Question 152: Electromagnetic Induction

Question

A conducting circular loop is rotated about its diameter at a constant angular speed of $100 \mathrm{rad} / \mathrm{s}$ in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by $30^{\circ}$ from the horizontal position, the induced EMF is 15.4 mV . The radius of the loop is $\_\_\_\_$ mm.
$$ \left(	ext { Take } \pi=\frac{22}{7}ight) $$

Accepted Answer

When a conducting coil rotates in a uniform magnetic field, the magnetic flux passing through the coil changes continuously with time. According to Faraday's Law of Electromagnetic Induction, this changing flux induces an electromotive force (EMF) in the loop.
As the coil rotates with a constant angular velocity $\omega$, the angle $	heta$ changes continuously with time $t$. If we assume the normal of the coil is parallel to the magnetic field at time $t=0$ (so $	heta=0$ ), the angle at any given time t is :
$$ 	heta=\omega t $$
So, the magnetic flux through the conducting loop is,
$$ \Phi_{\mathrm{B}}=\mathrm{B} \mathrm{~A} \cos (	heta)=\mathrm{B} \mathrm{~A} \cos (\omega \mathrm{t}) $$
For a coil consisting of N tightly wound turns, the total magnetic flux linkage is simply N times the flux through a single loop :
$$ \Phi_{	ext {total }}=\mathrm{N} \mathrm{~B} \mathrm{~A} \cos (\omega \mathrm{t}) $$
According to Faraday's Law of Electromagnetic Induction, the induced electromotive force (e) in a closed loop is equal to the negative rate of change of the total magnetic flux with respect to time :
$$ \mathrm{e}=-\frac{\mathrm{d} \Phi_{	ext {total }}}{\mathrm{dt}} $$
$$\Rightarrow $$ $e=-\frac{d}{d t}[N B A \cos (\omega t)]$
$$\Rightarrow $$ $e=-N B A \frac{d}{d t}[\cos (\omega t)]$
$$ e=N B A \omega \sin (\omega t) $$
So, the formula for the instantaneous induced EMF (E) in a rotating coil is:
The number of turns, $\mathrm{N}=1$
Angular speed, $\omega=100 \mathrm{rad} / \mathrm{s}$.
Magnetic field strength, $\mathrm{B}=0.5 \mathrm{~T}$.
Instantaneous induced EMF when angle of rotation $	heta=30^{\circ}$ is $\mathrm{e}=15.4 \mathrm{mV}=15.4 	imes 10^{-3} \mathrm{~V}$.
Let the radius of the circular loop is $r$, then the area (A) of a circular loop is $A=\pi r^2$
$\mathrm{e}=\mathrm{B}\left(\pi \mathrm{r}^2ight) \omega \sin (	heta)$
$$\Rightarrow $$ $\mathrm{r}^2=\frac{\mathrm{e}}{\mathrm{B} \pi \omega \sin (	heta)}$
$$\Rightarrow $$ $\mathrm{r}^2=\frac{15.4 	imes 10^{-3}}{0.5 	imes\left(\frac{22}{7}ight) 	imes 100 	imes \sin \left(30^{\circ}ight)}$
$$\Rightarrow $$ $\mathrm{r}^2=\frac{15.4 	imes 10^{-3} 	imes 7}{550}=1.96 	imes 10^{-4} \mathrm{~m}^2$
$$\Rightarrow $$ $\mathrm{r}=\sqrt{1.96 	imes 10^{-4}}=1.4 	imes 10^{-2} \mathrm{~m}=14 \mathrm{~mm}$

Electromagnetic Induction

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