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(B) The area of the square loop is $A = 25 \ cm^2 = 25 	imes 10^{-4} \ m^2$. The side length of the loop is $l = \sqrt{A} = 5 	imes 10^{-2} \ m$.The

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$1.0 	imes 10^{-3} \ J$

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(B) The area of the square loop is $A = 25 \ cm^2 = 25 	imes 10^{-4} \ m^2$. The side length of the loop is $l = \sqrt{A} = 5 	imes 10^{-2} \ m$.The magnetic field $B = 40 \ T$ and resistance $R = 10 \ \Omega$.When the loop is pulled out of the magnetic field with velocity $v$, the induced electromotive force $(EMF)$ is $\varepsilon = Blv$.The time taken to pull the loop out is $t = 1 \ s$. The velocity is $v = l/t = (5 	imes 10^{-2} \ m) / (1 \ s) = 5 	imes 10^{-2} \ m/s$.The induced $EMF$ is $\varepsilon = 40 	imes (5 	imes 10^{-2}) 	imes (5 	imes 10^{-2}) = 0.1 \ V$.The power dissipated as heat is $P = \varepsilon^2 / R = (0.1)^2 / 10 = 0.01 / 10 = 10^{-3} \ W$.The work done is equal to the heat dissipated: $W = P 	imes t = 10^{-3} \ W 	imes 1 \ s = 1.0 	imes 10^{-3} \ J$.

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