Question

A circular coil of radius 4 cm has 50 turns. In this coil a
current of 2 A is flowing. It is placed in a magnetic field of
0.1 weber/m2. The amount of work done in rotating it
through 180° from its equilibrium position will be
(a) 0.1J (b) 0.2 J (c) 0.4 J (d) 0.8 J​

Answer 1

Answer:

work done=MB(1-cosФ)

hence  

now M=NIA

M=50×2×3.14×16×10⁻⁴=0.5 amp-meter

as cos 180 is -1 the above formula becomes W=2MB  

Hence W=2×0.5×0.1= 0.1 joule

Explanation:

Answer 2

(a) The amount of work done in rotating the coil  from 180° from its equilibrium position will be 0.1 J.

Explanation:

Given the radius of circular coil, r = 4 cm = 0.04 m.

The number of turns, N =50

The current flowing in the coil, i = 2 A.

The magnetic fileld, B = 0.1 Wb/m^2.

 The magnetic moment of current carrying loop,

M = i A N

Here, A is the area of the coil.

substitute the given values, we get

$M=2A\times(\pi(0.04m)^2\times50$

M = 0.50 Am^2

Hence, the amount of work done in rotating the coil  from 180° from its equilibrium position

$W=MB(cos0^0-cos180^0)$

substitute the values, we get

$W=0.50A.m^2\times0.1Wb/m^2[1-(-1)]$

$W=0.50A.m^2\times0.1Wb/m^2\times2$

W = 0.1 J

#Learn More: magnetic moment.

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