Question

# A circular coil of wire consisting of 100 turns, each of radius 8.0 cmcarries a current of 0.40 A. What is the magnitude of the magneticfield B at the centre of the coil?

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Solution

## Given: The number of turns on the circular coil is 100, the radius of each turn is 8.0 cm, the current flowing in the coil is 0.40 A. The magnitude of the magnetic field at the centre of the coil is given as, | B → |= μ 0 NI 2r Where, the number of turns in the circular coil is N, the radius of each turn is r and the current in the coil is I. By substituting the given values in the above expression, we get | B → |= ( 4π× 10 −7 )×100×0.4 2×0.08 =3.14× 10 −4  T Thus, the magnitude of the magnetic field is 3.14× 10 −4  T.

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