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

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Question

(A) The area of the square loop is $A = (10 \; cm)^2 = 100 \; cm^2 = 10^{-2} \; m^2$.The angle $	heta$ between the area vector (perpendicular to the

Vedclass · Accepted Answer

$1.0$ mV,$2$ mA

Vedclass · Answer

(A) The area of the square loop is $A = (10 \; cm)^2 = 100 \; cm^2 = 10^{-2} \; m^2$.The angle $	heta$ between the area vector (perpendicular to the east-west plane) and the magnetic field (in the north-east direction) is $45^{\circ}$.The initial magnetic flux is $\phi_i = B A \cos 	heta = 0.10 	imes 10^{-2} 	imes \cos(45^{\circ}) = \frac{10^{-3}}{\sqrt{2}} \; Wb$.The final magnetic flux is $\phi_f = 0 \; Wb$.The change in flux is $\Delta \phi = |\phi_f - \phi_i| = \frac{10^{-3}}{\sqrt{2}} \; Wb$.The magnitude of the induced $emf$ is $\varepsilon = \frac{|\Delta \phi|}{\Delta t} = \frac{10^{-3}}{\sqrt{2} 	imes 0.70} \approx 1.01 	imes 10^{-3} \; V \approx 1.0 \; mV$.The magnitude of the induced current is $I = \frac{\varepsilon}{R} = \frac{1.01 	imes 10^{-3} \; V}{0.5 \; \Omega} \approx 2.02 	imes 10^{-3} \; A \approx 2.0 \; mA$.

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