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Moving Charges and Magnetism — NEET & JEE MCQs with answers
This chapter establishes that moving charges produce magnetic fields and that magnetic fields exert forces on moving charges and current-carrying conductors. It covers the Lorentz force, circular and helical motion of charges, the Biot-Savart law, Ampere's circuital law, the solenoid, the force between parallel conductors, the torque on a current loop (magnetic dipole), and the moving-coil galvanometer.
Topics in this chapter
- 4.1 Introduction
- 4.2 Magnetic Force
- 4.3 Motion in a Magnetic Field
- 4.4 Magnetic Field Due to a Current Element - Biot-Savart Law
- 4.5 Magnetic Field on the Axis of a Circular Current Loop
- 4.6 Ampere's Circuital Law
- 4.7 The Solenoid
- 4.8 Force Between Two Parallel Currents and the Ampere
- 4.9 Torque on Current Loop and Magnetic Dipole
- 4.10 The Moving Coil Galvanometer
Free practice questions
The magnetic field at the centre of a circular current loop of radius $R$ carrying current $I$ is given by:
- A. $B=\dfrac{\mu_0 I}{2\pi R}$
- B. $B=\dfrac{\mu_0 I}{2R}$
- C. $B=\dfrac{\mu_0 I}{4\pi R}$
- D. $B=\dfrac{\mu_0 I R}{2}$
Show answer
Answer: B — Setting $x=0$ in the axial formula gives $B=\dfrac{\mu_0 I}{2R}$ (Eq. 4.12).
NCERT Class 12 Physics Ch 4
Two long parallel wires carry currents $8\ \text{A}$ and $5\ \text{A}$, separated by $0.2\ \text{m}$. The magnitude of the force on a $1\ \text{m}$ segment of one wire due to the other is: $\left(\dfrac{\mu_0}{2\pi}=2\times10^{-7}\right)$
- A. $8 \times 10^{-5}\ \text{N}$
- B. $2 \times 10^{-5}\ \text{N}$
- C. $1.6 \times 10^{-4}\ \text{N}$
- D. $4 \times 10^{-5}\ \text{N}$
Show answer
Answer: D — $F = \dfrac{\mu_0 I_a I_b L}{2\pi d} = 2\times10^{-7}\times\dfrac{8\times5\times1}{0.2} = 4 \times 10^{-5}\ \text{N}$.
NCERT Class 12 Physics Ch 4
A $1\ \text{cm}$ element carrying $10\ \text{A}$ ($\theta=90^\circ$) gives a field of $4\times10^{-8}\ \text{T}$. The distance to the point is:
- A. $1\ \text{m}$
- B. $0.05\ \text{m}$
- C. $0.25\ \text{m}$
- D. $0.5\ \text{m}$
Show answer
Answer: D — 4×10⁻⁸ = 10⁻⁷ × 10 × 0.01 / r² = 10⁻⁸/r², so r² = 10⁻⁸/4×10⁻⁸ = 0.25, r = 0.5 m. (Example 4.4 in reverse.)
NCERT Class 12 Physics Ch 4
In a moving coil galvanometer the restoring (counter) torque is provided by
- A. the soft iron core
- B. the radial magnetic field
- C. a spring
- D. the shunt resistance
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Answer: C — A spring supplies a restoring torque $k\phi$ that balances the magnetic torque $NIAB$.
NCERT Class 12 Physics Ch 4
A $1\ \text{m}$ wire carrying $3\ \text{A}$ at $90^\circ$ to a field feels $0.9\ \text{N}$. The magnetic field is:
- A. $0.9\ \text{T}$
- B. $0.3\ \text{T}$
- C. $3\ \text{T}$
- D. $0.1\ \text{T}$
Show answer
Answer: B — $B=\dfrac{F}{IL}=\dfrac{0.9}{3\times1}=0.3\ \text{T}$.
NCERT Class 12 Physics Ch 4
A solenoid of length $0.5$ m has $500$ turns in total and carries $5$ A. The magnetic field inside is:
- A. $1.26 \times 10^{-2}$ T
- B. $3.14 \times 10^{-3}$ T
- C. $6.28 \times 10^{-3}$ T
- D. $3.14 \times 10^{0}$ T
Show answer
Answer: C — $n=N/L=500/0.5=1000$ turns/m (turns per unit length, NOT total turns); $B=\mu_0 n I=6.28 \times 10^{-3}$ T.
NCERT Class 12 Physics Ch 4
A thick wire of radius $a=0.03$ m carries $5$ A uniformly. The field at interior radius $r=0.01$ m is:
- A. $4.44 \times 10^{-5}$ T
- B. $1.11 \times 10^{-5}$ T
- C. $5.56 \times 10^{-6}$ T
- D. $2.22 \times 10^{-5}$ T
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Answer: B — For $r<a$, $B=\dfrac{\mu_0 I r}{2\pi a^2}=1.11 \times 10^{-5}$ T ($B\propto r$).
NCERT Class 12 Physics Ch 4
For a magnetic dipole, the flux of the total field (external + loop's own) is maximum in the orientation of:
- A. maximum torque
- B. unstable equilibrium
- C. $\mathbf{m}\perp\mathbf{B}$
- D. stable equilibrium
Show answer
Answer: D — Example 4.11(b): stable equilibrium ($\mathbf{A}\parallel\mathbf{B}$) gives maximum total flux.
NCERT Class 12 Physics Ch 4
A cross symbol in this chapter's convention is best pictured as:
- A. a compass needle
- B. the feathered tail of an arrow moving away from you
- C. the tip of an arrow pointed at you
- D. a point charge at rest
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Answer: B — A cross is like the feathered tail of an arrow moving away from you (field/current going into the page); a dot is the tip of an arrow pointed at you.
NCERT Class 12 Physics Ch 4
A proton ($m=1.6\times10^{-27}\ \text{kg}$) moves on a circle of radius $0.2\ \text{m}$ in a $0.4\ \text{T}$ field. Its speed $v=qBr/m$ is:
- A. $1.25\times10^{-7}\ \text{m s}^{-1}$
- B. $8\times10^{6}\ \text{m s}^{-1}$
- C. $4\times10^{6}\ \text{m s}^{-1}$
- D. $1.6\times10^{7}\ \text{m s}^{-1}$
Show answer
Answer: B — $v=\dfrac{qBr}{m}=\dfrac{(1.6\times10^{-19})(0.4)(0.2)}{1.6\times10^{-27}}=\dfrac{1.28\times10^{-20}}{1.6\times10^{-27}}=8\times10^{6}\ \text{m s}^{-1}$.
NCERT Class 12 Physics Ch 4
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