A circular coil has moment of inertia 0.8 kg m 2 around any diameter and is carrying current to produce a magnetic moment of 20 Am 2. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be:

Question

A circular coil has moment of inertia 0.8 kg m 2 around any diameter and is carrying current to produce a magnetic moment of 20 Am 2. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60° will be: (A) 10 rad s-1 (B) 20 π rad s-1 (C) 10 π rad s -1 (D) 20 rad s -1

Tardigrade Admin · Accepted Answer

I text dia =0.8 kg / m 2 M =20 Am 2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9b3e650f94e91a9bd84efc8cfe58ad6c-.png" /> U i + K i = U f + K f 0+0=- MB cos 30°+(1/2) I ω2 20 × 4 × (√3/2)=(1/2)(0.8) ω2 ω=√100 √3=10(3)1 / 4

$10$ rad $s^{- 1}$

$20 π$ rad $s^{- 1}$

$10 π$ rad $s^{- 1}$

$20$ rad $s^{- 1}$

$I_{dia} = 0.8 k g / m^{2}$
$M = 20 A m^{2}$

$U_{i} + K_{i} = U_{f} + K_{f}$
$0 + 0 = - MB cos 3 0^{\circ} + \frac{1}{2} I ω^{2}$
$20 \times 4 \times \frac{3}{2} = \frac{1}{2} (0.8) ω^{2}$
$ω = 1003 = 10 (3)^{1/4}$

10 rad s−1

20π rad s−1

10π rad s−1

20 rad s−1

I dia ​=0.8kg/m2 M=20Am2 Ui​+Ki​=Uf​+Kf​ 0+0=−MBcos30∘+21​Iω2 20×4×23​​=21​(0.8)ω2 ω=1003​​=10(3)1/4

$10$ rad $s^{- 1}$

$20 π$ rad $s^{- 1}$

$10 π$ rad $s^{- 1}$

$20$ rad $s^{- 1}$

$I_{dia} = 0.8 k g / m^{2}$
$M = 20 A m^{2}$

$U_{i} + K_{i} = U_{f} + K_{f}$
$0 + 0 = - MB cos 3 0^{\circ} + \frac{1}{2} I ω^{2}$
$20 \times 4 \times \frac{3}{2} = \frac{1}{2} (0.8) ω^{2}$
$ω = 1003 = 10 (3)^{1/4}$