a circular coil of wire consisting of hundred turn each of radius 8.0 centimetre carries a current of 0.4 T A what is the magnitude of magnetic field at the centre of the coil
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Answer:
B at the center of a current carrying coil is given by B= μ₀NI/2R
Explanation:
μ₀=4π×10⁻⁷
now if you substitute
B=4×3.14×10⁻⁷×100×0.4/8×10⁻²
B=6.28×10⁻⁴ T
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