Question

A proton of energy $E$ is moving along a circular path in a uniform magnetic field. If an alpha particle describes the same circular path, its energy should be :

• A. $4E$
• B. $2E$
• C. $E$
• D. $0.5E$

Answer 1

The correct option is C $E$

proton $\frac{m{v}^2}{2}=E$
$r=\frac{mv}{qB}$=$\frac{\sqrt{2Em}}{qB}$
for $\alpha$ particle, $M=4mandQ=2q$
$r=\frac{\sqrt{2E^{\prime }4m}}{2qB}$
$r=\frac{\sqrt{2E^{\prime }m}}{qB}$
$E^{\prime }=E$