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Electromagnetic Induction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
Question
A magnetic field in a certain region is given by $\text{B}=\text{B}_0\cos(\omega\text{t})\hat{\text{k}}$ and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (see Fig). Find the magnitude and the direction of the current at (a, 0, 0) at $\text{t}=\frac{\pi}{2\omega},\text{t}=\frac{\pi}{\omega} \text{ and }\text{t}=\frac{3\pi}{2\omega}$.
✓

Answer

Key concept:
First law: When ever the number of magnetic lines of force(magnetic flux) passing through a circuit changes, an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
Second law: The induced emf is given by the rate of change of magnetic flux linked with the circuit, i.e., $\text{e}=-\frac{\text{d}\phi}{\text{dt}}$. For n Turns $\text{e}=-\frac{\text{N d}\phi}{\text{dt}}$; Negative sign indicates that induced emf (e) apposes the change of flux.
First we need to find out the flux passing through the rign at any instant and that is given by
$\phi_\text{m}-\vec{\text{B}}.\vec{\text{A}}=\text{BA}\cos\theta$
And as we know both $\vec{\text{A}}$ (area vector) and $\vec{\text{B}}$ (magnetic field vector) are directed along z-axis. So, angle between them is 0.
So, $\cos\theta=1\ (\because\ \theta=0)$
$\Rightarrow\ \phi_\text{m}=\text{BA}$
Area of coil of radius $\text{a}=\pi\text{a}^2$
$\in=\text{B}_0(\pi\text{a}^2)\cos\omega\text{t}$
By Faraday's law of eletromagnetic induction,
Magnitude of induced emf is given by
$\in=\text{B}_0(\pi\text{a}^2)\omega\sin\omega\text{t}$
This causes flow of induced emf is given by
$\text{I}=\frac{\text{B}_0(\pi\text{a}^2)\omega\sin\omega\text{t}}{\text{R}}$
Now, the value of current at different instants,
  1. $\text{t}=\frac{\pi}{2\omega}$
$\text{I}=\frac{\text{B}_0(\pi\text{a}^2)\omega}{\text{R}}\text{ along}\hat{\text{j}}$

Because $\sin\omega\text{t}=\sin\Big(\omega\frac{\pi}{2\omega}\Big)=\sin\frac{\pi}{2}=1$
  1. $\text{t}=\frac{\pi}{\omega},\text{I}=\frac{\text{B}_0(\pi\text{a}^2)\omega}{\text{R}}=0$
$\sin\omega\text{t}=\sin\Big(\omega\frac{\pi}{\omega}\Big)=\sin{\pi}=0$
  1. $\text{t}=\frac{3}{2}\frac{\pi}{\omega}$
$\text{I}=\frac{\text{B}(\pi\text{a}^2)\omega}{\text{R}}\text{along}-\hat{\text{j}}$

$\sin\omega\text{t}=\sin\Big(\omega.\frac{3\pi}{2\omega}\Big)=\sin\frac{3\pi}{2}=-1$

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