Question

A proton and an $\alpha$ -particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 $\mu$ second to make 5 revolutions, then times period for the $\alpha$-particle would be

• A. 50 $\mu$ sec
• B. 25 $\mu$ sec
• C. 10 $\mu$ sec
• D. 5 $\mu$ sec

Answer 1

The correct option is C 10 $\mu$ sec

Time taken by proton to make one revolution $=\frac{25}{5}=5\mu sec$.

As, $T=\frac{2\pi m}{qB}$

So, $\frac{T_2}{T_1}=\frac{m_2}{m_1}\times \frac{q_1}{q_2}$

$T_2=T_1\frac{m_2{q}_1}{m_1{q}_2}=\frac{5\times 4m_1}{m_1}\times \frac{q}{2q}=10\mu sec.$