Question

23) A coil having 10 Am² magnetic moment is placed in a vertical plane & is free to
rotate about its horizontal axis coincides with its diameter. A uniform magnetic
field of 2T in the horizontal direction exists such that initially the axis of the coil
is in the direction of the field. The coil rotates through an angle of 90° under the
influence of magnetic field. The moment of Inertia of coil is 0.1 kg m². What will
be its angular speed?
(B) 10 rad/s
(A) 40 rad/s
(D) 5 rad/s
(G) 20 rad/s​

Answer 1

Answer:

20 rad/s Is the correct answer

I.e. option (G)

Answer 2

Answer:

The angular speed of the coil is equal to 20rad/s when it rotated through angle 90° under the influence of magnetic field.

Therefore, the option (G) is correct.

Explanation:

Given, the magnetic moment of the coil, $M=10Am^{-2}$

The uniform magnetic field is applied, $B=2T$

The moment of inertia of the coil, $I=0.1kgm^{2}$

Initially given $\vec_{B}$ coincides with axis of coil

⇒   Initially    $\vec_{M}$ $\parallel$ $\vec_{B}$

For a loop in the field, $\tau ={\vec_{M} }*{\vec_{B}}$

$\tau = |{\vec_{M}}|\;|{\vec_{B}}|\;sin\theta$

θ is the angle between magnetic moment and magnetic field.

The coil rotates through angle 90°.

Therefore, $\tau=(10)(2)sin90^{o}$

$\tau=20Nm$

From the rotation mechanics:

$\tau=I\alpha =I\frac{d\omega}{dt}$

$\tau=I\frac{d\omega }{d\theta}*\frac{d\theta}{dt}$

$\tau=I\omega\frac{d\omega}{d\theta}$                                                        ................(1)

Put the value of τ=20Nm and $I=0.1$ in eq.(1);

$20=I\int\limits^\omega_0 {\omega} \, d\omega$

$20=\frac{1}{2}I\omega^{2}$

$20=0.1*\frac{\omega^2}{2}$

$\omega^2=400$

$\omega =20rad/sec$

Therefore, the angular speed of the coil is equal to 20rad/s.