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Q.

A proton moving in a perpendicular magnetic filed of induction B posses kinetic energy E. If the magnetic field increased to 4B and the particle is constrained to move in the circular path of same radius, then its kinetic energy has to be

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a

2E

b

4E

c

16E

d

E

answer is B.

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Detailed Solution

r=2KmqB⇒K∝B2

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