Question

A particle of charge q enters a region of uniform magnetic field B with a constant speed. The field deflects the particle, a distance h above the original line of flight as shown in the Figure. The particle leaves the field region with momentum, if $h<<d$, of about :

• A. $\frac{qB{d}^2}{2h}$, making an angle $\frac{2h}{d}$ with initial direction
• B. $\frac{qB{h}^2}{2d}$, making an angle $\frac{2h}{d}$ with initial direction
• C. $\frac{qBd}{2}$, making an angle $\frac{2d}{h}$ with initial direction
• D. $\frac{qBh}{2}$, making an angle $\frac{2d}{h}$ with initial direction

Answer 1

The correct option is A $\frac{qB{d}^2}{2h}$, making an angle $\frac{2h}{d}$ with initial direction

The magnetic force acting on the particle gives it the centripetal acceleration.

Thus $qvB=\frac{m{v}^2}{R}$

$⟹R=\frac{mv}{qB}=\frac{p}{qB}$

Thus $p=qBR$

From figure, $R^2=d^2+(R-h{)}^2$

$⟹R=\frac{d^2+h^2}{2h}$

Since, $h<<d$, $R\approx \frac{d^2}{2h}$

$⟹p=\frac{qB{d}^2}{2h}$

$sin\theta =\frac{d}{R}=\frac{d}{d^2/2h}=\frac{2h}{d}$

Since $h<<d,sin\theta \approx \theta =\frac{2h}{d}$