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Numerical · Q22

Q.A circular coil of 100100 turns and radius 5 cm5\ \text{cm} carries a current of 0.5 A0.5\ \text{A}. Find the magnetic field at the centre of the coil.

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✓ Free question

Given N=100N=100, I=0.5 AI=0.5\ \text{A}, R=0.05 mR=0.05\ \text{m}:

B=μ0NI2R=(4π×10−7)(100)(0.5)2(0.05)=6.283×10−50.1=6.283×10−4 TB = \frac{\mu_0 N I}{2R} = \frac{(4\pi\times 10^{-7})(100)(0.5)}{2(0.05)} = \frac{6.283\times 10^{-5}}{0.1} = 6.283\times 10^{-4}\ \text{T}

✓Final answer

The field at the centre is B≈6.28×10−4 T=0.628 mTB \approx 6.28\times 10^{-4}\ \text{T} = 0.628\ \text{mT}.

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