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.

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

$[$1 Mark$]$

• A. $1:2$
• B. $2:1$
• C. $4:1$
• D. $1:4$

Answer 1

The correct option is A $1:2$

When $V$ and $B$ are perpendicular, the path will be circle.

The radius of the circle is given by $r=\frac{MV}{qB}$

For proton, let $M_P=m,q_P=q$

For $\alpha$ - particle, $M_{\alpha }=4m,q_{\alpha }=2q$

Radius of proton is
$\Rightarrow r_P=\frac{mV}{qB}$
Radius of alpha particle is
$r_{\alpha }=\frac{4mV}{2qB}=\frac{2mV}{qB}$

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