Name: Magnetic Flux
Uploaded: 2019-07-19
Duration: 6 min 44 s
Description: Magnetic flux is the total number of magnetic lines of force that are passing through a given surface. To learn more about this intriguing concept, watch the...

Question

A magnetic field in a certain region is given by $B = B_{0} \cos (ω t) \hat{k}$ 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 Figure). Find the magnitude and the direction of the current at $(a, 0, 0)$ at

$t = \frac{π}{2 ω}, t = \frac{π}{ω}$ and $t = \frac{3 π}{2 ω}$

Question Image

$t = \frac{π}{2 ω}, t = \frac{π}{ω}$ and $t = \frac{3 π}{2 ω}$

Accepted Answer

The direction of magnetic field is along $Z$ -axis. So the magnetic flux passing through circular region of coll of radius a placed In $X - Y$ plane.

$Φ = B \cdot A = B A \cos θ$

The direction area vector $(a, 0, 0)$ of coil is along $Z$ -direction and $B$ also in $Z$ -direction of $θ = 0 °$

$Φ = (B_{0} \cos ω t) . (π a^{2}) \cos 0 °$

$Φ = B_{0} π a^{2} \cos ω t$

By Faraday's law of electromagnetic induction.

$ε = \frac{d Φ}{d t} B_{0} π a^{2} \sin ω t \cdot ω$

$ε = B_{0} π a^{2} ω \sin ω t$

Flow of induced current in coil $I = \frac{ε}{R}$ where $R$ is resistance of coil.

$I = \frac{B_{0} π a^{2} ω}{R} \sin ω t$

For current at $t = \frac{π}{2 ω}, t = \frac{π}{ω}, t = \frac{3 π}{2 ω}$

At $t = \frac{π}{2 ω}$ , current wil be $I = \frac{B_{0} π a^{2} ω}{R} \sin ω \cdot \frac{π}{2 ω} = \frac{B_{0} π a^{2} ω}{R} \sin \frac{π}{2}$

Hence, at $t = \frac{π}{2 ω}$ , current is $I = \frac{B_{0} π a^{2} ω}{R} (along + \hat{j})$

For $t = \frac{π}{ω}$ current will be $I = \frac{B_{0} π a^{2} ω}{R} sinπ = \frac{B_{0} π a^{2} ω}{R} \times 0 = 0$

Hence, at $t = \frac{π}{ω}$ , induced current is $I = 0$ .

For $t = \frac{3 π}{2 ω}, current will be I = \frac{B_{0} π a^{2} ω}{R} \sin ω \frac{3 π}{2 ω}$

$\Rightarrow I = \frac{B_{0} π a^{2} ω}{R} \sin \frac{3 π}{2}$

$I = \frac{- B_{0} π a^{2} ω}{R}$

$I = \frac{- B_{0} {πa}^{2} ω}{R}$

Hence, at $t = \frac{3 π}{2 ω}$ , induced current will be $I = \frac{- B_{0} π a^{2} ω}{R} (along \hat{j})$

NCERT Solutions for Chapter: Electromagnetic Induction, Exercise 3: SA

NCERT Physics Solutions for Exercise - NCERT Solutions for Chapter: Electromagnetic Induction, Exercise 3: SA

Questions from NCERT Solutions for Chapter: Electromagnetic Induction, Exercise 3: SA with Hints & Solutions

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