A 100 -turn closely wound circular coil of radius 5 cm has a magnetic field of 3.14 × 10^ -3 T at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively

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Quantrex Academy · Accepted Answer

Correct answer: D. 2.5 A, 2 A m ^2

The magnetic field at the centre of a circular coil is given by:

B = \dfrac{\mu_0 N I}{2R}

Substituting the given values (

B = 3.14 \times 10^{-3}

T,

N = 100

,

R = 0.05

m,

\mu_0 = 4\pi \times 10^{-7}

T m/A):

3.14 \times 10^{-3} = \dfrac{4\pi \times 10^{-7} \times 100 \times I}{2 \times 0.05}

Taking

\pi = 3.14

:

3.14 \times 10^{-3} = \dfrac{4 \times 3.14 \times 10^{-5} \times I}{0.1}

10^{-3} = 4 \times 10^{-4} \times I

I = \dfrac{10^{-3}}{4 \times 10^{-4}} = 2.5 \text{ A}

The magnetic moment of the coil is given by:

M = N I A = N I (\pi R^2)

M = 100 \times 2.5 \times 3.14 \times (0.05)^2

M = 250 \times 3.14 \times 0.0025

M = 1.9625 \text{ A m}^2 \approx 2 \text{ A m}^2

Answer:

2.5

A,

2

A m

^2

A 100 -turn closely wound circular coil of radius 5 cm has a...

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