Question

A flat spiral coil has a large number of turns $N$. The turns are wound tightly and the inner and outer radii of the coil are $a$ and $b$ respectively. A uniform external magnetic field $(B)$ is applied perpendicular to the plane of the coil. Find the emf induced in the coil when the field is made to change at a rate $\frac{dB}{dt} = \alpha$.

Answer 1

-N \pi (b^2 - a^2) \alpha

Answer 2

The induced electromotive force (emf) in a coil is given by Faraday's law of induction: $\mathcal{E} = -N \frac{d\Phi_B}{dt}$ where $N$ is the number of turns and $\frac{d\Phi_B}{dt}$ is the rate of change of magnetic flux. The magnetic flux $\Phi_B$ through a single turn is $\Phi_B = B \cdot A$, where $A$ is the area of the coil. The area of the flat spiral coil is the area of the annulus: $A = \pi b^2 - \pi a^2 = \pi (b^2 - a^2)$ The rate of change of magnetic flux is: $\frac{d\Phi_B}{dt} = A \frac{dB}{dt}$ Given $\frac{dB}{dt} = \alpha$, we have: $\frac{d\Phi_B}{dt} = \pi (b^2 - a^2) \alpha$ Therefore, the induced emf is: $\mathcal{E} = -N \frac{d\Phi_B}{dt} = -N \pi (b^2 - a^2) \alpha$