Question

A circular coil of 25 turns and radius of 20 cm carrying a current of 1 A rests with its plane normal to an external field of magnitude 5.0 x 10-2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released its oscillates about its stable equilibrium with a frequency of 2 s-1. The moment of inertia of the coil about its axis of rotation

• A. 2.4 x 10-4 kg m2
• B. 1.2x10-4 kg m2
• C. 9.95 x 10-4 kg m2
• D. 9.25 x 10-4 kg m2

Answer 1

Answer: (C)

9.95 x 10-4 kg m2

Answer 2

: Here, , $\mathrm { n } = 25 , \mathrm { r } = 20 \mathrm {~cm} = 0.2 \mathrm {~m} , \mathrm { I } = 1 \mathrm {~A}$

$B = 5.0 \times 10 ^ { - 2 } T , v = 2.0 \mathrm {~s} ^ { - 1 }$

Then, magnetization , $\mathrm { M } = \mathrm { nIA } = \mathrm { nI } \pi \mathrm { r } ^ { 2 }$

$= 25 \times 1 \times 3.14 \times ( 0.2 ) ^ { 2 }$

$= 3.14 \mathrm { JT } ^ { - 1 }$

As frequency of coil,

$v = \frac { 1 } { 2 \pi } \sqrt { \frac { \mathrm { MB } } { \mathrm { I } } }$

$= \frac { 5 \times 10 ^ { - 2 } } { 16 \times 3.14 } = 9.95 \times 10 ^ { - 4 } \mathrm {~kg} \mathrm {~m} ^ { 2 }$