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Q.A particle of charge qq and mass mm is projected with velocity vv at an angle 6060^\circ to a uniform magnetic field BB. The radius of the helical path is

a
mvsin60qB\dfrac{mv \sin 60^\circ}{qB}
b
mvcos60qB\dfrac{mv \cos 60^\circ}{qB}
c
mvqBsin60\dfrac{mv}{qB \sin 60^\circ}
d
mvqB\dfrac{mv}{qB}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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