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Physics Question on Electromagnetic induction

A square loop of side 12 cm12 \,cm and resistance 0.60 Ω0.60 \, \Omega is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T0.10\, T is set up across the plane in north-east direction. The magnetic field is decreased to zero in 0.6 s0.6 \,s at a steady rate. The magnitude of current during this time interval is

A

1.42×10−3 A1.42 \times 10^{-3}\,A

B

2.67×10−3 A2.67 \times 10^{-3}\,A

C

3.41×10−3 A3.41 \times 10^{-3}\,A

D

4.21×10−3 A4.21 \times 10^{-3}\,A

Answer

2.67×10−3 A2.67 \times 10^{-3}\,A

Explanation

Solution

Here, Area A=l2=(12 cm)2A=l^{2}=(12\, cm)^{2} =1.4×10−2 m2=1.4\times 10^{-2}\,m^{2} R=0.60 ΩR=0.60\, \Omega, B1=0.10 T B_{1}=0.10\, T, θ=45∘\theta=45^{\circ} B2=0B_{2}=0, dt=0.6 sdt=0.6\, s Initial flux, ϕ1=B1A cos θ\phi_{1}=B_{1}A\, cos\, \theta =0.10×1.4×10−2×cos 45∘=0.10 \times1.4\times10^{-2}\times cos\, 45^{\circ} =9.8×10−4 Wb= 9.8 \times 10^{-4}\, Wb Final flux, ϕ2=0\phi_{2}=0 Induced emf, ε=∣dϕ∣dt=∣ϕ2−ϕ1∣dt\varepsilon=\frac{\left|d\phi\right|}{dt}=\frac{\left|\phi_{2}-\phi_{1}\right|}{dt} =∣9.8×10−4∣0.6 s=1.6×10−3 V=\frac{\left|9.8\times10^{-4}\right|}{0.6\,s}=1.6\times10^{-3}\,V Current, I=εR=1.6×10−30.6I=\frac{\varepsilon}{R}=\frac{1.6\times10^{-3}}{0.6} 2.67×10−3 A2.67\times10^{-3}\,A