Question

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

• A. $\frac{qB}{2}[\frac{a^2}{d}+d]$
• B. $\frac{qB}{2}$
• C. Zero
• D. Not possible to be determined as it keeps changing.

Answer 1

Answer: (A)

$\frac{qB}{2}[\frac{a^2}{d}+d]$

The vector product between $v$ and $B$ points upwards in the figure thus indicating that the charge of the particle is positive.

The Force acting on the moving charge $F=qvB$.

As a result of the magnetic force, the charged particle will follow a spherical trajectory.

The centripetal force $F_c=\frac{m{v}^2}{r}$.

Force acting on the moving charge $F=F_c$,

$qvB=\frac{m{v}^2}{r}$

$r=\frac{mv}{qB}=\frac{p}{qB}$,where $p$ is the momentum of the particle.

$\begin{array}{c}{r}^2=d^2+{\left(r-a\right)}^2\\ =\frac{d^2+a^2}{2d}\\ =\frac{1}{2}\left(d+\frac{a^2}{d}\right)\end{array}$

So,

$\begin{array}{c}p=qBr\\ =\frac{qB}{2}\left(d+\frac{a^2}{d}\right)\end{array}$