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Question #43011

Difficulty: 8/10 | 🟢 Free

The Question

A particle of rest mass $m_0$ and charge $q$ is accelerated from rest through a potential difference $V$ . It then enters a region of uniform magnetic field $B$ perpendicular to its velocity. Considering relativistic effects, what is the frequency of revolution of the particle in the magnetic field? (Here $c$ is the speed of light in vacuum)

Options

A.
$\frac{qB}{2\pi m_0}$
B.
$\frac{qB}{2\pi \left(m_0 + \frac{qV}{c^2}\right)}$
C.
$\frac{qB}{2\pi m_0 \sqrt{1 - \left(\frac{qV}{m_0 c^2}\right)^2}}$
D.
$\frac{qB}{2\pi \left(m_0 - \frac{qV}{c^2}\right)}$

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

Cyclotron Frequency Invariance|Centripetal Acceleration Vector|Wave-Particle Duality Constraint

Why This Question Is Tricky

Common Mistakes

This question tests the boundary conditions of classical electromagnetism and kinematics at relativistic speeds.

Students commonly fall into the trap of using the classical cyclotron frequency formula $f = qB/(2\pi m_0)$ , assuming mass is an invariant scalar.

However, the Work-Energy Theorem dictates that the potential difference imparts a kinetic energy $K = qV$ , which contributes to the particle's relativistic mass ( $m = m_0 + K/c^2$ ).

Because the magnetic force provides the centripetal acceleration ( $qvB = mv^2/r$ ), the revolution frequency $f = qB/(2\pi m)$ becomes inversely proportional to the total energy.

Failing to account for this relativistic mass variation leads to the incorrect classical constant-frequency assumption.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Moving Charges and Magnetism - Class 12
Concept	Cyclotron Frequency Invariance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free