Question

A uniform magnetic field is perpendicular to the plane of a circular wire loop of radius r. The magnitude of the field varies with time according to$\mathit{B}\mathbf{=}{\mathit{B}}_{\mathbf{0}}{\mathit{e}}^{\left(-\frac{t}{\tau }\right)}$, where${\mathit{B}}_{\mathbf{0}}$and$\mathit{\tau }$are constants. Find an expression for the emf in the loop as a function of time.

Short answer

Equation for the emf in the loop as a function of time will be

$\frac{B_0\pi r^2}{\tau }\left(e^{\left(-\frac{t}{\tau }\right)}\right)$

Step-by-step solution

Given

$B=B_0{e}^{\left(-\frac{t}{\tau }\right)}$

Understanding the concept

We use Lenz’s law for induced emf, we rearrange the formula to get the equation for the emf in the loop as a function of time.

Formula:

$\epsilon =-N\frac{d\varphi }{dt}$

$\varphi =BA$

Calculate the Equation for the emf in the loop as a function of time

We have,

$\epsilon =-\frac{d\varphi }{dt}$

We also have,

$\varphi =BA$

Here,

As the cross-section is circular,

$A=\pi r^2$

$\varphi =B\pi r^2$

We also have,

$$\begin{gathered} B=B_0{e}^{\left(-\frac{t}{\tau }\right)} \\ \varphi =\left(B_0{e}^{\left(-\frac{t}{\tau }\right)}\right)\pi r^2 \\ \epsilon =-\frac{d\left(\left(B_0{e}^{\left(-\frac{t}{\tau }\right)}\right)\pi r^2\right)}{dt} \\ \epsilon =B_0\pi r^2{e}^{\left(-\frac{t}{\tau }\right)}\left(-\frac{t}{\tau }\right) \\ \epsilon =\frac{B_0\pi r^{2}e}{\tau }\left(e^{\left(-\frac{t}{\tau }\right)}\right) \end{gathered}$$