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Question #43418

Difficulty: 8/10 | 🟢 Free

The Question

A smooth, non-conducting circular ring of radius $R$ is fixed in the horizontal $x-y$ plane in a gravity-free space. A small bead of mass $m$ and positive charge $q$ is threaded on the ring and is initially at rest. A spatially uniform but time-varying magnetic field $\vec{B} = B_0 (1 - \alpha t) \hat{k}$ is applied across the entire region, where $B_0$ and $\alpha$ are positive constants, and $t$ is time. The unit vector $\hat{k}$ is perpendicular to the $x-y$ plane. What is the magnitude of the normal force exerted by the ring on the bead at the instant $t = \frac{1}{2\alpha}$ ?

Options

A.
$\frac{3 q^2 B_0^2 R}{16 m}$
B.
$\frac{q^2 B_0^2 R}{16 m}$
C.
$\frac{q^2 B_0^2 R}{8 m}$
D.
$\frac{5 q^2 B_0^2 R}{16 m}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Moving Charge in Non-Uniform B|Lenz’s Law Energy Conservation

Why This Question Is Tricky

Common Mistakes

This question targets the common misconception that 'magnetic fields can never change the kinetic energy of a particle'.

While true for the static Lorentz force, a time-varying magnetic field induces a non-conservative electric field (Faraday's Law) that actively does work on the charge.

Students who blindly apply the zero-work heuristic will incorrectly assume the bead remains at rest ( $v=0$ ).

Furthermore, the problem tests the synthesis of kinematics and magnetic deflection: students must correctly evaluate the instantaneous magnetic field at $t = 1/(2\alpha)$ rather than using the initial $B_0$ , and properly assign the direction of the magnetic force (radially outward) to sum it with the required centripetal force.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Moving Charge in Non-Uniform B
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free