Question

A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicular to its path. The field deflects the particle into a circular arc of radius R. If the accelerating potential is tripled to 3V, what will be the radius of the circular

Answer 1

Answer:

$\sqrt{3}R$

Explanation:

Case 1:

$\Delta V$ = potential difference through which the charge is accelerated = V

$m$ = mass of the charge

$q$ = magnitude of charge

$v$ = speed gained by the charge due to potential difference

Using conservation of energy

Kinetic energy gained = Electric potential energy lost

$(0.5)mv^{2}=q\Delta V$

$v = \sqrt{\frac{2q\Delta V}{m}}$

$B$ = magnitude of magnetic field

$R$ = Radius of arc

Radius of arc is given as

$R = \frac{mv}{qB}$

$R = \frac{m}{qB}\sqrt{\frac{2q\Delta V}{m}}$

$R = \frac{1}{B}\sqrt{\frac{2m V}{q}}$                                eq-1

Case 2:

$\Delta V'$ = potential difference through which the charge is accelerated = 3V

$m$ = mass of the charge

$q$ = magnitude of charge

$v'$ = speed gained by the charge due to potential difference

Using conservation of energy

Kinetic energy gained = Electric potential energy lost

$(0.5)mv'^{2}=q\Delta V'$

$v' = \sqrt{\frac{2q\Delta V'}{m}}$

$B$ = magnitude of magnetic field

$R'$ = Radius of arc

Radius of arc is given as

$R' = \frac{mv'}{qB}$

$R' = \frac{m}{qB}\sqrt{\frac{2q\Delta V'}{m}}$

$R' = \frac{1}{B}\sqrt{\frac{2m\Delta V'}{q}}$

$R' = \frac{1}{B}\sqrt{\frac{2m (3V)}{q}}$

$R' = \sqrt{3}\frac{1}{B}\sqrt{\frac{2m V}{q}}$

Using eq-1

$R' = \sqrt{3}R$