Question

A proton and an $\alpha$ - particle (Helium nucleus) enters a uniform magnetic field with their direction of velocity perpendicular to the direction of magnetic field.

$\left(iii\right)$ If proton and $\alpha$ - particle enter the magnetic field, perpendicular to it with same momentum, then the ratio of the radii of curvature of proton and $\alpha$ - particle will be

​​​​​​​$[$1 Mark$]$

• A. $1:2$
• B. $2:1$
• C. $\sqrt{2}:2$
• D. $1:\sqrt{2}$

Answer 1

The correct option is B $2:1$

When $v$ and $B$ are perpendicular, the path will be circle.
The radius of the circle is given by $r=\frac{Mv}{qB}$ $\Rightarrow r=\frac{p}{qB}$

For proton, let $M_P=m,q_P=q$
Radius of proton is
$\Rightarrow r_P=\frac{p}{qB}$

For $\alpha$ - particle, $M_{\alpha }=4m,q_{\alpha }=2q$
Radius of alpha particle is
$r_{\alpha }=\frac{p}{2qB}$

Ratio of radii is
$\Rightarrow \frac{r_P}{r_{\alpha }}=\frac{2}{1}$