JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2020
- Medium
- MCQ
A small circular loop of conducting wire has radius \(a\) and carries current \(I\). It is placed in a uniform magnetic field \(\mathrm{B}\) perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period \(T\). If the mass of the loop is \(m\) then
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\mathrm{T}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{T}}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=-\mathrm{MB} \sin \theta\)
\(\mathrm{I} \alpha=-\mathrm{MB} \sin \theta\)
for small \(\theta\)
\(\alpha=-\frac{\mathrm{MB}}{\mathrm{I}} \theta\)
\(\omega=\sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}=\sqrt{\frac{(\mathrm{I})\left(\pi \mathrm{R}^{2}\right) \mathrm{B}}{\left(\frac{\mathrm{mR}^{2}}{2}\right)}}\)
\(\omega=\sqrt{\frac{2 \mathrm{I} \pi \mathrm{B}}{\mathrm{m}}}\)
\(\therefore \mathrm{T}=\frac{2 \pi}{\omega}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}\)
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