the same

T = 2πm/qB does not contain the speed: a faster electron moves on a proportionally larger circle and takes the same time.

Question 7
In a cyclotron the oscillator works at 12 MHz. The magnetic field needed for protons to go round in step with it is (e = 1.6 × 10⁻¹⁹ C, mass of proton = 1.67 × 10⁻²⁷ kg)
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Answer: (b) 0.79 T

ν_c = qB/2πm, so B = 2πmν_c/q = (2π × 1.67 × 10⁻²⁷ × 12 × 10⁶)/(1.6 × 10⁻¹⁹) = 0.79 T. Leaving out 2π gives 0.13 T.

Question 8
A proton enters a uniform magnetic field of 0.10 T with a speed of 4.0 × 10⁵ m s⁻¹ at 60° to the field. The pitch of its helical path is (e = 1.6 × 10⁻¹⁹ C, mass of proton = 1.67 × 10⁻²⁷ kg)
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Answer: (c) 13.1 cm

Only v∥ = v cos 60° = 2.0 × 10⁵ m s⁻¹ carries the proton along the field. p = 2πmv∥/qB = (2π × 1.67 × 10⁻²⁷ × 2.0 × 10⁵)/(1.6 × 10⁻¹⁹ × 0.10) = 0.131 m = 13.1 cm. Using v sin 60° gives 22.7 cm; using the full speed gives 26.2 cm.

Question 9
A short element of a wire carries a steady current. By the Biot–Savart law, its magnetic field is zero at a point that lies
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Answer: (a) on the line of the element itself

|dB| = (μ₀/4π) I dl sin θ/r². On the line of the element θ = 0° or 180°, so sin θ = 0 and the field there is zero.

Question 10
With ε₀ the permittivity and μ₀ the permeability of free space, and c the speed of light in vacuum, the product ε₀μ₀ equals
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Answer: (c) 1/c²

ε₀μ₀ = (4πε₀)(μ₀/4π) = [1/(9 × 10⁹)] × 10⁻⁷ = 1/(3 × 10⁸)² = 1/c².

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