Question

A proton is accelerated through a potential difference V, subjected to a
uniform magnetic field acting normal to the velocity of the proton. If the
potential difference is doubled, the radius of the circular path described by
the proton in the magnetic field will change how
​

Answer 1

if the potential difference is doubled then radius of the path becomes 1/√2 times of its initial.

given,

proton is accelerated through a potential difference , V subjected to a uniform magnetic field acting normal to the velocity of the proton, v.

due to magnetic proton moves in circular path.

so, centripetal force = magnetic fo

⇒mv²/r = Bqv

⇒r = mv/qB ........(1)

we also know, kinetic energy of proton = potential energy of proton

⇒1/2 mv² = qV

⇒(mv) = √(2qVm)

putting it in equation (1) we get,

r = √(2qVm)/qB

⇒r = 1/B √(2Vm/q)

⇒r ∝ 1/√V

hence, radius of path is inversely proportional to square root of potential difference.

now if the potential difference is doubled then radius of the path becomes 1/√2 times of its initial.

Answer 2

$\huge\bold\purple{Answer:-}$

The radius of the circular path will be :

r'=√2r

Explanation: Given, a proton is accelerated through a potential difference V, the direction of magnetic field is normal to the velocity of proton.

This means that the Potential energy of proton is converted into kinetic energy.

1/2 mpv^2 = eV

v=root ( 2eV/mp)

If the potential difference is doubled:

V'=2V

Therefore,

v=root ( 2e x 2V/mp )

v'=√2v

Radius of the circular path will be:

qvB = mpv/r

r=mpv/qvB

r'=mpx2v/qvB

r'=√2r