The figure shows a square loop of side 5 , cm | vedclass

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(A) Given: Speed $v = 1 \, cm/s = 0.01 \, m/s$, Side length $l = 5 \, cm = 0.05 \, m$, Magnetic field $B = 0.6 \, T$.$(a) \, 	ext{At } t = 2 \, s$: T

Vedclass · Accepted Answer

$(a) \, 3 	imes 10^{-4} \, V, \, (b) \, 	ext{Zero}, \, (c) \, 3 	imes 10^{-4} \, V$

Vedclass · Answer

(A) Given: Speed $v = 1 \, cm/s = 0.01 \, m/s$, Side length $l = 5 \, cm = 0.05 \, m$, Magnetic field $B = 0.6 \, T$.$(a) \, 	ext{At } t = 2 \, s$: The loop has entered the field by a distance $x = v 	imes t = 1 \, cm/s 	imes 2 \, s = 2 \, cm$. The induced $emf$ is $E = Bvl = 0.6 \, T 	imes 0.01 \, m/s 	imes 0.05 \, m = 3 	imes 10^{-4} \, V$.$(b) \, 	ext{At } t = 10 \, s$: The loop has moved $10 \, cm$ into the field. Since the entire loop is now inside the uniform magnetic field, the magnetic flux linked with the loop is constant. Therefore, the induced $emf$ is $E = -d\phi/dt = 0$.$(c) \, 	ext{At } t = 22 \, s$: The front edge has moved $22 \, cm$. Since the field is $20 \, cm$ wide, the front edge is $2 \, cm$ outside the field region. The trailing edge is still inside the field. The induced $emf$ is $E = Bvl = 0.6 \, T 	imes 0.01 \, m/s 	imes 0.05 \, m = 3 	imes 10^{-4} \, V$.

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