Question

A proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2mT$ perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by $90^{\circ }$. Then at this tune, the angle between the velocity vectors of proton and $\alpha$ - particle is

• A. $60^{\circ }$
• B. $90^{\circ }$
• C. $45^{\circ }$
• D. $180^{\circ }$

Answer 1

Answer: (C)

In a circular motion, a body changes its direction by $90^{\circ }$ in one - fourth of its time period.
Given, magnetic field, $B=2mT$
Let $T_p$ be the time-period of revolution of proton in the magnetic field.
$T_p=\frac{2\pi m_p}{eB}\dots$(i)
$m_{\alpha }\simeq 4mp$ and $q_{\alpha }=2q^p$
and $T_{\alpha }$ be the time-period of revolution of
$\alpha$ -particle in the magnetic field,
$T_{\alpha }=\frac{2\pi m_p}{qB}=\frac{2\pi \left(4m_p\right)}{(2e)B}\dots$(ii)
Now, the ratio of time-period of proton and
$\alpha$ -particle, by dividing Eq. (i) and (ii),
$\frac{T_p}{T_{\alpha }}=\frac{1}{2}$
$\Rightarrow T_{\alpha }=2T_p$
Hence, the time-period of $\alpha$ -particle is double of the proton, i.e. if proton covers $90^{\circ }$ of angle from its starting, then $\alpha$ -particle will cover $45^{\circ }$ of the angle.