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(C) The self-inductance $L$ of a solenoid is given by $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length, $A$ is the cross-sectiona

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$120$

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(C) The self-inductance $L$ of a solenoid is given by $L = \mu_0 n^2 A l$, where $n$ is the number of turns per unit length, $A$ is the cross-sectional area, and $l$ is the length of the solenoid.Given: $n = 10	ext{ turns/cm} = 1000	ext{ turns/m}$, $A = 5	ext{ cm}^2 = 5 	imes 10^{-4}	ext{ m}^2$, $l = 30	ext{ cm} = 0.3	ext{ m}$.$L = (4\pi 	imes 10^{-7} 	ext{ T m/A}) 	imes (1000 	ext{ m}^{-1})^2 	imes (5 	imes 10^{-4} 	ext{ m}^2) 	imes (0.3 	ext{ m})$.$L = 4\pi 	imes 10^{-7} 	imes 10^6 	imes 5 	imes 10^{-4} 	imes 0.3 = 0.6\pi 	imes 10^{-3} 	ext{ H}$.The induced e.m.f. $\epsilon$ is given by $\epsilon = L \frac{di}{dt}$.Here, $\frac{di}{dt} = \frac{4	ext{ A} - 2	ext{ A}}{3.14	ext{ s}} = \frac{2}{3.14} 	ext{ A/s}$.$\epsilon = (0.6\pi 	imes 10^{-3}) 	imes \frac{2}{3.14}$.Since $\pi \approx 3.14$, we have $\epsilon = 0.6 	imes 3.14 	imes 10^{-3} 	imes \frac{2}{3.14} = 1.2 	imes 10^{-3} 	ext{ V}$.$\epsilon = 120 	imes 10^{-5} 	ext{ V}$.Comparing this with $\alpha 	imes 10^{-5} 	ext{ V}$, we get $\alpha = 120$.

$A$ $30\text{ cm}$ long solenoid has $10$ turns per cm and area of $5\text{ cm}^2$. The current through the solenoid coil varies from $2\text{ A}$ to $4\text{ A}$ in $3.14\text{ s}$. The e.m.f. induced in the coil is $\alpha \times 10^{-5}\text{ V}$. The value $\alpha$ is . . . . . . .

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