Question

A rectangular coil has 60 turns and its length and width are $20cm$ and $10cm$ respectively. The coil rotates at a speed of 1800 rotations per minute in a uniform magnetic field of $0.5T$ about one of its diameter. Calculate the maximum induced e.m.f.

Answer 1

Hint: First mention the equation $\phi = \left( {AB\cos (wt)} \right)N$ , then substitute this value of $\phi $ in the equation $\phi = \left( {AB\cos \omega t} \right)N$ to get the following equation $e = \dfrac{{ - d\left( {NAB\cos \left( {\omega t} \right)} \right)}}{{dt}}$ . By solving this equation you get this equation $e = NAB\omega \sin \omega t$ . Since $\omega = 2\pi f$ the equation becomes $e = 2\pi fNAB\sin \omega t$ . Now for maximum induced e.m.f. the value of $\sin \omega t$ should be $1$ . Then the equation becomes ${e_{\max }} = 2\pi fNAB\sin \omega t$ . Substitute all the values given in the problem and calculate the result to reach the solution.

Complete answer:
We know that the total magnetic flux linked with a rectangular coil is
$\phi = \left( {AB\cos \theta } \right)N$
$\phi = $ The magnetic flux that is linked with the coil
$A$ and $B$ are the dimensions of the loops in the rectangular coil
$\theta = $ The angle that the normal to the plane of the coil makes with the magnetic field
$N = $ The number of loops in the rectangular coil
$\because \theta = \omega t$ ( $\omega = $ Angular speed)
$\therefore \phi = \left( {AB\cos \omega t} \right)N$ (Equation 1)
We also know that
$e = \dfrac{{ - d\phi }}{{dt}}$
Here, $e = $ The e.m.f. induced in the rectangular coil
$\dfrac{{ - d\phi }}{{dt}} = $ The rate of change in magnetic flux that links with the rectangular coil.
$\because \phi = \left( {AB\cos \omega t} \right)N$
$\therefore e = \dfrac{{ - d\left( {NAB\cos \theta } \right)}}{{dt}}$
$ \Rightarrow e = - AB\omega \left( { - \sin \omega t} \right)N$
$ \Rightarrow e = NAB\omega \sin \omega t$
$\because \omega = 2\pi f$ ( $f = $ Frequency)
$\therefore e = 2\pi fNAB\sin \omega t$
In this problem, we are required to calculate the maximum induced e.m.f. For the induced e.m.f. to be maximum the value of $\sin \omega t$ should be maximum and we know that the maximum value of $\sin \omega t$ can only be $1$ .
So, ${e_{\max }} = 2\pi fNAB$
In the problem, we are given
$\pi = 3.14$
$A = 20cm = 0.2m$
$B = 10cm = 0.1m$
$f = 1800rpm = 30$ rotations per second
$N = 60$ turns
So, ${e_{\max }} = 2 \times 3.14 \times 60 \times \left( {0.2 \times 0.1} \right) \times 0.5 \times 30$
${e_{\max }} = 113.04volt$
Hence, the maximum induced e.m.f that can be induced in this particular circuit is $113.04volt$.

Note: In the solution, we used the concept of induced e.m.f. (i.e. an e.m.f. is induced in a circuit when the magnetic flux passing through a circuit changes), this concept is used in AC generators that are used in most power hydro and water power plants to produce green energy by using renewable resources like air and wind.