Question

A proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2\, mT$ 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

The Correct Option is C

In a circular motion, a body changes its direction by $90^{\circ}$ in one - fourth of its time period. Given, magnetic field, $B=2 mT$ Let $T_{p}$ be the time-period of revolution of proton in the magnetic field. $T_{p} =\frac{2 \pi m_{p}}{e B} \dots$(i) $m_{\alpha} \simeq 4 m p $ and $ q_{\alpha}=2 q^{p}$ and $T_{\alpha}$ be the time-period of revolution of $\alpha$ -particle in the magnetic field, $T_{\alpha}=\frac{2 \pi m_{p}}{q B}=\frac{2 \pi\left(4 m_{p}\right)}{(2 e) B} \dots$(ii) Now, the ratio of time-period of proton and $\alpha$ -particle, by dividing E (i) and (ii), $ \frac{T_{p}}{T_{\alpha}} =\frac{1}{2} $ $\Rightarrow T_{\alpha}=2 T_{p}$ 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.