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Question 1

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer

Number of turns on the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given by the relation,

| B | = μ0 2πnI / 4π r

Where,

μ= Permeability of free space

= 4π × 10–7 T m A–1

| B | = 4π x 10-7 x 2π x 100 x 0.4 / 4π x 0.08

       = 3.14 x 10-4 T

Hence, the magnitude of the magnetic field is 3.14 × 10–4 T.

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