Question

A particle carrying a charge of +e travels in a circular path in a uniform magnetic field. If instead the particle carried a charge of +2e, the radius of the circular path would have been
A. 2R
B. 4R
C. 8R
D. R/2
E. R/4

Answer 1

Answer:

The radius of the circular path = $\frac{R}{2}$

Explanation:

The force on a charged particle s given by

F = q v B ------ (1)

This force is equal to the centripetal force acting on the particle because the particle travels in circular path.

The centripetal force

$F = m \frac{v^{2} }{r}$ ----- (2)

Equation (1) = Equation (2)

q v B = $m \frac{v^{2} }{r}$

$R = \frac{mv}{q B}$

From the above relation we are seeing that radius of the circular path is inversely proportional to the charge (q).

So when the charge doubles the radius of  circular path is halved.

So  the radius of the circular path = $\frac{R}{2}$