Figure 30-72a shows a rectangular conducting loop of resistance $\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{020}\mathbf{\Omega }\mathbf{}\mathbf{,}\mathbf{}\mathbf{height}\mathbf{}\mathbf{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{length}\mathbf{}\mathbf{D}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$, height , and length being pulled at constant speed through two regions of uniform magnetic field. Figure 30-72b gives the current i induced in the loop as a function of the position x of the right side of the loop. The vertical axis scale is set by is${\mathbf{i}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{mA}$. For example, a current equal to is is induced clockwise as the loop enters region 1. What are the (a) magnitude and (b) direction (into or out of the page) of the magnetic field in region 1? What are the (c) magnitude and (d) direction of the magnetic field in region 2?

$$\begin{gathered} R=0.020\Omega \\ {i}_s=3\times {10}^{-6}\mathrm{A} \\ H=1.5\mathrm{cm} \\ D=2.5\mathrm{cm} \\ V=40\mathrm{cm}/\mathrm{s}=0.40\mathrm{m}/\mathrm{s} \end{gathered}$$

By using the Ohm’s law, we can find the current in the loop, which depends on the emf. After that, we can find the total emf due to region 1 and region 2. Finally, we can find the magnitude of the magnetic field in each region. Also, from Lenz’s law, we can decide the direction of the magnetic field in each region.

Formula:

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ {\mathrm{\varphi }}_{\mathrm{B}}=\mathrm{BA} \end{gathered}$$

According to Ohm’s law, we can find the current as

$$\begin{gathered} i=\frac{\left|\epsilon \right|}{R} \\ \mathrm{Where}\mathrm{emf}\mathrm{is}\mathrm{as} \\ \left|\mathrm{\epsilon }\right|=\left|\frac{\mathrm{d\varphi }}{\mathrm{dt}}\right| \\ \mathrm{i}=\frac{1}{\mathrm{R}}\left|\frac{\mathrm{d\varphi }}{\mathrm{dt}}\right|.............................................................\left(1\right) \end{gathered}$$

As from figure, suppose we consider the length of the region 2 is x so that the length of the region 1 is D-x so that the total flux is

$$\begin{gathered} {\mathrm{\varphi }}_{\mathrm{B}}=\mathrm{BA} \\ {\mathrm{\varphi }}_{\mathrm{B}}=\left(\mathrm{D}-\mathrm{x}\right){\mathrm{HB}}_1+{\mathrm{xHB}}_2 \\ {\mathrm{\varphi }}_{\mathrm{B}}=\left({\mathrm{DHB}}_1-{\mathrm{xHB}}_1\right)+{\mathrm{cHB}}_2 \\ {\mathrm{\varphi }}_{\mathrm{B}}={\mathrm{DHB}}_1+\mathrm{xH}\left({\mathrm{B}}_1+{\mathrm{B}}_2\right) \end{gathered}$$

Differentiate this equation with respect to time and put in equation (1)

$$\begin{gathered} \frac{d{\mathrm{\varphi }}_{\mathrm{B}}}{dt}=\frac{dx}{dt}H\left(B_2-B_1\right) \\ \frac{d{\mathrm{\varphi }}_{\mathrm{B}}}{dt}=vH\left(B_2-B_1\right) \end{gathered}$$

Where, the v is the velocity,

Now equation (1) is

$i=\frac{1}{R}vH\left(B_2-B_1\right)$

Now, we have to find the magnitude of the magnetic field in the region 1. From figure 30-70(b) as $B_2=0$

$i=\frac{1}{R}vH\left(B_2-B_1\right)$

So that the current i is

$i=\frac{1}{R}vH{B}_1$

We have to find the magnitude of the magnetic field so that the equation can be written as

$B_1=\frac{iR}{vH}$

By substituting the value from figure,

$$\begin{gathered} B_1=\frac{3\times {10}^{-6}\times 0.020}{0.40\times 0.015} \\ {B}_1=1\times {10}^{-5}\mathrm{T} \end{gathered}$$

According to Lenz’s law, as the area increases in region 1, the induced current flows in the opposite direction. Therefore, we can conclude that the direction of magnetic field is out of the page.

Now, we have to find the magnitude of the magnetic field in the region 1. From figure 30-70(b) as

$$\begin{gathered} 30-70\left(\mathrm{b}\right)\mathrm{as}{\mathrm{B}}_1=1\times {10}^{-5}\mathrm{T} \\ \mathrm{i}=\frac{1}{\mathrm{R}}\mathrm{vH}\left({\mathrm{B}}_2-{\mathrm{B}}_1\right) \end{gathered}$$

So that the current i is

$i\frac{R}{vH}=B_2-B_1$

Where i is from the graph,

$i=-\frac{2}{3}{i}_s$

We have to find the magnitude of the magnetic field so that the equation can be written as,

$B_2=-\frac{2}{3}\frac{iR}{vH}+B_1$

By substituting the value from fig

$$\begin{gathered} B_2=-\frac{2\times {10}^{-6}\times 0.020}{0.40\times 0.015}+10\times {10}^{-6} \\ {B}_2=-6.67\times {10}^{-6}+10\times {10}^{-6} \\ {B}_2=3.3\times {10}^{-6}\mathrm{T} \end{gathered}$$

According to Lenz’s law, as the area decreases in region 2, so the induced current flows in the same direction. Therefore, we can conclude that the direction of the magnetic field is into the page.