Question

A charged particle accelerated by a potential $V$ moves in a circular path with velocity $v$ in a uniform magnetic field $B$ perpendicular to motion. Which of the following is/are correct if the value of $V$ is increased?

• A. Kinetic energy of the particle increases
• B. Radius of the circular path increases
• C. Time period of the motion increases
• D. Work done by the magnetic field increases

Hint:

Magnetic field only changes direction, not speed — so no work done. Radius ∝ √V, Time period independent of V.

Answer 1

The Correct Option is A, B

Step 1: Relationship between potential and velocity.
\[ \frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}} \] Thus, increasing $V$ increases $v$. Step 2: Magnetic radius formula.
\[ r = \frac{mv}{qB} \Rightarrow r \propto v \Rightarrow r \propto \sqrt{V} \] Hence, radius increases. Step 3: Time period of circular motion.
\[ T = \frac{2\pi m}{qB} \] Independent of $v$ or $V$; thus, no change. Step 4: Work by magnetic field.
The magnetic force does no work because it is perpendicular to motion. Step 5: Conclusion.
Hence, (A) and (B) are correct.