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Tems11 [23]
3 years ago
8

A charge is accelerated from rest through a potential difference V and then enters a uniform magnetic field oriented perpendicul

ar 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
Physics
1 answer:
lys-0071 [83]3 years ago
4 0

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

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