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  • A square loop of side 12 cm and resistance 0.60 is placed vertically in the east-west plane. A uniform magnetic field of 0.10 G is set up across the plane in north-east direction. The magnetic fiel

A square loop of side 12 cm and resistance 0.60 is placed vertically in the east-west plane. A uniform magnetic field of 0.10 G is set up across the plane in north-east direction. The magnetic field is decreased to zero in 0.6 s at a steady rate. The magnitude of current during this time interval is:

Option: 1

\mathrm{1.42 \times 10^{-3} \mathrm{~A}}


Option: 2

\mathrm{2.72 \times 10^{-3} \mathrm{~A}}


Option: 3

\mathrm{3.41 \times 10^{-3} \mathrm{~A}}


Option: 4

\mathrm{4.21 \times 10^{-3} \mathrm{~A}}


Answers (1)

best_answer

\mathrm{\text { Here, Area } A=l^2=(12 \mathrm{~cm})^2=1.4 \times 10^{-2} \mathrm{~m}^2, \mathrm{R}=0.60 \Omega, \mathrm{B}_1=0.10 \mathrm{~T}, \theta=45^{\circ}, \mathrm{B}_2=0, \mathrm{dt}=0.6 \mathrm{~s}}

\mathrm{Initial \: flux, \phi_1=B_1 A \cos \theta=0.10 \times 1.4 \times 10^{-2} \times \cos 45^{\circ}=9.8 \times 10^{-4} \mathrm{~Wb} }

\mathrm{Final\: flux, \phi_2=0}

\mathrm{Induced \: emf, \varepsilon=\frac{|\mathrm{d} \phi|}{\mathrm{dt}}=\frac{\left|\phi_2-\phi_1\right|}{\mathrm{dt}}=\frac{\left|9.8 \times 10^{-4}\right|}{0.6 \mathrm{~s}}=1.63 \times 10^{-3} \mathrm{~V} }

\mathrm{Current, I=\frac{\varepsilon}{R}=\frac{1.63 \times 10^{-3}}{0.6}=2.72 \times 10^{-3} \mathrm{~A} }

 

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manish painkra

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