Question

A circular coil of diameter 15 mm having 300 turns is placed in a magnetic field of 30 mT such that the plane of the coil is perpendicular to the direction of the magnetic field. The magnetic field is reduced uniformly to zero in 20 ms and again increased uniformly to 30 mT in 40 ms. If the EMFs induced in the two time intervals are \( e_1 \) and \( e_2 \) respectively, then the value of \( e_1 / e_2 \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

The induced EMF is proportional to the rate of change of magnetic flux. The faster the change, the greater the induced EMF.

Answer 1

The Correct Option is B

To solve the problem of finding \( \frac{e_1}{e_2} \), we start with the formula for the EMF induced in a coil: \[ e = -N \frac{d\Phi}{dt} \] where \( N \) is the number of turns, \(\Phi\) is the magnetic flux, and \(t\) is time. The magnetic flux \(\Phi\) is given by \(\Phi = B \cdot A\), where \(B\) is the magnetic field and \(A\) is the area of the coil. For a circular coil, the area \(A\) is \(\pi r^2\). Given the diameter of the coil is 15 mm (0.015 m), the radius \(r\) is 0.0075 m, so: \[ A = \pi (0.0075)^2 \] \[ A = \pi \times 5.625 \times 10^{-5} \] The initial magnetic field \(B\) is 30 mT (0.03 T). Thus, the initial magnetic flux is: \[ \Phi_i = 0.03 \times \pi \times 5.625 \times 10^{-5} \] The coil then goes through two changes:

• Case 1: \(B\) reduces uniformly to 0 in 20 ms (0.02 s). The change in flux \(\Delta\Phi_1\) is: \[ \Delta\Phi_1 = -\Phi_i \]

• Case 2: \(B\) increases uniformly to 0.03 T in 40 ms (0.04 s). The change in flux \(\Delta\Phi_2\) is: \[ \Delta\Phi_2 = \Phi_i \]

Let's calculate the induced EMFs \(e_1\) and \(e_2\):

• \(e_1\) when \(B\) reduces to 0: \[ e_1 = -N \frac{\Delta\Phi_1}{\Delta t_1} = -300 \times \frac{-\Phi_i}{0.02} = 300 \times \frac{\Phi_i}{0.02} \]

• \(e_2\) when \(B\) increases to 0.03 T: \[ e_2 = -N \frac{\Delta\Phi_2}{\Delta t_2} = -300 \times \frac{\Phi_i}{0.04} = 300 \times \frac{-\Phi_i}{0.04} \]

Taking the ratio \(\frac{e_1}{e_2}\): \[ \frac{e_1}{e_2} = \frac{300 \times \Phi_i / 0.02}{300 \times \Phi_i / 0.04} = \frac{1/0.02}{1/0.04} = \frac{0.04}{0.02} = 2 \] The correct value of \(\frac{e_1}{e_2}\) is 2.