Question

A coil having area \(2 \mathrm{~m}^{2}\) is placed in a magnetic field which changes from \(1 \mathrm{wb} / \mathrm{m}^{2}\) to $4 \mathrm{wb} / \mathrm{m}^{2}$ in an interval of 2 second. The emf induced in the coil of single turn is.... (a) \(4 \mathrm{v}\) (b) \(3 \mathrm{v}\) (c) \(1.5 \mathrm{v}\) (d) \(2 \mathrm{v}\)

Short answer

The short answer for the given problem is (b) $3 \mathrm{v}$.

Step-by-step solution

Find the initial and final magnetic flux

To find the initial and final magnetic flux, we need to multiply the magnetic field strength with the area of the coil. The magnetic field strength changes from \(1 \mathrm{wb/m^2}\) to \(4 \mathrm{wb/m^2}\). Let \( \Phi_{1} \) and \( \Phi_{2} \) be the initial and final magnetic flux, respectively. For the initial magnetic flux: \( \Phi_{1} = B_{1} \cdot A \) \( \Phi_{1} = (1 \mathrm{wb/m^2})(2 \mathrm{m^2}) = 2 \mathrm{wb} \) For the final magnetic flux: \( \Phi_{2} = B_{2} \cdot A \) \( \Phi_{2} = (4 \mathrm{wb/m^2})(2 \mathrm{m^2}) = 8 \mathrm{wb} \)

Calculate the change in magnetic flux

Next, we need to find the change in magnetic flux, which is the difference between the final magnetic flux and the initial magnetic flux. Change in magnetic flux: \( \Delta \Phi = \Phi_{2} - \Phi_{1} \) \( \Delta \Phi = 8 \mathrm{wb} - 2 \mathrm{wb} = 6 \mathrm{wb} \)

Calculate the induced emf using Faraday's law

Now, we can calculate the induced emf using Faraday's law of electromagnetic induction. The formula is: \( |emf| = \frac{|\Delta \Phi|}{\Delta t} \) where \( |emf| \) is the induced emf, \( |\Delta \Phi| \) is the change in magnetic flux, and \( \Delta t \) is the time interval. We have \( |\Delta \Phi| = 6 \mathrm{wb} \) and \( \Delta t =2 \mathrm{s} \). Plugging these values into the formula: \( |emf| = \frac{6 \mathrm{wb}}{2 \mathrm{s}} \) \( |emf| = 3 \mathrm{v} \)

Choose the correct option

The induced emf in the single-turn coil is calculated as 3 volts. Therefore, the correct answer is: (b) 3 volts