Question

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

• A. \( 1.32 \times 10^{-1} \, \text{V} \)
• B. \( 4 \times 10^{-1} \, \text{V} \)
• C. \( 1.16 \times 10^{-1} \, \text{V} \)
• D. \( 8 \times 10^{-2} \, \text{V} \)

Hint:

To find the induced emf, make sure to use the correct number of turns in the solenoid and the coil, and apply Faraday's law by calculating the change in magnetic flux due to the changing current.

Answer 1

The Correct Option is A

The induced emf (\( \varepsilon \)) in the secondary coil is given by Faraday's law of induction: \[ \varepsilon = - N \frac{d\Phi}{dt} \] where: - \( N \) is the number of turns in the coil (250 turns), - \( \Phi \) is the magnetic flux through a coil, and - \( \frac{d\Phi}{dt} \) is the rate of change of the magnetic flux. The magnetic flux \( \Phi \) is given by: \[ \Phi = B \times A \] where: - \( B \) is the magnetic field due to the solenoid, and - \( A \) is the area of cross-section of the solenoid. The magnetic field \( B \) inside the solenoid is given by: \[ B = \mu_0 \frac{N_s}{L} I \] where: - \( \mu_0 = 4\pi \times 10^{-7} \, \text{T.m/A} \) is the permeability of free space, - \( N_s \) is the number of turns in the solenoid (3000 turns), - \( L \) is the length of the solenoid (0.2 m), and - \( I \) is the current passing through the solenoid (2 A). Substitute the given values into the equation for \( B \): \[ B = (4\pi \times 10^{-7}) \times \frac{3000}{0.2} \times 2 \] \[ B = 1.884 \times 10^{-2} \, \text{T} \] Now, substitute into the equation for flux \( \Phi \): \[ \Phi = B \times A = (1.884 \times 10^{-2}) \times (1.4 \times 10^{-3}) = 2.6376 \times 10^{-5} \, \text{Wb} \] The rate of change of flux is: \[ \frac{d\Phi}{dt} = \frac{\Delta \Phi}{\Delta t} = \frac{2 \times 2.6376 \times 10^{-5}}{0.2} = 2.6376 \times 10^{-4} \, \text{Wb/s} \] Finally, the induced emf is: \[ \varepsilon = - N \frac{d\Phi}{dt} = - 250 \times 2.6376 \times 10^{-4} = 1.32 \times 10^{-1} \, \text{V} \] Thus, the induced emf produced is \( 1.32 \times 10^{-1} \, \text{V} \).