Question

A charged particle $q$ enters a region of uniform magnetic field $B$ (out of page) and is deflected distance $d$ after travelling a horizontal distance a. The magnitude of the momentum of the particle is

• A. $\frac{qB}{2}\left[\frac{a^2}{d}+d\right]$
• B. $\frac{qB}{2}$
• C. Zero
• D. $2qB$

Answer 1

Answer: (A)

From the figure given below we have, $AB=\sqrt{a^2+d^2}=h$
$AB=\sqrt{a^2+d^2}=h$
In $\Delta OAD,cos\theta =\frac{h}{2R}$
In $\Delta ACB,sin\left(\frac{\pi }{2}-\theta \right)=\frac{d}{h}$
$\Rightarrow \frac{h}{2R}=\frac{d}{h}$
$\Rightarrow h^2=2Rd$
$\Rightarrow a^2+d^2=2Rd$

$\Rightarrow R=\frac{a^2}{2d}+\frac{d}{2}$
$\therefore$ Momentum, $P=mv$
$=qBR=\frac{qB}{2}\left(\frac{a^2}{d}+d\right)\left[\because R=\frac{mv}{qB}\right]$