KCET · Physics · Magnetic Effects of Current
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 \(5(\sqrt{3} \hat{\mathbf{i}}-\hat{\mathbf{j}})\)
- D \(5(\sqrt{3} \hat{i}+\hat{j})\)
Answer & Solution
Correct Answer
(D) \(5(\sqrt{3} \hat{i}+\hat{j})\)
Step-by-step Solution
Detailed explanation
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 \(P\),
\(\mathbf{v}=10\left(\cos 30^{\circ} \hat{\mathbf{i}}+\sin 30^{\circ} \hat{\mathbf{j}}\right)\)
\(=10\left(\frac{\sqrt{3}}{2} \hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}\right)=5(\sqrt{3} \hat{\mathbf{i}}+\hat{\mathbf{j}})\)
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 \(P\),
\(\mathbf{v}=10\left(\cos 30^{\circ} \hat{\mathbf{i}}+\sin 30^{\circ} \hat{\mathbf{j}}\right)\)
\(=10\left(\frac{\sqrt{3}}{2} \hat{\mathbf{i}}+\frac{1}{2} \hat{\mathbf{j}}\right)=5(\sqrt{3} \hat{\mathbf{i}}+\hat{\mathbf{j}})\)
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