Question

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

Answer 1

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

Answer 2

The radius of the circular path described by the proton is r'= √2r

Explanation:

• Potential energy of proton is converted into kinetic energy.

1/2 mpv^2 = e.V

v= √ ( 2eV/mp)

• If the potential difference is doubled:

V' = 2 V

Therefore,

v= √ ( 2e x 2V/mp )

v'=√2v

• Now radius of the circular path will be:
• As we know that
• R ∝ √v
• The radius will become √2 times

qvB = mpv/r

r = mpv / qvB

r' = mp x 2v / qvB

r'= √2r

Thus the radius of the circular path described by the proton is r'= √2r

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