Question 57: Electromagnetic Induction

Question

A conducting circular loop of area $1.0 \mathrm{~m}^2$ is placed perpendicular to a magnetic field which varies as $B=\sin (100 t)$ Tesla. If the resistance of the loop is $100 \Omega$, then the average thermal energy dissipated in the loop in one period is $\_\_\_\_$ J.

Accepted Answer

According to Faraday's Law of Electromagnetic Induction, the induced $\operatorname{emf}(\varepsilon)$ is the negative rate of change of magnetic flux $(\phi)$ through the loop :
$$ \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}} $$
When the field is perpendicular to the plane of the loop then magnetic flux is
$$ \phi=\mathrm{B} \cdot \mathrm{~A} $$
The area of the loop is $\mathrm{A}=1.0 \mathrm{~m}^2$
The magnetic field in the region is $B=\sin (100 t)$
So, the magnetic flux is,
$$ \phi=1.0 \sin (100 t)=\sin (100 t) $$
Now on differentiating the flux with respect to time ( $t$ ) we get the induced emf :
$$ \varepsilon=-\frac{d}{d t}[\sin (100 t)] $$
$$\Rightarrow $$ $$ \varepsilon=-100 \cos (100 t) 	ext { Volts } $$
The instantaneous power $(\mathrm{P})$ dissipated as heat in a resistor is:
$$ \mathrm{P}=\frac{\varepsilon^2}{\mathrm{R}} $$
Given resistance of the loop is $\mathrm{R}=100 \Omega$ :
$$ P=\frac{[-100 \cos (100 t)]^2}{100} $$
$$\Rightarrow $$ $$ \mathrm{P}=\frac{10000 \cos ^2(100 \mathrm{t})}{100}=100 \cos ^2(100 \mathrm{t}) 	ext { Watts } $$
The average thermal energy dissipated in one period is equivalent to finding the average power over one full cycle ( T ) in time T .
The time period $(T)$ for the function $\sin (100 t)$ is,
$$ \mathrm{T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{100}=\frac{\pi}{50} 	ext { seconds } $$
The total energy (E) dissipated in one period is :
$$ \mathrm{E}=\int_0^{\mathrm{T}} \mathrm{Pdt}=\int_0^{\mathrm{T}}\left(100 \cos ^2(100 \mathrm{t})ight) \mathrm{dt} $$
$$\Rightarrow $$ $\mathrm{E}=100 \int_0^{\mathrm{T}} \cos ^2(100 \mathrm{t}) \mathrm{dt}=100 \int_0^{\mathrm{T}} \frac{1+\cos (200 \mathrm{t})}{2} \mathrm{dt}$
$$\Rightarrow $$ $E=50 \int_0^T 1+\cos (200 t) d t$
$$\Rightarrow $$ $E=50\left[t+\frac{\sin (200 t)}{200}ight]_0^{\pi / 50}$
$$\Rightarrow $$ $\mathrm{E}=50\left[\left(\frac{\pi}{50}+\frac{\sin \left(200 	imes \frac{\pi}{50}ight)}{200}ight)-(0+0)ight]$
$$\Rightarrow $$ $E=50\left[\frac{\pi}{50}+\frac{\sin (4 \pi)}{200}ight]$
$$\Rightarrow $$ $\mathrm{E}=50 	imes \frac{\pi}{50}=\pi$ Joules

Answer

$\pi^2$

Answer

$\frac{\pi}{2}$

Answer

$\pi$

Answer

$2 \pi$

Electromagnetic Induction

A conducting circular loop of area 1.0 \mathrm{~m}^2 is placed perpendicular to a magnetic field which varies as B=\sin (100 t) Tesla. If the resistance of the loop is 100 \Omega, then the average thermal energy dissipated in the loop in one period is \_\_\_\_ J.

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