Figure 30-39 shows a closed loop of wire that consists of a pair of equal semicircles, of radius3.7 cm, lying in mutually perpendicular planes. The loop was formed by folding a flat circular loop along a diameter until the two halves became perpendicular to each other. A uniform magnetic field$\overrightarrow{\mathbf{B}}$of magnitude 76 mTis directed perpendicular to the fold diameter and makes equal angles (of45°) with the planes of the semicircles. The magnetic field is reduced to zero at a uniform rate during a time interval of4.5 ms. During this interval, what are the (a) magnitude and (b) direction (clockwise or counterclockwise when viewed along the direction of $\overrightarrow{\mathbf{B}}$) of the emf induced in the loop?

1. The radius of the semi-circle,$r=3.7\text{cm}$.
2. The magnitude of the magnetic field,$\left|\overrightarrow{B}\right|=76\text{mT}$.
3. The angle made by the field with the plane of the semicircle,$\theta ={45}^0$.
4. The magnetic field is reduced to zero at a uniform rate within time,$\Delta t=4.5\text{ms}$.

The rate of change of magnetic field within a given time gives the induced emf in the coil which is the number of magnetic turns taken by the coil or the amount of magnetic flux entering the given area of the coil. Thus, the induced emf as per Lenz law is in the direction such that it opposes the change in the magnetic field.

Formulae:

The magnetic flux introduced by the magnetic field,${\Phi }_B=BA\cos \theta$ (i)

The emf introduced due to change in magnetic flux, $\epsilon =-\frac{d{\Phi }_B}{dt}$ (ii)

At first, the total flux introduced by the magnetic field due to two equal pairs of semicircles can be given using the data in equation (i) as follows:

$\begin{array}{rcl}{\Phi }_B& =& 2B\left(\pi r^2/2\right)\cos \theta \left(\because \text{Area of the semicircle,}A=\pi r^2/2\right)\\ & =& \pi B{r}^2\cos \left({45}^0\right)\\ & =& \frac{\pi B{r}^2}{\sqrt{2}}\\ & & \end{array}$

Now, the value of the emf induced in the given semicircular loop can be calculated using the above data and the given data in equation (ii) as follows:

$\begin{array}{rcl}\epsilon & =& -\frac{d}{dt}\left(\frac{\pi B{r}^2}{\sqrt{2}}\right)\\ & =& \frac{-\pi r^2}{\sqrt{2}}\left(\frac{\Delta B}{\Delta t}\right)\\ & =& \frac{-\pi {\left(3.7\times {10}^{-2}\text{m}\right)}^2}{\sqrt{2}}\left(\frac{0-76\times {10}^{-3}\text{T}}{4.5\times {10}^{-3}\text{s}}\right)\\ & =& 5.1\times {10}^{-2}\text{V}\\ & & \end{array}$

Hence, the value of the induced emf is $\begin{array}{rcl}& & 5.1\times {10}^{-2}\text{V}\end{array}$.

The direction of the induced current is clockwise when viewed in the direction of using Fleming’s left-hand rule. Thus, the induced emf would be in the counter-clockwise direction to oppose the increasing magnetic field.

Hence, the direction of induced emf is counterclockwise.