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

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

`R' = \sqrt3 R`

Step-by-step explanation:

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

`(1)/(2)mv^2 = qV`

now the speed of the charge particle is given as

`v = \sqrt{(2qV)/(m)}`

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

`qvB = (mv^2)/(R)`

now we have

`R = (mv)/(qB)`

now we have

`R = (m)/(qB) * \sqrt{(2qV)/(m)}`[/tex]

`R = (1)/(B) \sqrt{(2mV)/(q)}`

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

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

so we have

`(R)/(R') = (1)/(\sqrt3)`

`R' = \sqrt3 R`