Physics Problem: Kinetic Energy in Magnetic Field After 3 Seconds

Physics Problem: Kinetic Energy in a Magnetic Field

11. A particle of mass \( m \), charge \( Q \), and kinetic energy \( T \) enters a transverse uniform magnetic field of induction \( B \). After 3 seconds, the kinetic energy of the particle will be (2008):

(a) \( T \) (b) \( 4T \) (c) \( 3T \) (d) \( 2T \)

Correct Answer: (a) \( T \)

The kinetic energy remains \( T \) because the magnetic force, being perpendicular to the velocity, does no work on the particle. Thus, the kinetic energy is conserved.

Detailed Explanation

A charged particle in a magnetic field experiences a force \( F = Q v B \sin\theta \). With the field transverse to the velocity (\( \theta = 90^\circ \)), the force is \( F = Q v B \), causing circular motion with radius \( R = \frac{m v}{Q B} \).

Since the force is perpendicular to the velocity, the work done is:

\[ W = \vec{F} \cdot \vec{d} = 0 \]

By the work-energy theorem, \( \Delta KE = W = 0 \), so:

\[ KE_{\text{final}} = KE_{\text{initial}} = T \]

The time of 3 seconds doesn’t affect the kinetic energy, as the speed \( v \) remains constant in a uniform magnetic field with no other forces present.