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Question
A square-shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicular to a 1.0 T magnetic field. It is removed from the magnetic field in 0.25 s and restored in its original place in the next 0.25 s. Find the magnitude of the average emf induced in the loop during (a) its removal, (b) its restoration and (c) its motion.
Solution
(a) When the coil is removed from the magnetic field:-
Initial magnetic flux through the coil, ϕ1 = BA
∴ ϕ1 = 50 × 0.5 × 0.5 T-m2
= 12.5 T-m2
Now,
Initial magnetic flux through the coil, ϕ2 = 0
Time taken, t = 0.25 s
The average emf induced is given by
\[e = - \frac{∆ \phi}{∆ t} = \frac{\phi_1 - \phi_2}{dt}\]
\[ = \frac{12 . 5 - 0}{0 . 25} = \frac{125 \times {10}^{- 1}}{25 \times {10}^{- 2}} = 50 V\]
(b) When the coil is taken back to its original position:-
Initial magnetic flux through the coil, ϕ1 = 0
Initial magnetic flux through the coil, ϕ2 = 12.5 T-m2
Time taken, t = 0.25 s
The average emf induced is given by
\[e = \frac{12 . 5 - 0}{0 . 25} = 50 V\]
(c) When the coil is moving outside the magnetic field:-
Initial magnetic flux, ϕ1 = 0
Final magnetic flux, ϕ2 = 0
Because there is no change in the magnetic flux, no emf is induced.
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