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Physics Question on Electromagnetic induction

A circular coil of radius 10 cm10 \,cm, 500 turns500\, turns and resistance 2 Ω2 \,\Omega is placed with its plane perpendicular to the horizontal component of the earth?s magnetic field. It is rotated about its vertical diameter through 180∘^{\circ} in 0.25 s0.25\, s. The current induced in the coil is (Horizontal component of the earth?s magnetic field at that place is 3.0×10−53.0 \times 10^{-5} T)

A

1.9×10−3 A1.9 \times 10^{-3}\, A

B

2.9×10−3 A2.9 \times 10^{-3}\, A

C

3.9×10−3 A3.9 \times 10^{-3}\, A

D

4.9×10−3 A4.9 \times 10^{-3}\, A

Answer

1.9×10−3 A1.9 \times 10^{-3}\, A

Explanation

Solution

Initial magnetic flux through the coil, ϕi=BHAcosθ=3.0×10−5×(π×10−2)×cos0∘\phi_{i}=B_{H}Acos\theta=3.0\times10^{-5}\times\left(\pi\times10^{-2}\right)\times cos0^{\circ} =3π×10−7 Wb\quad =3\pi\times10^{-7}\,Wb Final magnetic flux after the rotation ϕi=3.0×10−5×(π×10−2)×cos180∘\phi_{i}=3.0\times10^{-5}\times\left(\pi\times10^{-2}\right)\times cos180^{\circ} =3π×10−7 Wb\quad =3\pi\times10^{-7}\,Wb Induced emf, ε=−Ndϕdt=−N(ϕf−ϕi)t\varepsilon=-N \frac{d\phi}{dt}=-\frac{N\left(\phi_{f} -\phi_{i}\right)}{t} =−500(−3π×10−7−3π×10−7)0.25=-\frac{500\left(-3\pi\times10^{-7}-3\pi\times10^{-7}\right)}{0.25} =500×(6π×10−7)0.25=3.8×10−3 V=\frac{500\times\left(6\pi\times10^{-7}\right)}{0.25}=3.8\times10^{-3}\,V I=εR=3.8×10−3 V2 Ω=1.9×10−3 AI=\frac{\varepsilon}{R}=\frac{3.8\times10^{-3}\,V}{2\,\Omega}=1.9\times10^{-3}\,A