Question

A charged particle moves in a magnetic field with velocity $ v $ perpendicular to the field $ B $. The radius of the circular path is $ r $. Which of the following expressions gives the charge $ q $ of the particle?

• A. $ q = \frac{mv}{Br} $
• B. $ q = \frac{mvB}{r} $
• C. $ q = \frac{mB}{vr} $
• D. $ q = \frac{Bvr}{m} $

Hint:

Use the relationship between the Lorentz force and centripetal force to derive the charge of the particle.

Answer 1

The Correct Option is A

Step 1: Understand the Motion.
When a charged particle moves perpendicular to a magnetic field, it experiences a centripetal force due to the Lorentz force: $$ F = qvB, $$ where: - $ q $ is the charge of the particle, - $ v $ is the velocity of the particle, - $ B $ is the magnetic field strength. This force provides the necessary centripetal acceleration for the particle to move in a circular path. The centripetal force is given by: $$ F_{\text{centripetal}} = \frac{mv^2}{r}, $$ where: - $ m $ is the mass of the particle, - $ r $ is the radius of the circular path. Equating the Lorentz force to the centripetal force: $$ qvB = \frac{mv^2}{r}. $$ Step 2: Solve for $ q $.
Rearrange the equation to solve for $ q $: $$ q = \frac{mv^2}{Br}. $$ Step 3: Verify the Options.
The options provided are: A) $ q = \frac{mv}{Br} $
B) $ q = \frac{mvB}{r} $
C) $ q = \frac{mB}{vr} $
D) $ q = \frac{Bvr}{m} $. From our derivation: $$ q = \frac{mv}{Br}. $$ Thus, the correct answer is: $$ \boxed{\text{A}} $$