Question

An electron projected in a perpendicular uniform magnetic field of $3\times {10}^{-3}T$, moves in a circle of radius $4mm$. The linear momentum of electron (in $kg-m/s$) is:

• A. $1.92\times {10}^{-21}$
• B. $1.2\times {10}^{-24}$
• C. $1.92\times {10}^{-24}$
• D. $1.2\times {10}^{-21}$

Answer 1

The correct option is C $1.92\times {10}^{-24}$

Centripetal force is provided by the force $evB$.
In this situation as the angle between $\overrightarrow{B}$ and $\overrightarrow{v}$ is $\theta =90$, so the force will be maximum $(=evB)$ and always perpendicular to motion (and also field) so, the path will be a circle (with its plane perpendicular to the field) as in a circle tangent and radius are always perpendicular to each other. As here centripetal force is provided by the force $evB$.
$\therefore \frac{m{v}^2}{r}=evB$
ie, $r=\frac{mv}{eB}$
As for a particle in motion, $p=mv$
$\therefore r=\frac{mv}{eB}=\frac{p}{eB}$
$\Rightarrow p=eBr$
Hence, $p=1.6\times {10}^{-19}\times 3\times {10}^{-3}\times 4\times {10}^{-3}$
$=1.92\times {10}^{-24}kg-m/s$