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प्रश्न
The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A s−1. Find the emf induced in it.
उत्तर
Given:-
Number of turns, N = 240
Radius of the solenoid, r = 2 cm
Length of the solenoid, l = 12 cm
The emf induced in the solenoid is given by
\[e = L\frac{di}{dt}\]
The self-inductance of the solenoid is given by
\[L = \frac{\mu_0 N^2 A}{l}\]
\[L = \frac{4\pi \times {10}^{- 7} \times {240}^2 \times \pi \times (2 \times {10}^{- 2} )^2}{12 \times {10}^{- 2}}\]
Thus, the emf induced in the solenoid is given by
\[e = \frac{4\pi \times {10}^{- 7} \times {240}^2 \times \pi \times (2 \times {10}^{- 2} )^2}{12 \times {10}^{- 2}} \times 0 . 8\]
\[ = 60577.3824 \times {10}^{- 8} = 6 \times {10}^{- 4} V\]
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