Question

A particle of mass \( m \) and charge \( q \) describes a circular path of radius \( R \) in a magnetic field. If its mass and charge were \( 2m \) and \( \frac{q}{2} \) respectively, the radius of its path would be:

• A. \( \frac{R}{4} \)
• B. \( \frac{R}{2} \)
• C. \( 2R \)
• D. \( 4R \)

Hint:

The radius of a charged particle's circular path in a magnetic field is proportional to its mass and inversely proportional to its charge. Doubling the mass and halving the charge increases the radius by a factor of 4.

Answer 1

The Correct Option is D

The formula for the radius \( r \) of a charged particle in a magnetic field is \( r = \frac{mv}{qB} \), where \( m \) is mass, \( v \) is velocity, \( q \) is charge, and \( B \) is magnetic field strength. For the initial problem, the radius is \( R = \frac{mv}{qB} \) (1). If the mass is doubled to \( 2m \) and the charge is halved to \( \frac{q}{2} \), the new radius \( r' \) is calculated as \( r' = \frac{(2m)v}{\frac{q}{2}B} = \frac{4mv}{qB} \). Comparing this to equation (1), we find \( r' = 4R \). Therefore, the new radius of the path is \( 4R \).