Question

A particle of mass m has an electric charge q. This particle is accelerated through a potential difference V and then entered normally in a uniform magnetic field B. It performs a circular motion of radius R. The ratio of its charge to the mass (q/m) is = ______ [(q/m) is also called specific charge.]
(A) 2V/B²R²
(B) V/2BR
(C) VB/2R
(D) mV/BR

Answer 1

The ratio of its charge to the mass (q/m) is $\bold{\frac{2V}{R^2B^2}}$

Given:

Mass of the particle = m

Charge of the particle = q

Velocity of the particle = v

Potential difference = V

Magnetic field = B

Radius of the motion = R

To find:

Ratio of charge to the mass = ?

Solution:

From question, the particle is electron which is accelerated through potential difference.

By energy conversion:

$\bold{qV=\frac{1}{2}mv^2}$

$\bold{v^2=\frac{2qV}{m}}$

$\bold{v=\sqrt{\frac{2qV}{m}}}}$

Since, the particle enters magnetic field, magnetic field is:

$\bold{qvB=\frac{mv^2}{R}}$

$\bold{\frac{q}{m}=\frac{v^2}{RvB}=\frac{v}{RB}}$

The velocity of the electron is given by the formula:

$\bold{v=\sqrt{\frac{2qV}{m}}}$

$\bold{\frac{q}{m}=\frac{1}{RB}\sqrt{\frac{2qV}{m}}}$

On squaring on both sides,

$\bold{\frac{q^2}{m^2}=\frac{1}{R^2B^2}\times\frac{2qV}{m}}$

$\bold{\therefore\frac{q}{m}=\frac{2V}{R^2B^2}}$