Question

A charged particle with charge \( q \) and velocity \( \vec{v} \) moves perpendicular to a magnetic field \( \vec{B} \). The radius of the circular path is \( r \). What is the expression for \( r \)?

• A. \( \frac{mv}{qB} \)
• B. \( \frac{qB}{m} \)
• C. \( \frac{m}{qBv} \)
• D. \( \frac{v}{mqB} \)

Hint:

For circular motion in a magnetic field, equate magnetic force \( qvB \) to centripetal force \( \frac{mv^2}{r} \). Always solve for \( r \) to get the radius.

Answer 1

The Correct Option is A

Step 1: Use Lorentz force for circular motion.
When a charged particle moves perpendicular to a magnetic field, it experiences a centripetal force due to magnetic force: \[ F = qvB = \frac{mv^2}{r} \] Step 2: Solve for \( r \). \[ qvB = \frac{mv^2}{r} \quad \Rightarrow \quad r = \frac{mv}{qB} \] So, the radius \( r \) is directly proportional to mass and velocity, and inversely proportional to charge and magnetic field.