Question

A coil of mean area $500c{{m}^{2}}$ and having $1000$ turns is held perpendicular to a uniform field of $0.4gauss$. The coil is turned through $180{}^\circ$ in $\dfrac{1}{10}s$. Find the average induced electromotive force.
$\begin{aligned} & A.0 \\\ & B.30mV \\\ & C.40mV \\\ & D.50mV \\\ \end{aligned}$

Answer 1

First of all, we have to find out the change in magnetic flux in the coil. For that the initial flux and final flux have to be calculated. After that the change in flux is found out.
Using this change in flux value, the induced electromotive force has been calculated.
These all may help you to solve this question.

Complete answer:

It is already mentioned in the question that,
Area of cross section of the coil,
$A=500c{{m}^{-2}}$
Number of turns in the coil is given as,
$N=1000$
The magnetic field over the region is given as per the equation,
$B=0.4G=0.4\times {{10}^{-4}}T$
Time taken by the coil in order to make turning is given as,
$\Delta t=\dfrac{1}{10}s=0.1s$
And also the angle at which coil has been made rotation is,
From,
$\theta =0{}^\circ$
To,
$\theta =180{}^\circ$
Using all these parameters, we can find out the change in magnetic flux
That is, change in magnetic flux is given by the equation,
$\phi =NBA\cos \theta$
Substituting the values in the equation at the initial state is given as,

& {{\phi }_{i}}=NBA\cos 0 \\\ & {{\phi }_{i}}=1000\times 0.4\times {{10}^{-4}}\times 0.05\cos 0 \\\ & =0.002Wb \\\ \end{aligned}$$ The final flux will be, $$\begin{aligned} & {{\phi }_{i}}=NBA\cos 180 \\\ & {{\phi }_{i}}=1000\times 0.4\times {{10}^{-4}}\times 0.05\cos 180 \\\ & =-0.002Wb \\\ \end{aligned}$$ Therefore the change in flux will be given as, $$\begin{aligned} & \Delta \phi ={{\phi }_{i}}-{{\phi }_{f}} \\\ & =0.002-\left( -0.002 \right)=0.004Wb \\\ \end{aligned}$$ Now let us find out the average induced electromotive force, which is given by the equation, ${{E}_{av}}=\dfrac{2NBA}{\Delta t}$ $${{E}_{avg}}=\dfrac{\Delta \phi }{\Delta t}$$ Substituting these values in the given equation will give, $$\begin{aligned} & {{E}_{avg}}=\dfrac{\Delta \phi }{\Delta t} \\\ & =\dfrac{0.004}{0.1}=40mV \\\ & {{E}_{avg}}=40mV \\\ \end{aligned}$$ Therefore the value of average emf. **Hence the correct answer is option C.** **Note:** There is some difference between the magnetic field and the magnetic flux. The magnetic field is the region around the magnet where the moving charge is experiencing a magnetic force. But the magnetic flux is defined as the quantity or the strength of magnetic lines developed by the magnet.