Consider an infinite one-dimensional wire carrying a uniform current $I$ as shown in the figure. A s...

The magnetic field at a distance $ r $ from an infinitely long wire carrying a current $ I $ can be determined using Ampere's Law:$ B = \frac{\mu_0 I}{2 \pi r} $where $ \mu_0 $ is the permeability of free space and $ r $ is the radial distance from the wire.As the square loop moves rightwards at a uniform speed $ v $, after a time $ t $, the center of the loop will be at a new distance $ R' = R + vt $.The change in magnetic flux through the loop is affected by this movement. The magnetic flux $\Phi_B$ can be calculated as:$ \Phi_B = \int B \, dA = \int B \, a \, dR' $where the limits of integration are from $ R' - \frac{a}{2} $ to $ R' + \frac{a}{2} $, covering the width of the square loop.$ \Phi_B = \frac{\mu_0 I a}{2 \pi} \int_{R' - \frac{a}{2}}^{R' + \frac{a}{2}} \frac{dR'}{R'} = \frac{\mu_0 I a}{2 \pi} \left( \ln \left( \frac{R' + \frac{a}{2}}{R' - \frac{a}{2}} \right) \right) $To find the induced electromotive force (emf) as a function of time, we use Faraday's Law of electromagnetic induction, which states:$ \varepsilon = -\frac{d\Phi_B}{dt} $Calculating the derivative, we find:$ \varepsilon = \frac{\mu_0 I a}{2 \pi} \frac{d}{dt} \left(\ln(R + vt + \frac{a}{2}) - \ln(R + vt - \frac{a}{2})\right) $$ \varepsilon = \frac{\mu_0 I a}{2 \pi} \left( \frac{v}{(R + vt + \frac{a}{2})} - \frac{v}{(R + vt - \frac{a}{2})} \right) $Simplifying, the induced emf as a function of time $ t $ becomes:$ \varepsilon = \frac{\mu_0 I a^2 v}{2 \pi \left((R + vt)^2 - \frac{a^2}{4}\right)} $

The magnetic fux $\phi_B(t)$ through a coil at a time $t$ is given by $\phi_B(t)=\phi_0 \cos \omega ...

The magnitude of the induced electromotive force (emf) in a coil can be determined using Faraday's Law of Induction:$ \varepsilon = -\frac{d\left(\phi_B(t)\right)}{d t} $The magnetic flux through the coil is given by:$ \phi_B(t) = \phi_0 \cos(\omega t) $To find the rate of change of the magnetic flux, we differentiate:$ \frac{d \phi_B(t)}{d t} = -\phi_0 \omega \sin(\omega t) $Thus, the induced emf becomes:$ \varepsilon = -\frac{d \phi_B(t)}{d t} = \phi_0 \omega \sin(\omega t) $Next, we determine the maximum value of the magnitude of the induced emf, $|\varepsilon|$:$ |\varepsilon| = \left|\phi_0 \omega \sin(\omega t)\right| $The sine function, $\sin(\omega t)$, achieves its maximum value of $1$ when:$ \omega t = \frac{\pi}{2} $Solving for time $t$, we get:$ t = \frac{\pi}{2 \omega} $Therefore, the time at which the magnitude of the induced emf reaches its maximum is:$ t = \frac{\pi}{2 \omega} $

Electromagnetic Induction - IAT-IISER Physics PYQs

The magnetic field at a distance

r

from an infinitely long wire carrying a current

I

can be determined using Ampere's Law:

B%20%3D%20%5Cfrac%7B%5Cmu_0%20I%7D%7B2%20%5Cpi%20r%7D

where

%5Cmu_0

is the permeability of free space and

r

is the radial distance from the wire.As the square loop moves rightwards at a uniform speed

v

, after a time

t

, the center of the loop will be at a new distance

R'%20%3D%20R%20%2B%20vt

.The change in magnetic flux through the loop is affected by this movement. The magnetic flux

%5CPhi_B

can be calculated as:

%5CPhi_B%20%3D%20%5Cint%20B%20%5C%2C%20dA%20%3D%20%5Cint%20B%20%5C%2C%20a%20%5C%2C%20dR'

where the limits of integration are from

R'%20-%20%5Cfrac%7Ba%7D%7B2%7D

R'%20%2B%20%5Cfrac%7Ba%7D%7B2%7D

, covering the width of the square loop.

%5CPhi_B%20%3D%20%5Cfrac%7B%5Cmu_0%20I%20a%7D%7B2%20%5Cpi%7D%20%5Cint_%7BR'%20-%20%5Cfrac%7Ba%7D%7B2%7D%7D%5E%7BR'%20%2B%20%5Cfrac%7Ba%7D%7B2%7D%7D%20%5Cfrac%7BdR'%7D%7BR'%7D%20%3D%20%5Cfrac%7B%5Cmu_0%20I%20a%7D%7B2%20%5Cpi%7D%20%5Cleft(%20%5Cln%20%5Cleft(%20%5Cfrac%7BR'%20%2B%20%5Cfrac%7Ba%7D%7B2%7D%7D%7BR'%20-%20%5Cfrac%7Ba%7D%7B2%7D%7D%20%5Cright)%20%5Cright)

To find the induced electromotive force (emf) as a function of time, we use Faraday's Law of electromagnetic induction, which states:

%5Cvarepsilon%20%3D%20-%5Cfrac%7Bd%5CPhi_B%7D%7Bdt%7D

Calculating the derivative, we find:

%5Cvarepsilon%20%3D%20%5Cfrac%7B%5Cmu_0%20I%20a%7D%7B2%20%5Cpi%7D%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft(%5Cln(R%20%2B%20vt%20%2B%20%5Cfrac%7Ba%7D%7B2%7D)%20-%20%5Cln(R%20%2B%20vt%20-%20%5Cfrac%7Ba%7D%7B2%7D)%5Cright)

%5Cvarepsilon%20%3D%20%5Cfrac%7B%5Cmu_0%20I%20a%7D%7B2%20%5Cpi%7D%20%5Cleft(%20%5Cfrac%7Bv%7D%7B(R%20%2B%20vt%20%2B%20%5Cfrac%7Ba%7D%7B2%7D)%7D%20-%20%5Cfrac%7Bv%7D%7B(R%20%2B%20vt%20-%20%5Cfrac%7Ba%7D%7B2%7D)%7D%20%5Cright)

Simplifying, the induced emf as a function of time

t

becomes:

%5Cvarepsilon%20%3D%20%5Cfrac%7B%5Cmu_0%20I%20a%5E2%20v%7D%7B2%20%5Cpi%20%5Cleft((R%20%2B%20vt)%5E2%20-%20%5Cfrac%7Ba%5E2%7D%7B4%7D%5Cright)%7D

The magnitude of the induced electromotive force (emf) in a coil can be determined using Faraday's Law of Induction:

%5Cvarepsilon%20%3D%20-%5Cfrac%7Bd%5Cleft(%5Cphi_B(t)%5Cright)%7D%7Bd%20t%7D

The magnetic flux through the coil is given by:

%5Cphi_B(t)%20%3D%20%5Cphi_0%20%5Ccos(%5Comega%20t)

To find the rate of change of the magnetic flux, we differentiate:

%5Cfrac%7Bd%20%5Cphi_B(t)%7D%7Bd%20t%7D%20%3D%20-%5Cphi_0%20%5Comega%20%5Csin(%5Comega%20t)

Thus, the induced emf becomes:

%5Cvarepsilon%20%3D%20-%5Cfrac%7Bd%20%5Cphi_B(t)%7D%7Bd%20t%7D%20%3D%20%5Cphi_0%20%5Comega%20%5Csin(%5Comega%20t)

Next, we determine the maximum value of the magnitude of the induced emf,

%7C%5Cvarepsilon%7C

%7C%5Cvarepsilon%7C%20%3D%20%5Cleft%7C%5Cphi_0%20%5Comega%20%5Csin(%5Comega%20t)%5Cright%7C

The sine function,

%5Csin(%5Comega%20t)

, achieves its maximum value of

1

when:

%5Comega%20t%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D

Solving for time

t

, we get:

t%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%20%5Comega%7D

Therefore, the time at which the magnitude of the induced emf reaches its maximum is:

t%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%20%5Comega%7D

Electromagnetic Induction

High-Yield Trend

Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Electromagnetic Induction

High-Yield Trend

Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Questions
3 MCQs