Question

A particle of specific charge \(\frac{q}{m}=\pi Ckg^{-1}\) is projected the origin towards positive x-axis with the velocity 10ms^-1 in a uniform magnetic field \(\overrightarrow{B}=-2\hat{k}T\). The velocity \(\overrightarrow{v}\) of particle after time \(t=\frac{1}{12}s\) will be ( in ms^-1 )

• A. \(5(\hat{i}+\hat{j})\)
• B. \(5(\hat{i}+\sqrt{3}\hat{j})\)
• C. \(5(\sqrt{3}\hat{i}-\hat{j})\)
• D. \(5(\sqrt{3}\hat{i}+\hat{j})\)

Answer 1

The Correct Option is D

Given:

• Specific charge: \( \frac{q}{m} = \pi \, \text{C/kg} \)
• Initial velocity: \( \vec{v}_0 = 10\hat{i} \, \text{m/s} \)
• Magnetic field: \( \vec{B} = -2\hat{k} \, \text{T} \)
• Time: \( t = \frac{1}{12} \, \text{s} \)

Step 1: Use Cyclotron Motion Equation

A charged particle in a uniform magnetic field undergoes circular motion with angular frequency:

\[ \omega = \frac{qB}{m} = \left( \frac{q}{m} \right) B = \pi \cdot 2 = 2\pi \, \text{rad/s} \]

Step 2: Angle Rotated in Time \( t \)

\[ \theta = \omega t = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6} \, \text{radians} = 30^\circ \]

The velocity vector rotates by \( 30^\circ \) counterclockwise (right-hand rule for negative \( \hat{k} \)) in the xy-plane.

Step 3: Resolve Final Velocity

Initial velocity magnitude is 10 m/s, so after rotating \( 30^\circ \), the velocity vector is:

\[ \vec{v} = 10 \cos(30^\circ) \hat{i} + 10 \sin(30^\circ) \hat{j} = 10 \cdot \frac{\sqrt{3}}{2} \hat{i} + 10 \cdot \frac{1}{2} \hat{j} = 5\sqrt{3} \hat{i} + 5 \hat{j} \]

The velocity of the particle at time \( t = \frac{1}{12} \, \text{s} \) is \( \boxed{5\sqrt{3}\hat{i} + 5\hat{j}} \, \text{m/s} \), so the correct answer is (D).

Answer 2

The time period \( T \) of the particle's motion is given by: \[ T = \frac{2\pi m}{qB} = \frac{2\pi}{m \times qB} \] Substitute the values \( m = 2 \, \text{kg} \), \( q = \pi \, \text{C} \), and \( B = 2 \, \text{T} \): \[ T = \frac{2\pi}{2 \times \pi \times 2} = 15 \, \text{s} \] The particle will be at point P after time \( t = \frac{1}{12} \, \text{s} \). Thus, the particle moves a fraction of the time period: \[ t = \frac{1}{12} \, \text{s} = \frac{1}{12} \times T \] The angle \( \theta \) between the magnetic field and the velocity of the particle after this time can be calculated as: \[ \theta = \frac{2\pi}{12} = 30^\circ \] Thus, the velocity of the particle at point P can be determined using the following vector sum: \[ \vec{v} = 10 \cos(30^\circ) \hat{i} + 10 \sin(30^\circ) \hat{j} \] Now, using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \), we get: \[ \vec{v} = 10 \times \frac{\sqrt{3}}{2} \hat{i} + 10 \times \frac{1}{2} \hat{j} \] \[ \vec{v} = 5\sqrt{3} \hat{i} + 5 \hat{j} \] Thus, the velocity of the particle at point P is: \[ \boxed{5 \sqrt{3} \hat{i} + \hat{j}} \]