A proton and an alpha particle enter a uniform magnetic field with the same velocity and move along circular paths. The time period of the alpha particle will be:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Two times that of proton
A charged particle moving perpendicular to a uniform magnetic field goes round a circle because the magnetic force \( qvB \) provides the centripetal force:
\( \displaystyle qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} \)
\( \displaystyle T = \frac{2\pi r}{v} = \frac{2\pi m}{qB} \)
The time period does not depend on the speed; it depends only on the ratio \( m/q \). For the two particles in the same field:
\( m_{\alpha} = 4m_p \)
\( q_{\alpha} = 2e = 2q_p \)
\( \displaystyle \frac{T_{\alpha}}{T_p} = \frac{m_{\alpha}/q_{\alpha}}{m_p/q_p} = \frac{4m_p/2e}{m_p/e} = 2 \)
So the time period of the alpha particle is two times that of the proton.