If charge is moving perpendicular to uniform magnetic field then its time period of revolution is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → independent of its radius of circular path
The time period (T) of revolution is -
$T=\frac{2πr}{v}=\frac{2π.\frac{mv}{qB}}{v}$
$⇒T=\frac{2πm}{qB}$
Hence, it is independent of it's Radius.