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: Understanding the Concept:
The particle gains kinetic energy from the potential difference: \(qV = \frac12mv^2\). In the magnetic field it moves in a circle, with \(\frac{mv^2}{r} = qvB\).

Step 2: Eliminate v:
From the circular motion, \(v = \frac{qBr}{m}\).
Substitute: \(qV = \frac12m\cdot\frac{q^2B^2r^2}{m^2} = \frac{q^2B^2r^2}{2m}\).

Step 3: Solve for m:
\[ m = \frac{q^2B^2r^2}{2qV} = \frac{qB^2r^2}{2V} \]
The other options have wrong powers of \(q\) or \(B\). For example, option A has \(q^2\), and option D has \(B\) to the first power, which would not fit the dimension of mass.

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