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(A) Given: Total length of wire $L = 30 \, cm = 0.3 \, m$. Since it is a square loop,the side length $a = L/4 = 0.3/4 = 0.075 \, m = 7.5 \, cm$.Diamet

Vedclass · Accepted Answer

$0.61$

Vedclass · Answer

(A) Given: Total length of wire $L = 30 \, cm = 0.3 \, m$. Since it is a square loop,the side length $a = L/4 = 0.3/4 = 0.075 \, m = 7.5 \, cm$.Diameter of wire $d = 4 \, mm$,so radius $r = 2 \, mm = 2 	imes 10^{-3} \, m$.Resistivity $ho = 1.23 	imes 10^{-8} \, \Omega m$.Rate of change of magnetic field $dB/dt = 0.032 \, T s^{-1}$.Area of the loop $A = a^2 = (0.075)^2 = 5.625 	imes 10^{-3} \, m^2$.Induced $EMF$ $\varepsilon = |d\phi/dt| = A(dB/dt) = (5.625 	imes 10^{-3}) 	imes 0.032 = 1.8 	imes 10^{-4} \, V$.Resistance of the wire $R = ho (L/A_{wire}) = ho (L / (\pi r^2)) = (1.23 	imes 10^{-8} 	imes 0.3) / (\pi 	imes (2 	imes 10^{-3})^2) = (3.69 	imes 10^{-9}) / (4\pi 	imes 10^{-6}) \approx 2.937 	imes 10^{-4} \, \Omega$.Induced current $i = \varepsilon / R = (1.8 	imes 10^{-4}) / (2.937 	imes 10^{-4}) \approx 0.613 \, A$.Thus,the induced current is close to $0.61 \, A$.

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