A square wire loop 20 cmon a side, with resistance 20 m$\mathit{\Omega }$ , has its plane normal to a uniform magnetic field of magnitude B = 2.0 T . If you pull two opposite sides of the loop away from each other, the other two sides automatically draw toward each other, reducing the area enclosed by the loop. If the area is reduced to zero in time$\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{20}\mathit{s}$ ,(a)What is the average emf?(b)What is the average current induced in the loop during$\mathbf{∆}\mathit{t}\mathbf{\text{'}}$?

1. The side of a square loop is$l=20\mathrm{cm}=20\times {10}^{-2}\mathrm{m}$
2. The resistance of a wire is$R=20\mathrm{m\Omega }=20\times {10}^{-3}\mathrm{\Omega }$
3. The uniform magnetic field is$B=2.0\mathrm{T}$
4. The time for the reducing area to zero is $∆t=0.20\mathrm{s}$

The average induced emf can be calculated using the equation of emf relating to change in flux over change in time. To find the average current induced we need to use the equation of Ohm’s law.

Formulae:

$$\begin{gathered} {\epsilon }_{avg}=\left|-\frac{d{\varnothing }_B}{dt}\right| \\ ∆{\varnothing }_B=BA \\ {\epsilon }_{avg}=i_{avg}R \end{gathered}$$

The average emf:

The area of the square loop is

$$\begin{gathered} A=l^2 \\ A={\left(20\times {10}^{-2}\mathrm{m}\right)}^2 \\ A=400\times {10}^{-4}{\mathrm{m}}^2 \end{gathered}$$

The expression of magnetic flux is

$$\begin{gathered} ∆{\varnothing }_B=BA \\ ∆{\varnothing }_B=2.0T\times 400\times {10}^{-4}{m}^2 \end{gathered}$$

The average induced emf is

$$\begin{gathered} {\epsilon }_{avg}=\left|-\frac{d{\varnothing }_B}{dt}\right| \\ {\epsilon }_{avg}=\left|-\frac{d{\varnothing }_B}{dt}\right| \\ {\epsilon }_{avg}=\frac{2.0T\times 400\times {10}^{-4}{\mathrm{m}}^2}{0.20\mathrm{s}} \\ {\epsilon }_{avg}=0.40\mathrm{V} \end{gathered}$$

The average current induced in the loop during $∆t$:

According to the Ohm’s law, the average induced current is

$$\begin{gathered} i_{avg}=\frac{{\epsilon }_{avg}}{R} \\ {i}_{avg}=\frac{0.40V}{20\times {10}^{-3}\Omega } \\ {i}_{avg}=20A \end{gathered}$$