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

• A

  $5(\hat{\mathbf{i}}+\hat{\mathbf{j}})$
• B

  $5(\hat{\mathbf{i}}+\sqrt{3 \mathbf{j}})$
• C

  $5(\sqrt{3} \hat{\mathbf{i}}-\hat{\mathbf{j}})$
• Correct answer
  D

  $5(\sqrt{3} \hat{i}+\hat{j})$

Solution

We know that time period, $T=\frac{2 \pi m}{q B}=\frac{2 \pi}{\frac{q}{m} B} 1 \mathrm{~s}$ Since, the particle will be at point $P$ after time, $t=\frac{1}{12} \mathrm{~s}=\frac{T}{12} \mathrm{~s}$ It will be deviated by an angle, $\theta=2 \pi / 12=30^{\circ}$
Hence, velocity at point $P$, $\mathbf{v}=10\left(\cos 30^{\circ} \hat{\mathbf{i}}+\sin 30^{\circ} \hat{\mathbf{j}}\right)$ $=10\left(\frac{\sqrt{3}}{2} \hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}\right)=5(\sqrt{3} \hat{\mathbf{i}}+\hat{\mathbf{j}})$

Hence, the correct answer is option: 'D'.