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 magnetic field deflects the particle into a circular arc of radius R. If the accelerating potential is tripled to 3V, the radius of the circular arc will now be:

Answer 1

Answer:

New radius of the charge particle when potential is increased by 3times of initial value

$R' = \sqrt3 R$

Explanation:

As we know that charge particle is accelerated due to potential difference V then we have

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

now the speed of the charge particle is given as

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

now in constant magnetic field which is perpendicular to the motion of charge  we have

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

now we have

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

now we have

$R = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}}$[/tex]

$R = \frac{1}{B} \sqrt{\frac{2mV}{q}}$

now if we changed the potential to three times of initial value then we have

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

so we have

$\frac{R}{R'} = \frac{1}{\sqrt3}$

$R' = \sqrt3 R$