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Question
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?
Solution
N = 100,
R = 8 × 10−2 m,
I = 0.4 A,
μ0 = 4π × 10−7 T m/A
B = `(mu_0 "NI")/"2R"`
= `((4pi xx 10^-7)(100)(0.4))/(2(8 xx 10^-2))`
= 3.14 × 10−4 T
Hence, the magnitude of the magnetic field is 3.14 × 10−4 T.
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