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 radius \( r \) of a charged particle moving in a magnetic field is given by the formula: \[ r = \frac{mv}{qB} \] where: - \( m \) is the mass of the particle, - \( v \) is the velocity of the particle, - \( q \) is the charge of the particle, - \( B \) is the magnetic field strength. For the given problem, the radius of the path is \( R \) for the particle with mass \( m \) and charge \( q \). So, the radius of the path is: \[ R = \frac{mv}{qB} \quad \text{(1)} \] Now, if the mass becomes \( 2m \) and the charge becomes \( \frac{q}{2} \), the new radius \( r' \) will be: \[ r' = \frac{(2m)v}{\frac{q}{2}B} = \frac{2m \cdot v}{\frac{q}{2} \cdot B} = \frac{4mv}{qB} \] Comparing this with equation (1), we get: \[ r' = 4R \] Hence, the new radius of the path is \( 4R \).