Question

A square loop of wire, with sides of length a , lies in the first quadrant of the xy plane, with one comer at the origin. In this region, there is a nonuniform time-dependent magnetic field $\mathbf{B}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{ky}}^{\mathbf{3}}{\mathbf{t}}^{\mathbf{2}}\hat{\mathbf{z}}$ (where k is a constant). Find the emf induced in the loop.

Short answer

The induced emf into loop is $-\frac{kt}{2}{a}^5.$

Step-by-step solution

Write the given data from the question.

The side of the square loop is a.

The square loop exists in xy plane.

The non-uniform time dependent magnetic field,$\mathbf{B}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{ky}}^{\mathbf{3}}{\mathbf{t}}^{\mathbf{2}}\hat{\mathbf{z}}$

Calculate the induced emf in the loop.

Let’s take small strip dy on the square loop which at distance y from the x-axis.

The magnetic flux is given by,

$\varphi =B(y,t).dA$

Here dA is the area of the strip.

Substitute $K{y}^3{t}^2$for $B(y,t)$and ady for dA into above equation.

$\varphi =k{y}^3{t}^2.ady$

Integrate the above equation to calculate the flus through the entire loop.

$$\begin{gathered} \varphi ={\int }_0^{a}k{y}^3{t}^2.ady \\ \varphi =ka{t}^2{\int }_0^a{y}^{3}dy \\ \varphi =ka{t}^2{\left(\frac{y^4}{4}\right)}_0^a \end{gathered}$$

Apply the limits,

$$\begin{gathered} \varphi =ka{t}^2\left(\frac{a^4}{4}-\frac{0^4}{4}\right) \\ \varphi =ka{t}^2\left(\frac{a^4}{4}\right) \\ \varphi =\frac{k{t}^2}{4}{a}^5 \end{gathered}$$

The induced emf in the loop is given by,

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

Substitute $\frac{k{t}^2}{4}{a}^5$for $\varphi$into above equation.

$$\begin{gathered} e=-\frac{d}{dt}\left(\frac{k{t}^2}{4}{a}^5\right) \\ e=-\frac{2kt}{4}{a}^5 \\ e=-\frac{kt}{2}{a}^5 \end{gathered}$$

Hence the induced emf into loop is $-\frac{kt}{2}{a}^5$.