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Question
Figure shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at t = 0. Find the total heat produced in the loop during the interval 0 to 30 s if the resistance of the loop is 4.5 mΩ.
Solution
Resistance of the coil, R = 45 mΩ = 4.5 × 10−3 Ω
The heat produced is found by taking the sum of the individual heats produced.
Thus, the net heat produced is given by
H = H1 + H2 + H3 + H4
(a) Heat developed for the first 5 seconds:-
Emf induced, e = 3 × 10−4 V
Current in the coil,
\[i = \frac{e}{R} = \frac{3 \times {10}^{- 4}}{4 . 5 \times {10}^{- 3}}\]= 6.7 × 10−2 A
H1 = (6.7 × 10−2)2 × 4.5 × 10−3 × 5
There is no change in the emf from 5 s to 20 s and from 25 s to 30 s.
Thus, the heat developed for the above mentioned intervals is given by
H2 = H4 = 0
Heat developed in interval t = 25 s to 30 s:-
The current and voltage induced in the coil will be the same as that for the first 5 seconds.
H3 = (6.7 × 10−2)2 × 4.5 × 10−3 × 5
Total heat produced:-
H = H1 + H3
= 2 × (6.7 × 10−2)2 × 4.5 × 10−2 × 5
= 2 × 10−4 J
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