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?
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,
μ0 = 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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Question is incomplete because the value of mean radius is not given.but anyway you use the formula to find magnetic field at the center of coil that B= munot NI divided by 2 multiply mean radius
Can u plz post the solution for this question that a circular coil has 35 turns and a mean radius of 35 turns carries a current of 1.2 A. Find magnetic field at center of coil.
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