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GUJCET · Physics · Moving Charges and Magnetism
A coil having \(10 Am ^2\) magnetic moment is placed in a vertical plane and is free to rotate about its horizontal axis coincides with its diameter. A uniform magnetic field of 2 T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coll rotates through an angle of \(90^{\circ}\) under the influence of magnetic field. The moment of inertia of coil is \(0.1 kg m ^2\). What will be its angular speed ?
- A \(40 rad / s\)
- B \(10 rad / s\)
- C \(20 rad / s\)
- D \(5 rad / s\)
Answer & Solution
Correct Answer
(C) \(20 rad / s\)
Step-by-step Solution
Detailed explanation
C
Change in potential energy of a coil
\(\begin{array}{l}\Delta U=U_f-U_i \\\Delta U=\left[-m B \cos 90^{\circ}-\left(-m B \cos 0^{\circ}\right)\right] \\\Delta U=m B\end{array}\)
Change in energy = gained angular kinetic energy
\(\begin{aligned}m B =\frac{1}{2} I \omega^2 \\\omega =\sqrt{\frac{2 m B}{I}} \\\omega =\sqrt{\frac{2(10)(2)}{0.1}} \\\omega =20 rad / s\end{aligned}\)
Change in potential energy of a coil
\(\begin{array}{l}\Delta U=U_f-U_i \\\Delta U=\left[-m B \cos 90^{\circ}-\left(-m B \cos 0^{\circ}\right)\right] \\\Delta U=m B\end{array}\)
Change in energy = gained angular kinetic energy
\(\begin{aligned}m B =\frac{1}{2} I \omega^2 \\\omega =\sqrt{\frac{2 m B}{I}} \\\omega =\sqrt{\frac{2(10)(2)}{0.1}} \\\omega =20 rad / s\end{aligned}\)
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