Physics - Magnetism Question with Solution | TestHub

The radius $r$ of a charged particle moving in a magnetic field $B$ is given by:

$$r=\frac{mv}{qB}$$

where $m$ is the mass, $v$ is the velocity, and $q$ is the charge. The kinetic energy $KE$ is given by:

$$KE=\frac{1}{2}m{v}^2$$

So, $v=\sqrt{\frac{2KE}{m}}$. Substituting this into the radius equation:

$$r=\frac{m}{qB}\sqrt{\frac{2KE}{m}}=\frac{\sqrt{2mKE}}{qB}$$

Since the kinetic energy $KE$ and the magnetic field $B$ are the same for both particles, we can write:

$$r\propto \frac{\sqrt{m}}{q}$$

For the proton: mass = $m$, charge = $e$, so $r_p\propto \frac{\sqrt{m}}{e}$

For the alpha particle: mass = $4m$, charge = $2e$, so $r_{\alpha }\propto \frac{\sqrt{4m}}{2e}=\frac{2\sqrt{m}}{2e}=\frac{\sqrt{m}}{e}$

Therefore, $r_p=r_{\alpha }$. The alpha particle and the proton will be bent in a circular path with the same radius.