Question

A charged particle of mass \( m \) and charge \( q \) moves with a velocity \( \vec{v} \) perpendicular to a magnetic field \( \vec{B} \). The radius of the circular path it follows is:

• A. \( \frac{qB}{mv} \)
• B. \( \frac{qv}{mB} \)
• C. \( \frac{mv}{qB} \)
• D. \( \frac{mB}{qv} \)

Hint:

For a charged particle in a magnetic field, the radius of the circular path is derived by equating the magnetic force \( qvB \) to the centripetal force \( \frac{m v^2}{r} \).

Answer 1

The Correct Option is C

- A charged particle moving perpendicular to a magnetic field experiences a Lorentz force \( F = qvB \), which provides the centripetal force for circular motion. - Centripetal force required: \( F = \frac{m v^2}{r} \). - Equate the forces: \[ q v B = \frac{m v^2}{r} \] - Solve for \( r \): \[ r = \frac{m v^2}{q v B} = \frac{m v}{q B} \] - This matches option (C).