A square loop of area 25 , cm^2 has a resista | vedclass

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(B) Given: Area of square loop $A = 25 \, cm^2 = 25 	imes 10^{-4} \, m^2$.Side length $l = \sqrt{A} = 5 \, cm = 0.05 \, m$.Resistance $R = 10 \, \Ome

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

Vedclass · Answer

(B) Given: Area of square loop $A = 25 \, cm^2 = 25 	imes 10^{-4} \, m^2$.Side length $l = \sqrt{A} = 5 \, cm = 0.05 \, m$.Resistance $R = 10 \, \Omega$, Time $t = 1 \, s$, Magnetic field $B = 40 \, T$.The velocity $v$ required to pull the loop out in $1 \, s$ is $v = l/t = 0.05/1 = 0.05 \, m/s$.The induced motional emf is $\varepsilon = Blv$.The induced current is $I = \varepsilon/R = (Blv)/R$.Substituting the values: $I = (40 	imes 0.05 	imes 0.05) / 10 = 0.01 \, A$.The magnetic force acting on the loop is $F = BIl$.Substituting the values: $F = 40 	imes 0.01 	imes 0.05 = 0.02 \, N$.The work done is $W = F 	imes l = 0.02 	imes 0.05 = 1 	imes 10^{-3} \, J$.

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