Question

A charged particle of specific charge $\alpha $ moves with a velocity $\vec v = {v_o}\hat i$ in a magnetic field $\vec B = \dfrac{{{B_o}}}{2}\left( {\hat j + \hat k} \right)$. Then (specific charge = charge per unit mass)
(A) path of the particle is a helix
(B) path of the particle is a circle
(C) distance moved by the particle in time $t = \dfrac{\pi }{{{B_o}\alpha }}$ is $\dfrac{{\pi {v_o}}}{{{B_o}\alpha }}$
(D) velocity of the particle after time $t = \dfrac{\pi }{{{B_o}\alpha }}$ is $\left( {\dfrac{{{v_o}}}{2}\hat i + \dfrac{{{v_o}}}{2}\hat j} \right)$

Answer 1

Hint: Since the velocity of the particle is perpendicular to the applied magnetic field given in the question, so the particle will follow a circular path. Now, since the velocity of the particle will remain constant so the distance travelled will be the product of the velocity and the time.

Complete Step by Step Solution:
In the given question, the velocity of the particle is given $\vec v = {v_o}\hat i$. So from the equation we can see that the velocity is along the X-axis, since the unit vector along X-axis is $\hat i$.
Now in the question we are also provided with the magnetic field in the region which is given by, $\vec B = \dfrac{{{B_o}}}{2}\left( {\hat j + \hat k} \right)$. From this equation, we see that the magnetic field is acting along the Y-Z plane as $\hat j$ is the unit vector along Y-axis and $\hat k$ is the unit vector along Z-axis, so the sum of these two unit vector will give resultant in the Y-Z plane.
Therefore we can conclude from here that the velocity of the particle is perpendicular to the applied magnetic field.
So there will be force acting on the particle which will cause the particle to follow a circular path. This force will be acting on the particle radially inward.
Now, the velocity of the particle in the magnetic field will be constant. So the distance travelled by the particle in a given time will be,
$S = vt$
Now in the question, the magnitude of the velocity is $\left| {\vec v} \right| = {v_o}$and the time is given, $t = \dfrac{\pi }{{{B_o}v}}$. So the distance travelled by the particle will be,
$S = {v_o} \times \dfrac{\pi }{{{B_o}v}}$
This gives us,
$S = \dfrac{{\pi {v_o}}}{{{B_o}v}}$
So therefore, the path of the particle will be circular and the distance followed by the particle will be $\Rightarrow \dfrac{{\pi {v_o}}}{{{B_o}v}}$.

Therefore, the correct answer will be both the options B. and C.

Note: When a charged particle is moving in a magnetic field, it experiences a force which is given by,
$\vec F = \vec v \times \vec B$
The magnitude of this force is zero when the velocity of the particle is in the same direction as that of the applied magnetic field. But when the magnetic field is applied in a perpendicular direction to the velocity of the particle, then the magnitude of this force is maximum and it causes the particle to move in a circular path