First slide

Force on a charged particle moving in a magnetic field

Question

A charged particle of mass m and charge q enters along AB at point A in a uniform magnetic field existing in the rectangular region of size $a \times b .$ The particle leaves the region exactly at corner point C. What is the speed v of the particle ?

Moderate

A

$\frac{qB (a^{2} + b^{2})}{2 mb}$
B

$\frac{qB (a^{2} + b^{2})}{2 m a}$
C

$\frac{qB (a^{2} - b^{2})}{2 m a}$
D

$\frac{qB (a^{2} - b^{2})}{2 m b}$

Solution

Let speed of the particle be v (speed will remain constant).
$r = AO = CO = \frac{mv}{qB}; Also \frac{a}{r} = \sin θ, \frac{r - b}{r} = \cos θ$
Solving the above equations, we get $v = \frac{qB (a^{2} + b^{2})}{2 mb}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App