Question

The magnetic flux through a coil of wire containing two loops changes at a constant rate from -58 Wb to 38 Wb in 0.34 s. What is the emf induced in the coil?

Answer 1

The emf induced in the coil is given by the Faraday's law:

$$\mathcal{E} = -\dfrac{\Delta\Phi_B}{\Delta t}$$

where $\Delta\Phi_B = 38 Wb - (-58 Wb) = 66Wb$ is the change in magnetic flux, and $\Delta t = 0.34s$ is the time elapsed. Thus, obtain:

$$\mathcal{E} = -\dfrac{66Wb}{0.34s} \approx 1.9\times 10^2V$$

Answer. $1.9\times 10^2V$.