The magnetic field B S at all points within the coloured circle shown in Fig. E29.15 has an initial magnitude of 0.750 T. (The circle could represent approximately the space inside a long, thin solenoid.) The magnetic field is directed into the plane of the diagram and is decreasing at the rate of -0.0350 T/s. (a) what is the shape of the field lines of the induced electric field shown in Fig. E29.15, within the coloured circle? (b) What are the magnitude and direction of this field at any point on the circular conducting ring with radius 0.100 m? (c) What is the current in the ring if its resistance is 4.00 Ω? (d) What is the emf between points a and b on the ring? (e) If the ring is cut at some point and the ends are separated slightly, what will be the emf between the ends?

We know that the induced electric field is in the direction of the induced current. The direction of the magnetic field into the page and is decreasing so the magnetic flux is decreasing too. Therefore, according Lorenz’s law, the direction of the induced magnetic field must be in the same direction of the original magnetic field. Using the right-hand thumb rule, we get the direction of the induced current that is in clockwise direction. The electric field since is tangent to the right hence its direction of clockwise.

The induced electric field at is given by Faraday’s law

Now using Faraday’s law we get:

$$\begin{gathered} \epsilon =\oint \overrightarrow{E}·\overrightarrow{dl} \\ =-\frac{d\varphi B}{dt} \end{gathered}$$

Now, the magnetic flux through the circle of radius :

$$\begin{gathered} \varphi =AB \\ =\pi r^{2}B \end{gathered}$$

Now integrating LHS we get:

$$\begin{gathered} \oint \overrightarrow{E}·\overrightarrow{dl}\backslash \mathrm{hfill} \\ =E\left(2\pi r\right)\backslash \mathrm{hfill} \end{gathered}$$

Absolute value of is:

$E\left(2\pi r\right)=\pi r^2\left|\frac{dB}{dt}\right|$

Or it can also be written as:

$E=\frac{1}{2}r\left|\frac{dB}{dt}\right|$

Putting the values, we get:

$$\begin{gathered} E=\frac{1}{2}(0.100m)(0.0350) \\ =1.75\times {{10}^{-}}^{3}V/s \end{gathered}$$

Therefore, the magnitude of the electric field is$1.75\times {{10}^{-}}^{3}V/s$

$$\begin{gathered} \epsilon =\oint \overrightarrow{E}·\overrightarrow{dl} \\ =E\left(2\pi \right) \\ =(1.75\times {{10}^{-}}^{3}V/m)\left(2\pi \right)(0.100m) \\ =1.100\times {{10}^{-}}^{3}V \end{gathered}$$

Therefore, the induced current is$1.100\times {{10}^{-}}^{3}V$

We are given $R=4.00\Omega$, then the induced current is:

$$\begin{gathered} I=\frac{\epsilon }{R} \\ =\frac{1.100\times {{10}^{-}}^{3}V}{4.00\Omega } \\ =2.75\times {{10}^{-}}^{4}A \end{gathered}$$

Therefore, the current is$2.75\times {{10}^{-}}^{4}A$$2.75\times {{10}^{-}}^{4}A$

To calculate the emf at point a and b

$$\begin{gathered} \epsilon =E\left(\pi r\right) \\ =(1.75\times {{10}^{-}}^{3}V/m)\left(\pi \right)(0.100m) \\ =5.50\times {{10}^{-}}^{4}V \end{gathered}$$

For emf between the end when the ring is separated

Even if the ring is cut and the ends are slightly separated then the induced emf will still be the same i,e.$5.50\times {{10}^{-}}^{4}V$