Question

A proton and alpha particle projected with the same velocity in uniform transverse magnetic field then?
A. Protons will have greater frequency than an alpha particle.
B. Alpha particles will have greater frequency than protons.
C. Both the particles will have the same frequency.
D. Their frequencies cannot be compared.

Answer 1

Hint when a charged particle is projected perpendicular to magnetic field then the force is $F = qvB$ in the direction perpendicular to both the field as well as velocity. This force will deflect the particle without changing the speed and the particle will move in a circle perpendicular to the field. The magnetic force provides the centripetal force.

Step by step solution: If R be the radius of the circle in which the charged particle.

Then, $qvB = m\dfrac{{{v^2}}}{R}$
Now we know, frequency $\upsilon = \dfrac{1}{T}$ ,where $T$ is time period.
$\because T = \dfrac{{2\pi R}}{v} = \dfrac{{2\pi m}}{{qB}} \\   \therefore \upsilon = \dfrac{1}{T} = \dfrac{{qB}}{{2\pi m}}......(1) \\     \\ $
This frequency is called Cyclotron frequency.
Now according to the question, speed is the same for both the particle but charge and mass differ.
Hence, For proton
Using equation (1), we get
${\upsilon _p} = \dfrac{{eB}}{{2\pi m}}......(2)$
Now for alpha particle ${q_\alpha } = 2e$ and ${m_\alpha } = 4m$
Hence, we get
${\upsilon _\alpha } = \dfrac{{(2e)B}}{{2\pi (4m)}}......(3)$
From equation (2) and (3) we get,
${v_\alpha } = \dfrac{{{v_p}}}{2}$
Hence protons will have greater frequency.

Option (A) is correct.

Note We see that the time period or say frequency is independent of the speed $v$. This is due to the fact if the particle moves faster, the radius is larger, it has to move a longer distance to complete the circle so that the time taken is the same. As the magnetic force on a particle is perpendicular to the velocity, it does not do any work on the particle. Hence, the kinetic energy or the speed of the particle does not change due to the magnetic force.