Question

A coil having magnetic moment 15 A placed in a uniform magnetic field of 4T in the horizontal direction exits such that initially the axis of coil is in the direction of the field. If the coil is rotated by 45° and the moment of inertia of the coil is 0.5 kg then the angular speed acquired by the coil is

• A. 20 rad
• B. 10 rad
• C. 8.34 rad
• D. 4.5 rad

Answer 1

Answer: (C)

8.34 rad

Answer 2

since $\tau = \mathrm { I } \frac { \mathrm { d } \omega } { \mathrm { dt } } = \mathrm { mB } \sin \theta$ … (i)

Now, ……(ii)

Then , from (i) and (ii)

Integrating both sides

$\frac { 1 } { 2 } \mathrm { I } \omega _ { \mathrm { f } } ^ { 2 } = - \mathrm { mB } \left[ \cos 45 ^ { \circ } - \cos 0 ^ { \circ } \right]$

$= - \mathrm { mB } \left[ \frac { 1 } { \sqrt { 2 } } - 1 \right] = 0.29 \mathrm { mB }$

$\omega _ { \mathrm { f } } ^ { 2 } = \frac { 2 \times 0.29 \mathrm { mB } } { \mathrm { I } } = \frac { 2 \times 0.29 \times 15 \times 4 } { 0.50 }$