Question

A charged particle of charge $q$ and mass $m$ enters in a uniform magnetic field $\vec{B} = \left(B_0\hat{k}\right)$ T with velocity $\vec{V} = (3\hat{i} - 4\hat{j})$ m/s. Select the correct option among the given one.

• A. Trajectory of particle will be helical
• B. Net initial force on the particle is $qB_0\left(4\hat{i} - 3\hat{j}\right)$ N
• C. Period of revolution is $\frac{\pi m}{qB_0}$
• D. Radius of the curve of revolution is $\frac{5m}{qB_0}$

Answer 1

Answer: (D)

Radius of the curve of revolution is $\frac{5m}{qB_0}$

Answer 2

The problem describes the motion of a charged particle in a uniform magnetic field. We need to analyze the trajectory, force, period, and radius of the path to determine the correct option.

Given:

• Charge of particle: $q$
• Mass of particle: $m$
• Magnetic field: $\vec{B} = B_0\hat{k}$ T
• Velocity of particle: $\vec{V} = (3\hat{i} - 4\hat{j})$ m/s

1. Determine the nature of the trajectory: The velocity vector $\vec{V}$ is in the x-y plane, and the magnetic field $\vec{B}$ is along the z-axis. This means the velocity is entirely perpendicular to the magnetic field (since $\vec{V} \cdot \vec{B} = (3\hat{i} - 4\hat{j}) \cdot (B_0\hat{k}) = 0$).

When a charged particle enters a uniform magnetic field with its velocity perpendicular to the field, it undergoes uniform circular motion in a plane perpendicular to the magnetic field. Therefore, the trajectory will be circular, not helical.

2. Calculate the initial force on the particle: The magnetic force on a charged particle is given by $\vec{F} = q(\vec{V} \times \vec{B})$. Let's compute the cross product $\vec{V} \times \vec{B}$: $\vec{V} \times \vec{B} = (3\hat{i} - 4\hat{j}) \times (B_0\hat{k})$ Using the properties of unit vector cross products ($\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$): $\vec{V} \times \vec{B} = (3B_0)(\hat{i} \times \hat{k}) - (4B_0)(\hat{j} \times \hat{k})$ $\vec{V} \times \vec{B} = (3B_0)(-\hat{j}) - (4B_0)(\hat{i})$ $\vec{V} \times \vec{B} = -4B_0\hat{i} - 3B_0\hat{j}$ Now, multiply by $q$ to get the force: $\vec{F} = q(-4B_0\hat{i} - 3B_0\hat{j}) = -qB_0(4\hat{i} + 3\hat{j})$

3. Calculate the period of revolution: For a charged particle moving in a circle in a uniform magnetic field, the magnetic force provides the necessary centripetal force: $qvB = \frac{mv^2}{r}$ The radius of the circular path is $r = \frac{mv}{qB}$. The angular frequency of revolution is $\omega = \frac{v}{r} = \frac{v}{(mv/qB)} = \frac{qB}{m}$. The period of revolution is $T = \frac{2\pi}{\omega}$. Substituting $\omega$: $T = \frac{2\pi m}{qB}$ In this problem, $B = B_0$. So, the period is: $T = \frac{2\pi m}{qB_0}$

4. Calculate the radius of the curve of revolution: The magnitude of the velocity is $v = |\vec{V}| = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ m/s. Using the formula for the radius of the circular path, $r = \frac{mv}{qB}$: $r = \frac{m(5)}{qB_0} = \frac{5m}{qB_0}$