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NCERT Class XII Chapter
Moving Charges and Magnetism
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Question : 1 of 28
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A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.4 A. What is the magnitude of the magnetic field B at the centre of the coil?
Solution:  
The magnetic field at the centre of a circular coil having 100 turns.
B = (μ0I2r)\left(\frac{\mu_0 I}{2r}\right) N or B = [4π×10−7×0.42×8×102]\left[\frac{4\pi \times 10^{-7} \times 0.4}{2 \times 8 \times 10^{2}}\right] × 100
B = 3.14 × 10−410^{-4} T
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