Electromagnetic Induction

$\phi=\left(5 t^2+4 t+2\right) \times 10^{-3}$ weber

Induced emf, $e=\frac{d \phi}{d t}$

$ \begin{aligned} & =\frac{d}{d t}\left(5 t^2+4 t+2\right) \times 10^{-3} \\ & =(10 t+4) \times 10^{-3} \mathrm{volt} \\ \text { At } t & =6 \mathrm{~s}, e=(10 \times 6+4) \times 10^{-3} \\ & =64 \times 10^{-3} \mathrm{volt} \\ & =6.4 \times 10^{-2} \mathrm{volt} \end{aligned} $

∴ Induced current

$ \begin{aligned} I & =\frac{e}{R}=\frac{6.4 \times 10^{-2}}{16} \\ & =0.4 \times 10^{-2} \mathrm{~A}=4 \times 10^{-3} \mathrm{~A}=4 \mathrm{~mA} \end{aligned} $

The small energy losses in transformers due to eddy currents can be reduced by laminating the core. The thin laminating core effectively increases the resistance to the flow of eddy currents within the core.

We know that the angle $\theta$ between the area of the coil and the direction of the magnetic field is $90^\circ$ when the coil is parallel to the field.

The formula for magnetic flux is $\phi = NBA\cos\theta$, where $N$ is the number of turns, $B$ is the magnetic field, $A$ is the area, and $\theta$ is the angle.

Substitute $\theta = 90^\circ$ into the formula. We get:

$\phi = N B A \cos 90^\circ = 0$

This is because $\cos 90^\circ = 0$.

$ \text { } \begin{aligned} |e| & =L \frac{d I}{d t} \\ 2.8 \times 10^{-3} & =L \frac{(8-3)}{0.2} \\ \Rightarrow \quad L & =\frac{2.8}{25} \times 10^{-3}=0.112 \times 10^{-3} \mathrm{H} \\ & =112 \times 10^{-6} \mathrm{H}=112 \mu \mathrm{H} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \text { Induced emf, } e=-N \frac{\Delta \phi}{\Delta t} \\ & =-\frac{900\left(\phi_2-\phi\right)}{\Delta t} \\ & =\frac{-900\left(B A \cos 180^{\circ}-B A \cos 0^{\circ}\right)}{\Delta t} \\ & =-900\left(\frac{-2 B A}{\Delta t}\right)=900 \times \frac{2 B A}{\Delta t} \\ & =\frac{900 \times 2 \times 3.5 \times 10^{-5} \times 3 \times 10^{-2}}{0.5} \\ & =3.78 \times 10^{-3} \mathrm{~V} \\ & \therefore \quad F=\frac{e}{R}=\frac{3.78 \times 10^{-3}}{1.8} \\ & =2.1 \times 10^{-3} \mathrm{~A}=2.1 \mathrm{~mA} \end{aligned} \end{aligned} $

Induced emf

$ \begin{aligned} E & =-\frac{N \Delta \phi}{\Delta t}=\frac{-N\left(\phi_2-\phi_1\right)}{\Delta t} \\ & =\frac{-N\left(B A \cos \theta_2-B A \cos \theta_1\right)}{\Delta t} \\ & =-250 \frac{\left(B A \cos 90^{\circ}-B A \cos 60^{\circ}\right)}{100 \times 10^{-3}} \\ & =\frac{-250\left(-\frac{B A}{2}\right)}{100 \times 10^{-3}} \\ & =2500 \times \frac{B A}{2}=1250 \times B A \\ & =1250 \times 2 \times 120 \times 10^{-4}=30 \mathrm{volt} \end{aligned} $

$\therefore$ Induced current

$ I=\frac{E}{R}=\frac{30}{8}=3.75 \mathrm{~A} $

$\phi=2 t^2+3 t-2$

$ \begin{aligned} & \therefore e=\frac{d \phi}{d t}=\frac{d}{d t}\left(2 t^2+3 t-2\right)=4 t+3 \\ & \text { at } t=3 \mathrm{~s}, e=4 \times 3+3 \\ & \quad e=15 \mathrm{volt} \end{aligned} $

Induced power, $P=\frac{e^2}{R}=\frac{15^2}{7.5}=30 \mathrm{~W}$

To find the energy stored in a coil with an inductance of 80 mH carrying a current of 2.5 A, we use the formula for the energy stored in an inductive coil:

$ U = \frac{1}{2} L i^2 $

Given:

Inductance, $ L = 80 \, \text{mH} = 80 \times 10^{-3} \, \text{H} $

Current, $ i = 2.5 \, \text{A} $

We substitute these values into the formula:

$ \begin{align*} U & = \frac{1}{2} \times 80 \times 10^{-3} \times (2.5)^2 \\ & = \frac{1}{2} \times 80 \times 10^{-3} \times 6.25 \\ & = 40 \times 10^{-3} \times 6.25 \\ & = 25 \times 10^{-2} \, \text{J} \\ U & = 0.25 \, \text{J} \end{align*} $

Therefore, the energy stored in the coil is $ 0.25 \, \text{J} $.

The mutual inductance between two coils is given as 8 mH, which can be converted to henries as:

$ M = 8 \, \text{mH} = 8 \times 10^{-3} \, \text{H} $

The current in one of the coils changes over time according to the equation:

$ I = 12 \sin(100t) $

where $ I $ is in amperes and $ t $ is in seconds.

To find the maximum emf induced in the second coil, we use Faraday's law of mutual induction, which states:

$ e = M \frac{dI}{dt} $

Calculating the derivative of the current with respect to time, we have:

$ \frac{dI}{dt} = \frac{d}{dt}[12 \sin(100t)] = 12 \times 100 \cos(100t) $

Substituting this into the formula for emf, we get:

$ \begin{align*} e &= M \cdot \frac{dI}{dt} \\\\ &= 8 \times 10^{-3} \times 12 \times 100 \cos(100t) \\\\ &= 9.6 \cos(100t) \end{align*} $

The expression $ e = 9.6 \cos(100t) $ is in the form of the standard equation for emf:

$ e = e_0 \cos(\omega t) $

From this, we can identify the maximum emf, $ e_0 $, as:

$ e_0 = 9.6 \, \text{V} $

Given,

Area of coil, $ A=200 \mathrm{~cm}^{2} $

Number of turn, $ N=50 $

Angular speed, $ \omega=40 \mathrm{rad} / \mathrm{s} $

Magnetic field, $ B=2 \times 10^{-2} \mathrm{~T} $

Maximum emf induced is given as $ \mathrm{emf}=N A B \omega \sin \omega t $

For maximum value $ \sin \omega t=1 $

$ \begin{aligned} (\mathrm{emf})_{\max } & =N A B \omega \\\\ & =50 \times 200 \times 10^{-4} \times 2 \times 10^{-2} \times 40 \\\\ & =0.8 \mathrm{~V} \end{aligned} $

Given, $l=1.66 \mathrm{~m}$

$ \begin{aligned} & \text { Speed }=90 \mathrm{kmph}=25 \mathrm{~m} / \mathrm{s} \\\\ & B_v=0.2 \times 10^{-4} \mathrm{~T} \\\\ & \text { Induced emf }=\text { Blv } \\\\ &=0.2 \times 10^{-4} \times 1.66 \times 25 \\\\ &=8.3 \times 10^{-4} \mathrm{~V} \\\\ &=0.83 \mathrm{mV} \end{aligned} $

Here, coil 1 has a radius of $r_1$ and coil 2 has a radius of $r_2$.

Given, $r_1 \ll r_2$

Let current $I_2$ flow in the coil 2 and magnetic field at its centre

$ B=\frac{\mu_0 I_2}{2 r_2} $

Since, $r_1 \ll r_2$

Flux through the coil, $\phi=\frac{\mu_0 I_2}{2 r_2} \times \pi \pi_1^2$

By definition of mutual inductance,

$ \phi=M_{12} I_2=\frac{\mu_0 I_2}{2 r_2} \pi r_1^2 \Rightarrow M_{12}=\frac{\mu_0}{2 r_2} \pi r_1^2 $

Hence, mutual inductance of the arrangement is $\frac{\mu_0}{2 r_2} \pi r_1^2$.

Given,

Velocity of conducting rod $=v$

Magnetic field $=B$

Induced current $=I$

For induced current direction, we use Fleming's right hand rule.

It states that if thumb, forefinger and middle finger of right hand are mutually perpendicular to each other, then if fore finger indicate the direction of external magnetic field ( $B$ ) and thumb shows direction of motion of conductor, then middle finger will show the direction of induced current and hence direction of emf.

Hence, from diagram and applying the Fleming's right hand rule, the direction of magnetic field is perpendicular to plane of paper and into the paper.

Metal detector works on the principle of electromagnetic induction. A metal detector produces a time varying magnetic field from the coil inside. When the hidden metal is exposed to varying magnetic field, it experience a change in flux. As a result EMF in induced and eddy current is produced. Since, charge in motion create magnetic field, eddy current produces its own magnetic field. This field is detected by metal detector.

Given,

Copper disc radius, $r=0.1 \mathrm{~m}$

Angular rotation, $n=10 \mathrm{rev} / \mathrm{s}$

Magnetic field, $B=0.1 \mathrm{~T}$

Let a small radial portion of $d x$ at a distance $X$ from centre of disc.

$ \begin{aligned} &\text { Velocity of this element is }\\ &v=\omega X \end{aligned} $

Emf induced across this element.

$ d e=v B d x $

or $\omega x B d x$

Total emf induced across radius $r$,

$ \begin{aligned} E & =\frac{1}{2} B \omega r^2, \text { where } \omega=2 \pi n=2 \pi \times 10 \\ & =20 \pi \mathrm{rad} / \mathrm{s} \end{aligned} $

Put the values in above equation,

$ \begin{aligned} E & =\frac{1}{2} \times 0.1 \times 20 \pi \times 0.1 \times 0.1 \\ & =10 \pi \times 10^{-3} \mathrm{~V} \\ E & =10 \pi \mathrm{mV} \end{aligned} $

Self inductance of a coil is given as

$ L=\frac{\mu_0 \pi n^2 r}{2} $

Where, $\mu$ is permeability $n$ is no. of turns and $A$ is area of coil Thus, self inductance of a coil depends on size, shape of the coil, and number of turns in it.

$ \begin{aligned} &\text { Induced emf is given by }\\ &\begin{aligned} & \begin{aligned} e & =\frac{-d \phi}{d t}=-\frac{d}{d t} \cdot(B A \cos \theta) \\ \Rightarrow \quad e & =+B A \sin \theta \Rightarrow \frac{e_1}{e_2}=\frac{\sin \theta_1}{\sin \theta_2} \\ \text { Here, } \theta_1 & =\left(90^{\circ}-60^{\circ}\right)=30^{\circ} \\ \theta_2 & =180^{\circ}-120^{\circ}=60^{\circ} \\ \frac{e_1}{e_2} & =\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}=\frac{1}{\sqrt{3}} \end{aligned} \end{aligned} \end{aligned} $

Speed of aeroplane,

$ \begin{aligned} v & =540 \mathrm{~km} / \mathrm{h} \\ & =540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s} \\ & =150 \mathrm{~m} / \mathrm{s} \end{aligned} $

Length of wing span, $l=20 \mathrm{~m}$

$ \begin{aligned} B_H & =25 \sqrt{3} \times 10^{-4} \mathrm{~T} \\ \delta & =30^{\circ} \end{aligned} $

Potential difference developed between the ends of the wing

$ \begin{aligned} \varepsilon & =B_V v l=\left(B_H \tan \delta\right) v l \\ & =2.5 \sqrt{3} \times 10^{-4} \times \tan 30 \times 150 \times 20 \\ & =2.5 \sqrt{3} \times 10^{-4} \times \frac{1}{\sqrt{3}} \times 3 \times 10^3 \\ & =0.75 \mathrm{~V} \end{aligned} $

Given the problem, let's break down the information provided and the calculations involved.

Given Data:

Speed of the aeroplane: $360 \, \text{km/h}$

Convert the speed to meters per second:

$ 360 \, \text{km/h} = 100 \, \text{m/s} $

Wing span of the aeroplane: $4 \, \text{m}$

Vertical component of Earth's magnetic field: $0.5 \times 10^{-4} \, \text{T}$

Formula Used:

The induced electromotive force (emf) generated across the wing span of the aeroplane due to its movement in the Earth's magnetic field is calculated using the formula:

$ e = B \cdot L \cdot v $

Where:

$e$ is the induced emf

$B$ is the magnetic field component ($0.5 \times 10^{-4} \, \text{T}$)

$L$ is the wing span ($4 \, \text{m}$)

$v$ is the speed of the aeroplane ($100 \, \text{m/s}$)

Calculation:

Substitute the values into the formula:

$ e = 0.5 \times 10^{-4} \times 4 \times 100 $

Solving this gives:

$ e = 2 \times 10^{-2} \, \text{V} = 20 \times 10^{-3} \, \text{V} $

Result:

Thus, the induced emf across the ends of the wings is $20 \times 10^{-3} \, \text{V}$.

Explanation:

To determine the induced electromotive force (emf) in the rim of the bicycle wheel, we need to consider the following information:

Radius of the Rim, $ r $: The radius is given as $ 40 \, \text{cm} $, which converts to $ 0.4 \, \text{m} $.

Angular Speed, $ \omega $: The wheel is rotating with an angular speed of $ 20 \, \text{rad/s} $.

Horizontal Component of Earth's Magnetic Field: Given as $ 0.26 \, \text{G} $, which is $ 0.26 \times 10^{-4} \, \text{T} $.

The formula for the induced emf is given by:

$ e = l(\mathbf{v} \times \mathbf{B}) $

Where:

$ e $ is the induced emf.

$ l $ is the length of the conductor or loop.

$ \mathbf{v} $ is the velocity of the loop.

$ \mathbf{B} $ is the magnetic field.

Since the plane of the wheel's revolution is parallel to the horizontal component of Earth's magnetic field, the cross product $ \mathbf{v} \times \mathbf{B} $ becomes zero. This is because the velocity vector of the wheel's rim is parallel to the direction of the magnetic field.

Thus, the induced emf is zero:

$ e = 0 $

Therefore, no emf is induced in this specific orientation of the wheel because the necessary conditions for inducing emf (a changing magnetic field perpendicular to the motion) are not met.

When the current $i$ through a solenoid increases at a constant rate, the analysis of the induced current involves understanding the behavior of magnetic field, magnetic flux, and their rates of change:

Magnetic Field through the Solenoid:

The expression for the magnetic field $ B $ within a solenoid is given by:

$ B = \mu_0 n I $

where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length, and $I$ is the current.

Rate of Change of Magnetic Field:

The rate of change of the magnetic field with respect to time is:

$ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} $

Since $\frac{dI}{dt}$ is constant (the current is increasing at a constant rate), $\frac{dB}{dt}$ is also constant.

Magnetic Flux through the Solenoid:

Magnetic flux $\phi$ through the solenoid is given by:

$ \phi = B \cdot A $

where $A$ is the cross-sectional area of the solenoid.

Rate of Change of Magnetic Flux:

The rate of change of magnetic flux is expressed as:

$ \frac{d\phi}{dt} = A \frac{dB}{dt} $

Given that $\frac{dB}{dt}$ is constant, $\frac{d\phi}{dt}$ is also constant.

Induced Current:

According to Lenz’s law, the induced current will act to oppose the change in magnetic flux. Therefore, even though the induced current is constant, it arises due to the constant rate of change of flux, working in opposition to the increase in the original current $i$.

Thus, the induced current is constant and occurs in a direction opposite to the original increasing current $i$.

To determine the mutual inductance of a pair of adjacent coils, we start with the given information:

The change in current ($\Delta I$) is from 10 A to 2 A, so $\Delta I = 10 \, \text{A} - 2 \, \text{A} = 8 \, \text{A}$.

The time interval ($\Delta t$) for this change is 0.2 s.

The induced electromotive force (emf) is 120 V.

We use Faraday's law of mutual induction to find the mutual inductance $M$. The formula for the induced emf ($E$) can be derived from Faraday's law as:

$ E = \frac{M \cdot \Delta I}{\Delta t} $

Substitute the known values into the equation:

$ 120 = \frac{M \times 8}{0.2} $

Solving for $M$:

$ 120 = \frac{M \times 8}{0.2} $

$ 120 = M \times 40 $

$ M = \frac{120}{40} = 3 \, \text{H} $

Thus, the mutual inductance of the pair of coils is 3 H.

Given, length of each spokes of wheel, $l=40 \mathrm{~cm}=0.4 \mathrm{~m}$

Angular velocity, $\omega=2 \pi f$

$ \begin{aligned} & =2 \pi \times \frac{180}{60} \mathrm{rad} / \mathrm{s} \\ & =6 \pi \mathrm{rad} / \mathrm{s} \end{aligned} $

Horizontal component of earth's magnetic field,

$ H_{\mathrm{e}}=0.4 \mathrm{G}=0.4 \times 10^{-4} \mathrm{~T} $

$ \begin{aligned} & \therefore \text { Induced emf, } e=\frac{1}{2} B \omega L^2 \\ & \qquad \begin{aligned} = & \frac{1}{2} \times 0.4 \times 10^{-4} \times 6 \pi \times(0.4)^2 \\ = & 1.2 \pi \times 10^{-4} \times 0.16 \\ = & 0.192 \pi \times 10^{-4} \mathrm{~V} \\ = & 192 \pi \times 10^{-7} \mathrm{~V} \end{aligned} \end{aligned} $

Radius of metal disc,

$ R=30 \mathrm{~cm}=0.3 \mathrm{~m} $

Angular velocity, $\omega=100 \mathrm{rad} / \mathrm{s}$

Magnetic field, $B=4 \mathrm{mT}$

$ =4 \times 10^{-3} \mathrm{~T} $

Magnitude of potential difference between the centre and the rim of the disc.

$ \begin{aligned} e & =\frac{1}{2} B \omega R^2 \\ & =\frac{1}{2} \times 4 \times 10^{-3} \times 100 \times(0.3)^2 \\ & =2 \times 10^{-3} \times 100 \times 0.09 \\ & =18 \times 10^{-3} \mathrm{~V}=18 \mathrm{mV} \end{aligned} $

The given situation is shown below.

Magnetic field,

$ B=2 \mathrm{~T} $

Area of the triangular loop,

$ \begin{aligned} A & =\frac{1}{2} \times P Q \times P R \\ & =\frac{1}{2} \times 2 \times 1=1 \mathrm{~cm}^2 \\ & =10^{-4} \mathrm{~m}^2 \end{aligned} $

∴ Magnetic flux through triangular loop,

$ \phi=B A=2 \times 10^{-4} \mathrm{~Wb} $

Area of wire loop, $A=0.2 \mathrm{~m}^2$

Resistance, $R=20 \Omega$

Magnetic field, $B=0.25 \mathrm{~T}$

According to Faraday's law of electromagnetic induction.

$ \begin{aligned} & \text { Induced emf, } e=\frac{-d \phi}{d t} \\ & \qquad \begin{aligned} & =\frac{-d}{d t} B A=-A \frac{d B}{d t} \\ & =-A \frac{\Delta B}{\Delta t}=-A \frac{\left(B_2-B_1\right)}{10^{-4}} \\ & =-0.2 \frac{(0-0.25)}{10^{-4}}=\frac{0.05}{10^{-4}} \\ \Rightarrow \quad e & =500 \mathrm{~V} \end{aligned} \\ & \begin{aligned} \end{aligned} \\ & \end{aligned} $

and induced current ,

$ I=\frac{e}{R}=\frac{500}{20}=25 \mathrm{~A} $

Number of turns in flat circular coil, $N=100$

Radius, $r=10 \mathrm{~cm}=0.1 \mathrm{~m}$

Rate of change of magnetic field,

$ \frac{d B}{d t}=0.1 \mathrm{~T} / \mathrm{s} $

∴ Induced emf in the coil,

$ \begin{aligned} |e| & =N \frac{d \phi}{d t}=N \cdot \frac{d}{d t} B A \quad[\because \phi=B A] \\ & =N \frac{d}{d t} B \cdot \pi r^2=N \cdot \pi r^2 \cdot \frac{d B}{d t} \\ & =100 \times \pi \times(0.1)^2 \times 0.1=\frac{\pi}{10} \mathrm{~V} \end{aligned} $

Given, number of turns per cm in long solenoid, $n=20$ turns $/ \mathrm{cm}=2000$ turns/m

$ \begin{aligned} & \text { Area of loop, } A=\frac{4}{\pi} \mathrm{~cm}^2=\frac{4}{\pi} \times 10^4 \mathrm{~m}^2 \\ & \Delta I=3-1 \neq 2 \mathrm{~A} \quad \Delta t=0.2 \mathrm{~s} \end{aligned} $

∴ Magnetic field inside the solenoid,

$ B=\mu_0 n I \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Magnetic flux linked with loop kept inside the solenoid normal to it,

$ \phi=B A=\mu_0 n I A\,\,\,\,\,\,\, [from Eq. (i)]$

∴ According to Faraday's law of electromagnetic induction, magnitude of induced emf is given as

$ \begin{aligned} & |e|=\frac{d \phi}{d t}=\frac{d}{d t}\left(\mu_0 n I A\right)=\mu_0 n A \frac{d I}{d t} \\ & \Rightarrow|e|=\mu_0 n A \cdot \frac{\Delta I}{\Delta t}=4 \pi \times 10^{-7} \times 2000 \times \frac{4}{\pi} \times 10^{-4} \times \frac{2}{0.2} \\ & =3.2 \times 10^{-6} \mathrm{~V}=3.2 \mu \mathrm{~V}\left(\because 10^{-6} \mathrm{~V}=1 \mu \mathrm{~V}\right) \end{aligned} $

According to given condition, Magnetic energy stored in solenoid $X=$ Magnetic energy stored in solenoid $Y$

$ \begin{array}{ll} \Rightarrow & U_X=U_Y \\ \Rightarrow & \frac{B_X^2 V_X}{2 \mu_0}=\frac{B_Y^2 V_Y}{2 \mu_0} \\ & {\left[\because U=\frac{B^2 V}{2 \mu_0}, V=\text { volume }\right]} \\ & \\ & B_X^2 V_X=B_Y^2 V_Y \end{array} $

$ \begin{aligned} & \Rightarrow \quad B_X^2 A_X L_X=B_Y^2 A_Y L_Y \\ & \Rightarrow \quad \frac{B_X^2}{B_Y^2}=\frac{A_Y}{A_X} \times \frac{L_Y}{L_X}=\frac{2 A_X}{A_X} \times \frac{2 L_X}{L_X} \\ & \Rightarrow \quad \frac{B_X^2}{B_Y^2}=4 \Rightarrow \frac{B_X}{B_Y}=\sqrt{4} \\ & \Rightarrow \quad \frac{B_X}{B_Y}=2 \Rightarrow B_X: B_Y=2: 1 \end{aligned} $

Self-induced emf in coil is given by

$ E=L(d I / d t) $

Here, $E=200 \mathrm{~V}, \frac{d I}{d t}=\left(\frac{10-0}{1.5}\right) \mathrm{A} / \mathrm{s}$

So, self-inductance of coil,

$ \begin{aligned} L & =\frac{E}{d I / d t}=\frac{200}{\left(\frac{10-0}{1.5}\right)} \\ \Rightarrow L & =\frac{200 \times 1.5}{10}=30 \mathrm{H} \end{aligned} $