As seen in Figure, a square loop of wire has sides of length 2.0 cm. A magnetic field is directed out of the page; its magnitude is given by $\mathit{B}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}{\mathit{t}}^{\mathbf{2}}\mathit{y}$, where B is in Tesla, t is in seconds, and y is in meters. At t = 2.5 s, (a) what is the magnitude of the emf induced in the loop? (b) what is the direction of the emf induced in the loop?

1. Length of each side l = 2 cm
2. Magnetic field $B=4t^{2}y$
3. Time t = 2.5 s

Substitute the given value of magnetic field in the formula for flux and integrate it over the given limit. Then substitute this value in Faraday’s law to find the emf induced in the loop. Using Lenz’s law, find the direction of the emf induced in the loop.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Lenz's law states that the current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force that opposes the motion.

_{Formulae are as follows:}

$\varphi =\int \overrightarrow{\mathrm{B}}.d\overrightarrow{A}$

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

Where,$\varphi$is magnetic flux, B is magnetic field, A is area,𝜀 is emf.

Magnitude of emf induced in the loop :

The magnetic flux is given as,

$\varphi =\int \mathrm{Bd}A$

Consider the small strip of height dy and width $l=2cm=0.200m$

The area of the small strip is then, dA = ldy.

The magnetic field is given as,

$B=4t^{2}y$

The limits of y are from$0\to l$.

Thus, the magnetic flux is,

$\varphi ={\int }_0^{l}4t^{2}yldy$

$\varphi =4t^{2}l{\left[\frac{Y^2}{2}\right]}_0^l=4t^{2}l\left(\frac{l^2}{2}\right)=2t^2{l}^3$

Now, by using faradays law,

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

So the magnitude of the emf induced is given by,

$$\begin{gathered} \left|\epsilon \right|=\left|\frac{d\varphi }{dt}\right|=\frac{d}{dt}\left(2t^2{l}^3\right) \\ \left|\epsilon \right|=4t{l}^3 \\ t=2.5s,l=2cm=0.020m \end{gathered}$$

Thus,

$$\begin{gathered} \left|\epsilon \right|=4x(2.5s)\times {\left(0.020m\right)}^3 \\ \left|\epsilon \right|=8\times {10}^{-5}V \end{gathered}$$

Hence, the magnitude of the emf induced in the loop is $8\times {10}^{-5}V$.

Direction of induced emf in loop :

Find the direction of the induced emf by using Lenz’s law.

From the bottom to top, the flux into the loop increases and its direction is out of the page. To oppose this magnetic flux, the current in the loop must be clockwise, and thus, the direction of the induced emf in the loop must be clockwise. Induced emf in the loop produces a magnetic field into the page and opposite to the original field.

Hence, the direction of the emf induced in the loop is clockwise.

Therefore, find the magnitude of the induced emf by using the magnetic flux formula and Faraday’s law. By using Lenz’s law, find the direction of the induced emf.