Question 19: Electromagnetic Induction

Question

Around the central part of an air cored solenoid of length $$20 \mathrm{~cm}$$ and area of cross section $$1.4 	imes 10^{-3} \mathrm{~m}^2$$ and 3000 turns, another coil of 250 turns is closely wound. A current $$2 \mathrm{~A}$$ in the solenoid is reversed in $$0.2 \mathrm{~s}$$, then the induced emf produced is

Accepted Answer

The magnetic field inside the solenoid is given by
$$B = \mu_0 nI = \mu_0 \frac{N}{L} I$$
where, $$N$$ is the number of turns, $$L$$ is the length of the solenoid, $$I$$ is the current flowing through the solenoid, and $$\mu_0$$ is the permeability of free space.
The magnetic flux through the smaller coil is given by
$$\phi = BA = \mu_0 \frac{N}{L} I A$$
where $$A$$ is the area of the smaller coil.
The induced emf in the smaller coil is given by Faraday's law of electromagnetic induction as
$$e = -\frac{d\phi}{dt}$$
Substituting the value of $$\phi$$ in the above equation, we get
$$e = -\frac{d}{dt} (\mu_0 \frac{N}{L} I A) = -\mu_0 \frac{N}{L} A \frac{dI}{dt}$$
Given that the current is reversed in $$0.2 s$$, the rate of change of current is
$$\frac{dI}{dt} = \frac{2-(-2)}{0.2} = 20 \mathrm{~A/s}$$
Substituting the values, we get
$$e = -4 \pi 	imes 10^{-7} 	imes \frac{3000}{0.2} 	imes 1.4 	imes 10^{-3} 	imes 20 = \boxed{1.32 	imes 10^{-1} \mathrm{~V}}$$
Therefore, the induced emf produced is $$1.32 	imes 10^{-1} \mathrm{~V}$$.

Answer

$$1.32 	imes 10^{-1} \mathrm{~V}$$

Answer

$$4 	imes 10^{-1} \mathrm{~V}$$

Answer

$$1.16 	imes 10^{-1} \mathrm{~V}$$

Answer

$$8 	imes 10^{-2} \mathrm{~V}$$

Electromagnetic Induction

Around the central part of an air cored solenoid of length $20 \mathrm{~cm} and area of cross section 1.4 \times 10^{-3} \mathrm{~m}^2 and 3000 turns, another coil of 250 turns is closely wound. A current 2 \mathrm{~A} in the solenoid is reversed in 0.2 \mathrm{~s}$, then the induced emf produced is

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