Question

A charge q, m enters a finite B - field with a speed v as shown (such that the charge enters the magnetic field region perpendicularly at a distance L from the top left corner in the figure). Find its deviation if $v=\frac{2qBL}{m}$

• A. $\delta =\frac{\pi }{6}$
• B. $\delta =\frac{\pi }{3}$
• C. $\delta =\frac{\pi }{4}$
• D. $\delta =\pi$

Answer 1

The correct option is A $\delta =\frac{\pi }{6}$

Here, when the charge enters the magnetic field, the force acting on the charge is F = q$\overrightarrow{v}\times \overrightarrow{B}$ such that the magnitude of this force is constant and is always perpendicular to the velocity of the charge. So, the charge follows a circular trajectory in the magnetic field region.

Let r be the radius of the circular path. Then,
$\Rightarrow \frac{m{v}^2}{r}=qvB$

$\Rightarrow r=\frac{mv}{qB}$

Substituting the value of v in the question, r = 2L

From the geometry of the figure if the angle of deviaion is $\delta$ ,then:
2Lsin$\delta$ = L

$\therefore$ The deviation is $\frac{\pi }{6}$