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Numericals · Q16

Q.A circular coil of wire is made up of 100 turns, each of radius 8.0 cm. If a current of 0.40 A passes through it, what will be the magnetic field at the centre of the coil?

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For NN turns, the centre-of-loop formula generalises to B=μ0NI2RB=\dfrac{\mu_0NI}{2R}. Substituting N=100N=100, I=0.40I=0.40 A, R=8.0 cm=0.08R=8.0\ \text{cm}=0.08 m, μ0=4π×10−7\mu_0=4\pi\times10^{-7} T.m/A: $B=\dfrac{(4\pi\times10^{-7})(100)(0.40)}{2(0.08)}=\dfr …

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