Answer to Question #330234 in Physics for Mercy

Question

A closely wound 0.20m × .20 m square coil of 100 turns is exposed to an external magnetic field of 0.74 T that is perpendicular to the plane of the coil. The coil is then rotated for 10 s until it makes 60 degrees with the magnetic field. Find the average induced electromotive force during the 10 s interval.

Answer 1

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

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

where $\Delta\Phi$ is the change in magnetic flux, and $\Delta t = 10s$ is the time elapsed. The initial magnetic flux through the coil is:

$$\Phi_i = NBA$$

where $N = 100$, $B= 0.74T, A = 0.20m × 0.20 m=0.04m^2$. The final flux is:

$$\Phi_f = NBA\cos\theta$$

where $\theta=60\degree$. Thus, the change is:

$$\Delta\Phi = \Phi_f - \Phi_i = NBA(\cos\theta -1)$$

and the emf:

$$\mathcal{E} = \dfrac{NBA(1-\cos\theta)}{\Delta t}\\ \mathcal{E} = \dfrac{100\cdot 0.74\cdot 0.04\cdot (1-\cos60\degree)}{10} = 0.148V$$

Answer. 0.148V.