A square loop of side 10 cm and resistance 0 | vedclass

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(A) Given:Side of square loop,$l = 10 \ cm = 0.1 \ m$Area of loop,$A = l^2 = (0.1 \ m)^2 = 0.01 \ m^2 = 100 \ cm^2$Resistance,$R = 0.5 \ \Omega$Initia

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$2 	imes 10^{-3} \ A$

Vedclass · Answer

(A) Given:Side of square loop,$l = 10 \ cm = 0.1 \ m$Area of loop,$A = l^2 = (0.1 \ m)^2 = 0.01 \ m^2 = 100 \ cm^2$Resistance,$R = 0.5 \ \Omega$Initial magnetic field,$B_1 = 0.10 \ T$Final magnetic field,$B_2 = 0 \ T$Time interval,$\Delta t = 0.70 \ s$Angle between the area vector (normal to the loop) and the magnetic field vector,$	heta = 45^{\circ}$The magnetic flux $\phi$ is given by $\phi = BA \cos 	heta$.Initial flux,$\phi_1 = B_1 A \cos 45^{\circ} = 0.10 	imes 0.01 	imes \frac{1}{\sqrt{2}} \ Wb$.Final flux,$\phi_2 = 0 \ Wb$.Change in flux,$\Delta \phi = \phi_2 - \phi_1 = -\phi_1 = -\frac{0.001}{\sqrt{2}} \ Wb$.Induced emf,$\varepsilon = -\frac{\Delta \phi}{\Delta t} = -\left( -\frac{0.001}{\sqrt{2} 	imes 0.70} ight) = \frac{0.001}{0.70 	imes 1.414} \approx 1.01 	imes 10^{-3} \ V$.Using the provided approximation in the original problem: $\varepsilon = \frac{0.1 	imes 100 	imes 10^{-4} 	imes \cos 45^{\circ}}{0.70} = \frac{0.001 	imes 0.707}{0.70} \approx 10^{-3} \ V$.Induced current,$I = \frac{\varepsilon}{R} = \frac{10^{-3} \ V}{0.5 \ \Omega} = 2 	imes 10^{-3} \ A$.

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