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.

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Solution

$H = \frac{μ_{0} n^{2} A d i}{6} \frac{d i}{d t} = \frac{4 π \times 10^{- 7} \times (240)^{2} \times π \times (2 \times 10^{- 2})^{2}}{12 \times 10^{- 2}} \times 0.8 = \frac{4 π \times (24)^{2} \times π \times 4 \times 8}{12} \times 10^{- 8} = 60577.3824 \times 10^{- 8} = 6 \times 10^{- 4} V .$

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The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of0.8As−1 Find the emf induced in it.

H=μ0n2Adi6didt=4π×10−7×(240)2×π×(2×10−2)212×10−2×0.8=4π×(24)2×π×4×812×10−8=60577.3824×10−8=6×10−4V.

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.