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Question

Physics Question on Electromagnetic induction

A closely wound flat circular coil of 2525 turns of wire has diaineter of 10 cm10\, cm which carries current of 4 A4\, A, the flux density at the centre of a coil will be

A

2.28×10−6T2.28 \times 10^{-6} T

B

1.679×10−6T1.679 \times 10^{-6} T

C

1.256×10−3T1.256 \times 10^{-3} T

D

1.572×10−5T1.572 \times 10^{-5} T

Answer

1.256×10−3T1.256 \times 10^{-3} T

Explanation

Solution

Number of turns in the coil, n=25n=25 Diameter of coil, d=10 cmd=10\, cm =0.1 m=0 .1 \,m ∴\therefore Radius, r=0.12=0.05 mr=\frac{0.1}{2}=0.05\, m Current i=4 Ai=4\, A Hence, flux density of the coil at the centre is given by B=μ04π×2πnirB=\frac{\mu_{0}}{4 \pi} \times \frac{2 \pi n i}{r} =10−7×2π×25×40.05=10^{-7} \times \frac{2 \pi \times 25 \times 4}{0.05} =1.256×10−3T=1.256 \times 10^{-3} T (∵μ0=4π×10−7NA−1m−1)\left(\because \mu_{0}=4 \pi \times 10^{-7} NA ^{-1} m ^{-1}\right)