Question

A charged particle \(q\) is accelerated by a potential difference of \(V\) enters a region of uniform magnetic field of induction \(B\) at right angles to the direction of the field. The charged particle completes semi circle of the radius \(r\) inside the magnetic field. The mass of the charged particle is (all quantities are in SI units)

• A. \(q^2r^2B^2/2V\)
• B. \(qr^2B^2/2V\)
• C. \(qBr/V\)
• D. \(q^2r^2B/2V\)

Hint:

Combine \(qV=\frac12mv^2\) with \(r=\frac{mv}{qB}\) and eliminate \(v\).

Answer 1

The Correct Option is B

Step 1: Plan:
Use momentum $p = qBr$ and kinetic energy in terms of $p$.

Step 2: Steps:
The radius of the path gives momentum $p = mv = qBr$. Kinetic energy is $\frac{p^2}{2m} = qV$.
So $m = \frac{p^2}{2qV} = \frac{q^2B^2r^2}{2qV} = \frac{qB^2r^2}{2V}$.

Final Answer:
The mass is $\frac{qr^2B^2}{2V}$, option (B). \[ \boxed{\frac{qr^2B^2}{2V}} \]