Question

Solve previous year Question of iit jee
chapter :- magnetic effect of current​

Answer 1

Answer:

option d is correct answer

Hope it will help you

Answer 2

Solution :

For a charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to the magnetic field B

$\frac{m {v}^{2} }{r} = Bqv$

$\frac{mv}{r} = Bq$

$v = \frac{Bqr}{m}$

Since, the particle completes one revolution

$\therefore \: v = \frac{2\pi \: r}{t}$

$\frac{2\pi \: r}{t} = \frac{Bqr}{m}$

$t = \frac{2\pi \: m}{qB}$

So, Option D) is correct