In Figure, a wire loop of lengths L = 40.0 cmand W = 25.0 cmlies in a magnetic field $\overrightarrow{\mathbf{B}}$.(a)What is the magnitude $\epsilon$if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}}\mathbf{)}\mathbf{y}\hat{\mathbf{k}}$?(b)What is the direction (clockwise or counterclockwise—or “none” if 0) of the emf induced in the loop if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}}\mathbf{)}\mathbf{y}\hat{\mathbf{k}}$?(c)What is the$\mathbf{\epsilon }$if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{s}}\mathbf{)}\mathbf{t}\hat{\mathbf{k}}$(d)what is the direction if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{s}}\mathbf{)}\mathbf{t}\hat{\mathbf{k}}$(e)What is the$\mathbf{\epsilon }$if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{)}\mathit{y}\mathbf{t}\hat{\mathbf{k}}$(f)What is the direction if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{s}}\mathbf{)}\mathbf{t}\hat{\mathbf{k}}$(g)What is the$\mathbf{\epsilon }$if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{)}\mathit{x}\mathbf{t}\hat{\mathbf{k}}$(h)What is the direction if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{)}\mathit{x}\mathbf{t}\hat{\mathbf{k}}$(i)What is the if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{)}\mathit{y}\mathbf{t}\hat{\mathbf{k}}$(j)What is the direction if $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\frac{\mathbf{T}}{\mathbf{m}\mathbf{.}\mathbf{s}}\mathbf{)}\mathit{y}\mathbf{t}\hat{\mathbf{k}}$

1. Length of loop, L = 40 cm = 0.4 m
2. Width of the loop, w = 25.0 cm=0.25 m

By using the concept of the magnetic flux, Faraday’s law and Lenz’s law, find the magnitude and the direction of the induced emf in all the cases.

Faraday's law 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 which opposes the motion.

Formulae are as follow:

1. Faraday's law$\epsilon =\frac{d\phi }{dt}$
2. Magnetic flux$\phi =\varphi BdA$

Where,𝜀 is emf, dt is time, $\mathrm{\Phi }$is magnetic flux, B is magnetic field, A is area.

The magnetic flux passing through the coil is given by,

$$\begin{gathered} \phi =\varphi B.dA=\varphi BdA\cos \theta \\ \phi =BA\cos \theta \end{gathered}$$

The emf induced in coil is given by,

$$\begin{gathered} \epsilon =-\frac{d\phi }{dt} \\ \epsilon =-\frac{d}{dt}\left[BA\cos \theta \right].........................................................\left(1\right) \\ \epsilon =BA\sin \theta \end{gathered}$$

Now, for the rectangular coil,

$$\begin{gathered} \mathrm{A}=0.4\times 0.25 \\ \mathrm{A}=0.1{\mathrm{m}}^2 \\ \therefore \mathrm{\epsilon }=0.1\mathrm{B}\mathrm{sin\theta }....................................................................\left(2\right) \end{gathered}$$

For $B=4\times {10}^{-2}T/m)y\hat{k}$

As the unit vector $\hat{k}$, it is along z axis and it is perpendicular to the coil. But it is not changing with time. Hence, from equation 1,

$$\begin{gathered} \epsilon =-\frac{d}{dt}\left[BA\cos \theta \right] \\ \epsilon =0 \end{gathered}$$

Hence, the emf induced in the coil is zero.

From a),

Hence, the direction of the emf is none.

For$\mathrm{B}=6\times {10}^{-2}\mathrm{T}/\mathrm{s})\mathrm{t}\hat{\mathrm{k}}$

It is perpendicular to the coil and changes with time. From equation 2,

$$\begin{gathered} \mathrm{\epsilon }=0.1\mathrm{Bsin\theta } \\ \mathrm{\epsilon }=0.1\times 6\times {10}^{-2}\sin 90 \\ \mathrm{\epsilon }=0.1\times 6\times {10}^{-2}\times 1 \\ \mathrm{\epsilon }=0.1\times 6\times {10}^{-2}\mathrm{V} \\ \mathrm{\epsilon }=6\times {10}^{-3}\mathrm{V} \\ \mathrm{\epsilon }=6\mathrm{mV} \end{gathered}$$

Hence, emf induced in the loop will be 6 mV.

Since, the magnetic field is directed into the page,

Hence, the direction of the induced emf is clockwise.

For$\mathrm{B}=8\times {10}^{-2}\mathrm{T}/\mathrm{m}.\mathrm{s})\mathrm{yt}\hat{\mathrm{k}}$,

which is in the direction of y axis, that is, along the width of the coil.

$$\begin{gathered} \phi =\varphi B.dA \\ \phi =\varphi 8\times {10}^{-2}yt\times d\left(l\times w\right) \end{gathered}$$

w is along x axis.

$$\begin{gathered} \phi =l\times 8\times {10}^{-2}t{\int }_0^{W}ydy \\ \phi =0.4\times 8\times {10}^{-2}t\times \frac{w^2}{2} \end{gathered}$$

$$\begin{gathered} \phi =0.4\times 8\times {10}^{-2}\times t\times \frac{{\left(0.25\right)}^2}{2} \\ \phi =0.1t\times {10}^{-2} \end{gathered}$$

The emf induced can be calculated as,

$$\begin{gathered} \mathrm{\epsilon }=\frac{\mathrm{d\phi }}{\mathrm{dt}} \\ \mathrm{\epsilon }=\frac{\mathrm{d}}{\mathrm{dt}}\left[0.1\mathrm{t}\times {10}^{-2}\right] \\ \mathrm{\epsilon }=1\times {10}^{-3}\mathrm{V} \\ \mathrm{\epsilon }=1\mathrm{mV} \end{gathered}$$

Hence, the emf induced in the coil is 1 mV.

From e),

Hence, the direction of the induced emf will be clockwise.

For$\mathrm{B}=3\times {10}^{-2}\mathrm{T}/\mathrm{m}.\mathrm{s})\mathrm{xt}\hat{\mathrm{j}}$ , the magnetic field is directed along the y axis i.e. parallel to the coil.

The flux through the coil i.e.$\mathrm{\phi }=0$

$\epsilon =0$

Hence, the emf induced in the coil is zero.

From g),

Hence, the direction of emf will be none.

For$\mathrm{B}=5\times {10}^{-2}\mathrm{T}/\mathrm{m}.\mathrm{s})\mathrm{yt}\hat{\mathrm{i}}$ , the magnetic field is directed along the x axis i.e. parallel to the coil.

The flux through the coil i.e.$\mathrm{\phi }=0$

$\epsilon =0$

Hence, the emf induced in the coil is zero.

From i),

Hence, the direction of emf will be none.

Therefore, by using the concept of the magnetic flux, Faraday’s law and Lenz’s law, find the magnitude and the direction of the induced emf in all the cases.