Question 136: Electromagnetic Induction

Question

A circular loop of radius 20 cm and resistance $2 \Omega$ is placed in a time varying magnetic field $\vec{B}=\left(2 t^2+2 t+3\right) T$. At $t=0$, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at $t=3 \mathrm{~s}$ is $\frac{\alpha}{50} \mathrm{~A}$. The value of $\alpha$ is $\_\_\_\_$ . (Take $\pi=22 / 7$ )

Accepted Answer

Magnetic flux is given by $\Phi=\vec{B} \cdot \vec{A}$, where $\vec{B}$ is the magnetic field vector in the region and $\vec{A}$ is the area vector of the closed loop.
Since the plane of the loop is perpendicular to the field, the angle between the area vector and the field is $0^{\circ}$.
$$ \Phi=\mathrm{B} \cdot \mathrm{~A}=\mathrm{B} \cdot\left(\pi \mathrm{r}^2ight) $$
The radius of circular loop is $\mathrm{r}=20 \mathrm{~cm}=0.2 \mathrm{~m}$
So, the area of loop is $\mathrm{A}=\pi(0.2)^2=0.04 \pi \mathrm{~m}^2$
The time varying magnetic field in the region is,
$$ B(t)=2 t^{\wedge} 2+2 t+3 $$
So, the magnetic flux is,
$$ \Phi(t)=\left(2 t^2+2 t+3ight)(0.04 \pi) $$
According to Faraday's Law, the magnitude of induced EMF is the rate of change of flux:
$$ \epsilon=\frac{d \Phi}{d t}=\frac{d}{d t}\left[0.04 \pi\left(2 t^2+2 t+3ight)ight] $$
$$ \epsilon=0.04 \pi(4 t+2) $$
Att $=3 \mathrm{~s}$ :
$$ \epsilon=0.04 \pi(4(3)+2)=0.04 \pi(14)=0.56 \pi V $$
Using Ohm's law $\left(\mathrm{I}=\frac{\epsilon}{\mathrm{R}}ight)$ with resistance $\mathrm{R}=2 \Omega$ :
$$ I=\frac{0.56 \pi}{2}=0.28 \pi \mathrm{~A} $$
Substituting $\pi=22 / 7$ :
$$ I=0.28 	imes \frac{22}{7}=0.04 	imes 22=0.88 \mathrm{~A}=\frac{\alpha}{50} \mathrm{~A}: $$
$$\Rightarrow $$ $\frac{\alpha}{50}=0.88$
$$\Rightarrow $$ $$ \alpha=0.88 	imes 50=44 $$
Therefore, the value of $\alpha$ is 44 .

Electromagnetic Induction

Integer / Numeric Answer Type