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

Solution :

Self inductance of solenoid :

$$L = \mu_0 n^2Al$$

Given :

Length of solenoid :

$$l = 30\text{ cm}$$

$$= 0.3\text{ m}$$

Turns per cm :

$$10\text{ turns/cm}$$

Therefore,

$$n = 1000\text{ turns/m}$$

Area :

$$A = 5\text{ cm}^2$$

$$= 5 \times 10^{-4}\text{ m}^2$$

Using,

$$\mu_0 = 4\pi \times 10^{-7}$$

Therefore,

$$L=(4\pi \times 10^{-7})(1000)^2(5 \times 10^{-4})(0.3)$$

$$=4\pi \times 10^{-7}\times 10^6 \times 1.5 \times 10^{-4}$$

$$=6\pi \times 10^{-5}\text{ H}$$

Induced emf :

$$e = L\frac{dI}{dt}$$

Given,

$$\frac{dI}{dt} = \frac{4-2}{3.14}$$

$$= \frac{2}{3.14}$$

Therefore,

$$e=6\pi \times 10^{-5}\times \frac{2}{3.14}$$

Using,

$$\pi \approx 3.14$$

$$e = 12 \times 10^{-5}\text{ V}$$

Hence,

$$\alpha = 12$$

Final Answer :

$$12$$