Dynamic EMI and Periodic EMI Questions in English

(C) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is proportional to the rate of change of magnetic flux through the coil.
When the magnet is moved in and out with a frequency $\nu$,the magnetic flux changes periodically,inducing an alternating current in the coil.
The galvanometer needle has a certain moment of inertia and damping,which limits its ability to respond to high-frequency changes.
If the frequency $\nu$ is low (e.g.,$1 \, Hz$ or $2 \, Hz$),the galvanometer needle can follow the changes in the induced current,and oscillations in the deflection will be observed clearly.
If the frequency $\nu$ is high (e.g.,$50 \, Hz$),the needle cannot keep up with the rapid oscillations due to its inertia and will remain at the zero position (or show a very small,steady deflection),making the oscillations invisible.

(C) The peak value of the induced electromotive force $(e.m.f.)$, denoted as $e_0$, is given by the formula: $e_0 = N B A \omega = N B A (2 \pi \nu)$.
Given values:
Number of turns, $N = 300$
Magnetic field strength, $B = 4 \times 10^{-2}\;T$
Area of the coil, $A = 25\;cm \times 10\;cm = 250\;cm^2 = 250 \times 10^{-4}\;m^2 = 2.5 \times 10^{-2}\;m^2$
Frequency of rotation, $\nu = 50\;cps$ (cycles per second)
Substituting these values into the formula:
$e_0 = 300 \times (4 \times 10^{-2}) \times (2.5 \times 10^{-2}) \times (2 \pi \times 50)$
$e_0 = 300 \times 0.04 \times 0.025 \times 100\pi$
$e_0 = 12 \times 0.025 \times 100\pi$
$e_0 = 0.3 \times 100\pi = 30\pi\;V$
Thus, the peak value of the induced $e.m.f.$ is $30\pi\;V$.

(B) The maximum induced $e.m.f.$ $(e_0)$ in a rotating coil is given by the formula: $e_0 = N B A \omega$.
Here,$N = 4000$,$B = 0.5 \, \text{gauss} = 0.5 \times 10^{-4} \, T$,$r = 7 \, cm = 0.07 \, m$,and the frequency $\nu = 1800 \, \text{rpm} = \frac{1800}{60} \, \text{Hz} = 30 \, \text{Hz}$.
The angular velocity $\omega = 2 \pi \nu = 2 \times \pi \times 30 = 60 \pi \, \text{rad/s}$.
The area $A = \pi r^2 = \pi \times (0.07)^2 = 0.0049 \pi \, \text{m}^2$.
Substituting these values into the formula:
$e_0 = 4000 \times (0.5 \times 10^{-4}) \times (0.0049 \pi) \times (60 \pi)$
$e_0 = 4000 \times 0.5 \times 10^{-4} \times 0.0049 \times 60 \times \pi^2$
$e_0 = 0.2 \times 0.0049 \times 60 \times 9.8696 \approx 0.58 \, V$.

(D) The peak value of the induced $EMF$ $(e_0)$ in an $AC$ generator is given by the formula: $e_0 = N B A \omega$.
Here,$N = 5000$ (number of turns),$B = 0.2\,Wb/m^2$ (magnetic field),$A = 0.25\,m^2$ (area),and the frequency $\nu = 100\,Hz$.
The angular velocity $\omega = 2 \pi \nu = 2 \times 3.14 \times 100 = 628\,rad/s$.
Substituting the values: $e_0 = 5000 \times 0.2 \times 0.25 \times 628$.
$e_0 = 1000 \times 0.25 \times 628 = 250 \times 628 = 157000\,V$.
Converting to $kV$: $157000\,V = 157\,kV$.

(C) The magnetic flux through the coil at any time $t$ is given by $\phi = NBA \cos(\omega t)$,where $\omega$ is the angular frequency.
The induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.
The amplitude of the induced $EMF$ is $e_0 = NBA\omega$.
The amplitude of the induced current is $i_0 = \frac{e_0}{R} = \frac{NBA\omega}{R}$.
Given: $B = 10^{-2} \, T$,$r = 0.3 \, m$,$A = \pi r^2 = \pi(0.3)^2 = 0.09\pi \, m^2$,$R = \pi^2 \, \Omega$,and frequency $f = \frac{200}{60} = \frac{10}{3} \, Hz$.
Angular frequency $\omega = 2\pi f = 2\pi \times \frac{10}{3} = \frac{20\pi}{3} \, rad/s$.
Assuming $N=1$ (single turn coil):
$i_0 = \frac{1 \times 10^{-2} \times (0.09\pi) \times (20\pi/3)}{\pi^2} = \frac{10^{-2} \times 0.03 \times 20 \times \pi^2}{\pi^2} = 0.006 \, A = 6 \, mA$.

(C) The maximum induced $emf$ $(e_0)$ in a rotating coil is given by the formula: $e_0 = N B A \omega$, where $\omega = 2\pi \nu$.
Given:
Number of turns $(N)$ = $300$
Magnetic field $(B)$ = $4 \times 10^{-2} \; T$
Area $(A)$ = $25 \; cm \times 10 \; cm = 250 \; cm^2 = 250 \times 10^{-4} \; m^2 = 2.5 \times 10^{-2} \; m^2$
Frequency $(\nu)$ = $50 \; rps$
Angular speed $(\omega)$ = $2\pi \times 50 = 100\pi \; rad/s$
Substituting the values:
$e_0 = 300 \times (4 \times 10^{-2}) \times (2.5 \times 10^{-2}) \times 100\pi$
$e_0 = 300 \times 0.04 \times 0.025 \times 100\pi$
$e_0 = 300 \times 0.001 \times 100\pi$
$e_0 = 30\pi \; V$.

(B) The magnetic flux linked with a coil rotating in a uniform magnetic field is given by $\phi = BA \cos(\omega t)$,where $\omega$ is the angular velocity.
According to Faraday's law of electromagnetic induction,the induced $e.m.f.$ is given by $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$: $e = -\frac{d}{dt}(BA \cos(\omega t)) = BA\omega \sin(\omega t)$.
Using the trigonometric identity $\sin(\theta) = \cos(\theta - \pi/2)$,we can write the induced $e.m.f.$ as $e = BA\omega \cos(\omega t - \pi/2)$.
Comparing the phase of the flux $(\omega t)$ and the phase of the induced $e.m.f.$ $(\omega t - \pi/2)$,the phase difference is $\pi/2$.

(B) The magnetic flux linked with the coil is given by $\phi = N A B \cos(\omega t)$.
The induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = N A B \omega \sin(\omega t)$.
The amplitude of the induced $EMF$ is $e_0 = N A B \omega$.
Given values: $N = 100$,$R = 10 \Omega$,$r = 10 \text{ cm} = 0.1 \text{ m}$,$B = 10 \text{ mT} = 10^{-2} \text{ T}$,and frequency $f = 100 \text{ Hz}$.
Angular frequency $\omega = 2 \pi f = 2 \pi \times 100 = 200 \pi \text{ rad/s}$.
Area $A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \text{ m}^2$.
The amplitude of the current is $I_0 = \frac{e_0}{R} = \frac{N A B \omega}{R}$.
Substituting the values: $I_0 = \frac{100 \times (0.01 \pi) \times 10^{-2} \times (200 \pi)}{10}$.
$I_0 = \frac{100 \times 0.01 \times 200 \times \pi^2 \times 10^{-2}}{10} = \frac{200 \times 10}{10} = 2 \text{ A}$.

(D) The induced $e.m.f.$ in a rotating coil is given by the formula $\varepsilon = N A B \omega \sin(\omega t)$.
Here,$N$ is the number of turns,$A$ is the area of the coil,$B$ is the magnetic field strength,and $\omega$ is the angular speed.
From the formula,it is clear that the induced $e.m.f.$ depends on $N$,$A$,$B$,and $\omega$.
It does not depend on the resistance of the coil,as resistance only affects the induced current $(I = \varepsilon / R)$,not the induced $e.m.f.$ itself.
Therefore,the correct option is $D$.

(A) The magnetic flux $\phi$ linked with the coil at any time $t$ is given by $\phi = N B A \cos(\omega t)$.
According to Faraday's law of electromagnetic induction,the induced emf $e$ is given by $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$: $e = -\frac{d}{dt}(N B A \cos(\omega t))$.
$e = -N B A \frac{d}{dt}(\cos(\omega t)) = -N B A (-\omega \sin(\omega t)) = N B A \omega \sin(\omega t)$.
The maximum value of emf $(e_{\max})$ occurs when $\sin(\omega t) = 1$.
Therefore,$e_{\max} = N B A \omega$.

(D) Given: Number of turns $N = 100$,Area $A = 0.10$ $m^2$,Magnetic field $B = 0.01$ $T$,Frequency $f = 0.5$ $rev/s$.
Angular frequency $\omega = 2 \pi f = 2 \times 3.14 \times 0.5 = 3.14$ $rad/s$.
The maximum induced electromotive force (voltage) is given by the formula $e_0 = N B A \omega$.
Substituting the values: $e_0 = 100 \times 0.01 \times 0.10 \times 3.14$.
$e_0 = 1 \times 0.10 \times 3.14 = 0.314$ $V$.
Thus,the maximum voltage generated in the coil is $0.314$ $V$.

(B) The area of the semicircle is $A = \frac{1}{2} \pi r^{2}$.
The magnetic flux linked with the loop at time $t$ is $\phi = B A \cos(\omega t) = B \left(\frac{1}{2} \pi r^{2}\right) \cos(\omega t)$.
The induced electromotive force $(EMF)$ is given by Faraday's law: $\varepsilon = -\frac{d\phi}{dt} = \frac{1}{2} B \pi r^{2} \omega \sin(\omega t)$.
The instantaneous power generated is $P(t) = \frac{\varepsilon^{2}}{R} = \frac{(\frac{1}{2} B \pi r^{2} \omega \sin(\omega t))^{2}}{R} = \frac{B^{2} \pi^{2} r^{4} \omega^{2} \sin^{2}(\omega t)}{4 R}$.
The mean power over a full period is $\langle P \rangle = \frac{B^{2} \pi^{2} r^{4} \omega^{2}}{4 R} \langle \sin^{2}(\omega t) \rangle$.
Since the mean value of $\sin^{2}(\omega t)$ over a period is $\frac{1}{2}$,we have $\langle P \rangle = \frac{B^{2} \pi^{2} r^{4} \omega^{2}}{4 R} \cdot \frac{1}{2} = \frac{(B \pi r^{2} \omega)^{2}}{8 R}$.

(C) The magnetic field $B$ at the center of the large circular coil of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2R}$.
The magnetic flux $\phi$ through the smaller coil of radius $r$ at an angle $\theta = \omega t$ with the magnetic field is $\phi = B A \cos(\omega t) = \left(\frac{\mu_0 I}{2R}\right) (\pi r^2) \cos(\omega t)$.
According to Faraday's law of electromagnetic induction,the induced $emf$ is $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$: $e = -\frac{d}{dt} \left[ \frac{\mu_0 I \pi r^2}{2R} \cos(\omega t) \right]$.
$e = -\frac{\mu_0 I \pi r^2}{2R} \cdot (-\omega \sin(\omega t)) = \frac{\mu_0 I}{2R} \omega \pi r^2 \sin(\omega t)$.

(B) The magnetic flux linked with the loop is given by $\Phi = BA \cos(\omega t)$.
The induced $emf$ is given by $e = -\frac{d\Phi}{dt} = BA\omega \sin(\omega t)$.
The direction of the induced $emf$ depends on the sign of $\sin(\omega t)$.
Since $\sin(\omega t)$ changes its sign after every $\pi$ radians (which corresponds to half a revolution),the direction of the induced $emf$ reverses after every $1/2$ revolution.

(B) The magnetic flux linked with a rotating coil is given by $\phi = \phi_0 \cos(\omega t)$.
According to Faraday's law of electromagnetic induction,the induced $emf$ is $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$,we get $e = -\frac{d}{dt}(\phi_0 \cos(\omega t)) = \phi_0 \omega \sin(\omega t)$.
Since the induced current $i = e/R$,the current is $i = \frac{\phi_0 \omega}{R} \sin(\omega t)$.
Both the induced $emf$ $(e)$ and the induced current $(i)$ are proportional to $\sin(\omega t)$.
Therefore,there is no phase difference between the induced $emf$ and the induced current,meaning the phase difference is zero.

(C) The magnetic flux linked with a rotating coil is given by $\phi = BA \cos(\omega t)$,where $\omega$ is the angular speed of rotation.
According to Faraday's law of induction,the induced emf $E$ is given by $E = -\frac{d\phi}{dt}$.
$E = -\frac{d}{dt}(BA \cos(\omega t)) = -BA(-\omega \sin(\omega t)) = BA\omega \sin(\omega t)$.
We can rewrite this as $E = BA\omega \cos(\omega t - \pi / 2)$.
Comparing the phase of the flux $\phi$ (which is $\omega t$) and the phase of the induced emf $E$ (which is $\omega t - \pi / 2$),the phase difference is $\pi / 2$.
Therefore,the induced emf lags behind the magnetic flux by $\pi / 2$.

(D) The magnetic flux $\phi$ through the loop is given by $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta = BA \cos(\omega t)$.
The induced emf $e$ is given by Faraday's law: $|e| = |\frac{d\phi}{dt}| = |BA\omega \sin(\omega t)|$.
The induced emf $|e|$ is maximum when $\sin(\omega t) = 1$,which occurs at $\omega t = \frac{\pi}{2}$.
Given the period $T = 10 \; s$,the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{10} = \frac{\pi}{5} \; rad/s$.
Substituting $\omega t = \frac{\pi}{2}$ gives $(\frac{\pi}{5})t = \frac{\pi}{2}$,which results in $t = 2.5 \; s$.
The induced emf $|e|$ is minimum when $\sin(\omega t) = 0$,which occurs at $\omega t = \pi$ (or $0, 2\pi, \dots$).
Substituting $\omega t = \pi$ gives $(\frac{\pi}{5})t = \pi$,which results in $t = 5.0 \; s$.
Therefore,the induced emf is maximum at $2.5 \; s$ and minimum at $5.0 \; s$.

(N/A) Principle: The principle of an $A.C.$ generator is based on electromagnetic induction. When a coil is rotated in a uniform magnetic field,the magnetic flux linked with the coil changes continuously,which induces an alternating $emf$ in the coil.
Construction: An $A.C.$ generator consists of a rectangular coil (armature) placed between the pole pieces of a strong permanent magnet. The coil is mounted on a rotor shaft and can be rotated. The ends of the coil are connected to two slip rings,which rotate with the coil. Two stationary carbon brushes press against these slip rings to conduct the induced current to the external circuit.
Derivation of induced $emf$:
Let the area of the coil be $A$,the number of turns be $N$,and the magnetic field be $B$. If the coil rotates with an angular velocity $\omega$,the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$ at any time $t$ is $\theta = \omega t$.
The magnetic flux $\phi$ linked with the coil is given by:
$\phi = N B A \cos(\theta) = N B A \cos(\omega t)$
According to Faraday's law of electromagnetic induction,the induced $emf$ $(\varepsilon)$ is:
$\varepsilon = -\frac{d\phi}{dt}$
$\varepsilon = -\frac{d}{dt} (N B A \cos(\omega t))$
$\varepsilon = -N B A (-\sin(\omega t)) \cdot \omega$
$\varepsilon = N B A \omega \sin(\omega t)$
Let $\varepsilon_0 = N B A \omega$ be the peak value of the $emf$. Then:
$\varepsilon = \varepsilon_0 \sin(\omega t)$

(N/A) The formula for the induced $emf$ $(\varepsilon)$ in an $AC$ generator is given by:
$\varepsilon = \varepsilon_{0} \sin(\omega t) = \varepsilon_{0} \sin(2 \pi \nu t) \quad \dots (1)$
where $\varepsilon_{0}$ is the peak value of the $emf$,$\omega$ is the angular frequency,and $\nu$ is the frequency of rotation.
Equation $(1)$ represents the instantaneous value of the $emf$. The $emf$ varies periodically between $+\varepsilon_{0}$ and $-\varepsilon_{0}$ as the coil rotates.
Based on the rotation of the coil in the magnetic field $\vec{B}$,we can analyze different stages:
$1$. Stage $1$ $(\omega t = 0^{\circ})$: The plane of the coil is perpendicular to the magnetic field $\vec{B}$. The magnetic flux is maximum,but the rate of change of flux is zero. Thus,$\varepsilon = \varepsilon_{0} \sin(0^{\circ}) = 0$.
$2$. Stage $2$ $(\omega t = 90^{\circ})$: The plane of the coil is parallel to the magnetic field $\vec{B}$. The rate of change of flux is maximum. Thus,$\varepsilon = \varepsilon_{0} \sin(90^{\circ}) = \varepsilon_{0}$.
$3$. Stage $3$ $(\omega t = 180^{\circ})$: The plane of the coil is again perpendicular to $\vec{B}$. Thus,$\varepsilon = \varepsilon_{0} \sin(180^{\circ}) = 0$.
$4$. Stage $4$ $(\omega t = 270^{\circ})$: The plane of the coil is parallel to $\vec{B}$,but the coil has rotated to the opposite side. Thus,$\varepsilon = \varepsilon_{0} \sin(270^{\circ}) = -\varepsilon_{0}$.
$5$. Stage $5$ $(\omega t = 360^{\circ})$: The coil returns to its initial position. Thus,$\varepsilon = \varepsilon_{0} \sin(360^{\circ}) = 0$.

(N/A) In an $AC$ generator,a coil rotates in a uniform magnetic field $B$. The magnetic flux $\phi$ linked with the coil at any time $t$ is given by $\phi = NBA \cos(\omega t)$,where $N$ is the number of turns,$A$ is the area of the coil,and $\omega$ is the angular velocity.
According to Faraday's law of induction,the induced $emf$ is $\varepsilon = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.
Let $\varepsilon_0 = NBA\omega$ be the peak value of the $emf$. Then $\varepsilon = \varepsilon_0 \sin(\omega t)$.
Characteristics:
$1$. The induced $emf$ varies sinusoidally with time.
$2$. At $\omega t = 0, 180^\circ, 360^\circ$,the $emf$ is zero because the plane of the coil is perpendicular to the magnetic field.
$3$. At $\omega t = 90^\circ$,the $emf$ is maximum positive $(\varepsilon_0)$ because the rate of change of flux is maximum.
$4$. At $\omega t = 270^\circ$,the $emf$ is maximum negative $(-\varepsilon_0)$ as the coil moves in the opposite direction relative to the field lines.

(N/A) In an $AC$ generator,the magnetic flux $\phi$ linked with the coil rotating with angular velocity $\omega$ in a magnetic field $B$ is given by $\phi = NBA \cos(\omega t)$,where $N$ is the number of turns and $A$ is the area of the coil.
According to Faraday's law of induction,the induced $emf$ is $\epsilon = -\frac{d\phi}{dt}$.
Substituting $\phi$,we get $\epsilon = -\frac{d}{dt}(NBA \cos(\omega t)) = NBA\omega \sin(\omega t)$.
The maximum value of $emf$ $(\epsilon_{max})$ occurs when $\sin(\omega t) = 1$.
Therefore,the equation for maximum $emf$ is $\epsilon_{max} = NBA\omega$.

(D) Given: Radius $r = 10\, cm = 0.1\, m$,Magnetic field $B = 3.0 \times 10^{-5}\, T$.
Time for half rotation is $0.2\, s$,so the time period $T = 0.4\, s$.
The magnetic flux through the coil at any time $t$ is $\phi = BA \cos(\omega t)$,where $A = \pi r^2$ is the area of the coil.
The induced $EMF$ is given by Faraday's law: $|\varepsilon| = |\frac{d\phi}{dt}| = |BA\omega \sin(\omega t)|$.
The maximum induced $EMF$ is $\varepsilon_{\max} = BA\omega$.
Substituting $\omega = \frac{2\pi}{T}$:
$\varepsilon_{\max} = B \times (\pi r^2) \times \frac{2\pi}{T} = \frac{2\pi^2 B r^2}{T}$.
Substituting the values: $\varepsilon_{\max} = \frac{2 \times \pi^2 \times 3.0 \times 10^{-5} \times (0.1)^2}{0.4}$.
Using $\pi^2 \approx 10$:
$\varepsilon_{\max} = \frac{2 \times 10 \times 3.0 \times 10^{-5} \times 0.01}{0.4} = \frac{6 \times 10^{-4}}{0.4} = 15 \times 10^{-6}\, V = 15\, \mu V$.

(A) The magnetic flux through the loop at time $t$ is $\phi = BA \cos(\omega t)$,where $A = \pi ab$ is the area of the elliptical loop.
The induced electromotive force $(EMF)$ is given by Faraday's law: $\epsilon = -\frac{d\phi}{dt} = AB\omega \sin(\omega t)$.
The instantaneous power loss due to Joule heating is $P = \frac{\epsilon^{2}}{R} = \frac{(AB\omega \sin(\omega t))^{2}}{R} = \frac{A^{2}B^{2}\omega^{2}}{R} \sin^{2}(\omega t)$.
The average power loss over a complete cycle is $P_{avg} = \langle P \rangle = \frac{A^{2}B^{2}\omega^{2}}{R} \langle \sin^{2}(\omega t) \rangle$.
Since the average value of $\sin^{2}(\omega t)$ over a cycle is $\frac{1}{2}$,we have $P_{avg} = \frac{A^{2}B^{2}\omega^{2}}{R} \cdot \frac{1}{2}$.
Substituting $A = \pi ab$,we get $P_{avg} = \frac{(\pi ab)^{2} B^{2} \omega^{2}}{2R} = \frac{\pi^{2} a^{2} b^{2} B^{2} \omega^{2}}{2R}$.

(C) The maximum induced $emf$ $(\varepsilon_{max})$ in a rotating coil is given by the formula: $\varepsilon_{max} = N \omega A B$,where $N$ is the number of turns,$\omega$ is the angular speed,$A$ is the area of the coil,and $B$ is the magnetic field strength.
Given: $N = 20$,$\omega = 50 \, rad \, s^{-1}$,$B = 3.0 \times 10^{-2} \, T$,and radius $r = 8.0 \, cm = 0.08 \, m$.
The area $A = \pi r^2 = \pi \times (0.08)^2 = 0.0064 \pi \, m^2$.
Substituting the values: $\varepsilon_{max} = 20 \times 50 \times (0.0064 \pi) \times (3.0 \times 10^{-2})$.
$\varepsilon_{max} = 1000 \times 0.0064 \times \pi \times 3.0 \times 10^{-2} = 6.4 \times \pi \times 3.0 \times 10^{-2} = 19.2 \pi \times 10^{-2}$.
Using $\pi \approx 3.14159$,$\varepsilon_{max} \approx 19.2 \times 3.14159 \times 10^{-2} \approx 60.319 \times 10^{-2} \, V$.
Rounding to the nearest integer,we get $60$.

(D) The maximum induced electromotive force $(EMF)$ in a rotating coil is given by the formula: $\varepsilon_{\max} = BAN\omega$.
Given:
$B = 0.07 \, T$ (Magnetic field)
$A = 1 \, m^{2}$ (Area of the coil)
$N = 1000$ (Number of turns)
$f = 1 \, \text{rev/s}$ (Frequency)
Angular velocity $\omega = 2\pi f = 2\pi(1) = 2\pi \, \text{rad/s}$.
Substituting the values:
$\varepsilon_{\max} = 0.07 \times 1 \times 1000 \times 2\pi$
$\varepsilon_{\max} = 70 \times 2\pi = 140\pi$.
Using $\pi \approx 3.14159$:
$\varepsilon_{\max} = 140 \times 3.14159 \approx 439.82 \, V$.
Rounding to the nearest integer, we get $440 \, V$.

(C) The maximum induced electromotive force (emf) in an $AC$ generator is given by the formula: $\varepsilon_{\max} = NAB\omega$.
Given:
Number of turns,$N = 100$.
Area of the coil,$A = 14 \times 10^{-2} \, m^2$.
Magnetic field,$B = 3.0 \, T$.
Frequency of rotation,$f = 360 \, rev/min = \frac{360}{60} \, rev/s = 6 \, Hz$.
Angular velocity,$\omega = 2\pi f = 2 \times \frac{22}{7} \times 6 \, rad/s$.
Substituting these values into the formula:
$\varepsilon_{\max} = 100 \times (14 \times 10^{-2}) \times 3.0 \times (2 \times \frac{22}{7} \times 6)$
$\varepsilon_{\max} = 100 \times 0.14 \times 3.0 \times (12 \times \frac{22}{7})$
$\varepsilon_{\max} = 14 \times 3.0 \times \frac{264}{7}$
$\varepsilon_{\max} = 42 \times \frac{264}{7}$
$\varepsilon_{\max} = 6 \times 264 = 1584 \, V$.

(D) Given: $N = 600$,$A = 70 \times 10^{-4} \, m^2$,$B = 0.4 \, T$,$f = \frac{500}{60} \, Hz$.
Angular velocity $\omega = 2 \pi f = 2 \pi \times \frac{500}{60} = \frac{50 \pi}{3} \, rad/s$.
The instantaneous emf is given by $e = N A B \omega \sin \theta$,where $\theta$ is the angle between the area vector $\vec{A}$ and the magnetic field $\vec{B}$.
Since the plane of the coil is inclined at $60^{\circ}$ with the field,the angle between the normal to the plane (area vector) and the field is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
$e = 600 \times (70 \times 10^{-4}) \times 0.4 \times (\frac{50 \pi}{3}) \times \sin(30^{\circ})$.
$e = 600 \times 0.007 \times 0.4 \times \frac{50}{3} \times \frac{22}{7} \times 0.5$.
$e = 4.2 \times 0.4 \times \frac{50}{3} \times \frac{22}{7} \times 0.5$.
$e = 1.68 \times \frac{50}{3} \times \frac{22}{7} \times 0.5 = 44 \, V$.

(C) The magnetic flux linked with the coil is given by $\phi = NAB \cos(\omega t)$.
The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\phi}{dt} = NAB\omega \sin(\omega t)$.
The maximum voltage generated is $\varepsilon_{\max} = NAB\omega$.
Given:
Number of turns $N = 200$
Area $A = 0.20 \ m^2$
Magnetic field $B = 0.01 \ T$
Frequency $f = 0.5 \ rev/s$,so angular frequency $\omega = 2\pi f = 2\pi(0.5) = \pi \ rad/s$.
Substituting the values:
$\varepsilon_{\max} = 200 \times 0.20 \times 0.01 \times \pi$
$\varepsilon_{\max} = 40 \times 0.01 \times \pi = 0.4\pi = \frac{4\pi}{10} = \frac{2\pi}{5} \ V$.
Comparing this with $\frac{2\pi}{\beta}$,we get $\beta = 5$.

(B) The magnetic flux $\phi$ through a coil rotating in a magnetic field is given by $\phi = NAB \cos(\omega t)$,where $\theta = \omega t$ is the angle between the area vector and the magnetic field vector $\vec{B}$.
The induced electromotive force (emf) $\varepsilon$ is given by Faraday's law of induction: $\varepsilon = -\frac{d\phi}{dt}$.
Calculating the derivative: $\varepsilon = -\frac{d}{dt}(NAB \cos(\omega t)) = NAB \omega \sin(\omega t)$.
When the magnetic field $\vec{B}$ is parallel to the plane of the coil,the angle between the area vector (which is perpendicular to the plane) and the magnetic field $\vec{B}$ is $90^\circ$ or $\frac{\pi}{2}$ radians.
At this instant,$\theta = \omega t = \frac{\pi}{2}$.
Substituting this into the expressions:
$\phi = NAB \cos(\frac{\pi}{2}) = NAB(0) = 0$.
$\varepsilon = NAB \omega \sin(\frac{\pi}{2}) = NAB \omega(1) = NAB \omega$.
Therefore,the magnetic flux is $0$ and the induced emf is $NAB \omega$.

(A) Given: Number of turns $N = 50$,Area $A = 2.5 \ m^2$,Magnetic field $B = 0.3 \ T$,Angular velocity $\omega = 60 \ rad \ s^{-1}$,Resistance $R = 500 \ \Omega$.
The maximum induced $emf$ $(V_m)$ is given by the formula:
$V_m = N B A \omega$
$V_m = 50 \times 0.3 \times 2.5 \times 60$
$V_m = 50 \times 0.3 \times 150 = 2250 \ V = 2.25 \ kV$.
The maximum current $(I_m)$ is given by Ohm's law:
$I_m = \frac{V_m}{R}$
$I_m = \frac{2250 \ V}{500 \ \Omega} = 4.5 \ A$.
Thus,the maximum induced $emf$ is $2.25 \ kV$ and the maximum current is $4.5 \ A$.

(B) Given: Resistance $R = 10 \ \Omega$,Radius $r = 10 \ cm = 0.1 \ m$,Number of turns $N = 100$,Frequency $f = 100 \ Hz$,Magnetic field $B = 10 \ mT = 10^{-2} \ T$.
Angular velocity $\omega = 2 \pi f = 2 \pi (100) = 200 \pi \ rad/s$.
Area of the loop $A = \pi r^2 = \pi (0.1)^2 = 0.01 \pi \ m^2$.
The peak electromotive force $(EMF)$ is given by $e_0 = N B A \omega$.
Substituting the values: $e_0 = 100 \times 10^{-2} \times 0.01 \pi \times 200 \pi = 1 \times 0.01 \times 200 \times \pi^2$.
Given $\pi^2 = 10$,so $e_0 = 0.01 \times 200 \times 10 = 20 \ V$.
The amplitude of the current $I_0 = \frac{e_0}{R} = \frac{20 \ V}{10 \ \Omega} = 2 \ A$.

(A) The induced electromotive force $(e)$ in a rotating coil is given by $e = NAB \omega \sin(\omega t)$.
The peak value of the induced $EMF$ is $e_0 = NAB \omega$.
The peak current is $i_0 = \frac{e_0}{R} = \frac{NAB \omega}{R}$.
The average power dissipated in an $AC$ circuit is given by $P_{av} = \frac{e_0 i_0}{2}$ or $P_{av} = \frac{e_0^2}{2R}$.
Substituting the value of $e_0$:
$P_{av} = \frac{(NAB \omega)^2}{2R} = \frac{N^2 A^2 B^2 \omega^2}{2R}$.

(A) The magnetic flux through the coil at any time $t$ is given by $\phi = N A B \cos(\omega t)$.
According to Faraday's law,the induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = N A B \omega \sin(\omega t)$.
The peak $EMF$ is $e_0 = N A B \omega$.
The peak current is $i_0 = \frac{e_0}{R} = \frac{N A B \omega}{R}$.
The average power dissipated in a complete cycle is given by $P_{avg} = \frac{1}{2} e_0 i_0$.
Substituting the values,$P_{avg} = \frac{1}{2} (N A B \omega) \left( \frac{N A B \omega}{R} \right) = \frac{N^2 A^2 B^2 \omega^2}{2 R}$.

(C) Given: Number of turns $N = 1000$,Area $A = 0.10 \ m^2$,Frequency $\nu = 0.5 \ Hz$,Magnetic field $B = 0.01 \ T$.
The angular velocity $\omega$ is given by $\omega = 2 \pi \nu$.
$\omega = 2 \pi (0.5) = \pi \ rad/s$.
The maximum induced emf $\varepsilon_{\max}$ is given by the formula $\varepsilon_{\max} = N A B \omega$.
Substituting the values: $\varepsilon_{\max} = 1000 \times 0.10 \times 0.01 \times \pi$.
$\varepsilon_{\max} = 1 \times \pi = 3.14 \ V$.

(B) The induced electromotive force $(\varepsilon)$ in a rotating coil is given by the formula: $\varepsilon = N B A \omega \sin(\omega t)$.
For the maximum induced emf $(\varepsilon_{\max})$, we set $\sin(\omega t) = 1$.
Thus, the formula becomes: $\varepsilon_{\max} = N B A \omega$.
Given values:
Number of turns $(N)$ = $100$
Magnetic field strength $(B)$ = $0.3 \ T$
Area $(A)$ = $2.5 \ m^2$
Angular velocity $(\omega)$ = $60 \ rad \ s^{-1}$
Substituting these values into the formula:
$\varepsilon_{\max} = 100 \times 0.3 \times 2.5 \times 60$
$\varepsilon_{\max} = 30 \times 2.5 \times 60$
$\varepsilon_{\max} = 75 \times 60 = 4500 \ V$.
Converting to $kV$:
$\varepsilon_{\max} = 4.5 \ kV$.

(C) When a rectangular coil is rotated in a uniform magnetic field about an axis passing through its centre and perpendicular to the field,the magnetic flux $\phi$ associated with the coil is given by:
$\phi = B A \cos \theta = B A \cos \omega t$ ...$(i)$
According to Faraday's law of electromagnetic induction,the induced emf $e$ is given by:
$e = -\frac{d \phi}{d t} = -\frac{d}{d t} (B A \cos \omega t)$
$e = -B A \frac{d}{d t} (\cos \omega t)$
$e = -B A (-\omega \sin \omega t)$
$e = B A \omega \sin \omega t$
Let $e_0 = B A \omega$ be the peak value of the induced emf.
Then,$e = e_0 \sin \omega t$ ...(ii)
From equation (ii),it is clear that the induced emf varies sinusoidally with time.

(C) The magnetic flux $\phi$ linked with a coil rotating in a magnetic field $B$ is given by $\phi = BA \cos(\omega t)$.
According to Faraday's law of electromagnetic induction,the induced emf $e$ is given by $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$: $e = -\frac{d}{dt} (BA \cos(\omega t)) = BA\omega \sin(\omega t)$.
Using the trigonometric identity $\sin(\theta) = \cos(\theta - \frac{\pi}{2})$,we can write $e = BA\omega \cos(\omega t - \frac{\pi}{2})$.
Comparing the phase of the flux $(\omega t)$ and the phase of the induced emf $(\omega t - \frac{\pi}{2})$,the phase difference is $\frac{\pi}{2}$.

(C) The magnetic field inside a long solenoid is given by $B = \mu_0 N I$. Substituting $I = I_0 \sin(\omega t)$,we get $B = \mu_0 N I_0 \sin(\omega t)$.
The square loop has its corners touching the solenoid,meaning the diagonal of the square is equal to the diameter of the solenoid. Let $l$ be the side of the square. Then,the diagonal $d = l\sqrt{2} = 2R$.
Thus,$l = \frac{2R}{\sqrt{2}} = R\sqrt{2}$.
The area of the square loop is $A = l^2 = (R\sqrt{2})^2 = 2R^2$.
The magnetic flux linked with the loop is $\phi = B \cdot A = (\mu_0 N I_0 \sin(\omega t)) \cdot (2R^2) = 2 \mu_0 N I_0 R^2 \sin(\omega t)$.
The induced emf is $e = -\frac{d\phi}{dt} = -\frac{d}{dt} [2 \mu_0 N I_0 R^2 \sin(\omega t)]$.
Taking the magnitude,$|e| = 2 \mu_0 N I_0 R^2 \omega \cos(\omega t)$.

(D) The magnetic flux linked with the coil is $\phi = NBA \cos(\omega t)$.
By Faraday's law,the induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.
The maximum $EMF$ is $e_{\max} = NBA\omega$.
The root-mean-square $(RMS)$ value of the induced $EMF$ is $e_{\text{rms}} = \frac{e_{\max}}{\sqrt{2}} = \frac{NBA\omega}{\sqrt{2}}$.
The average power loss due to Joule heating is given by $P_{\text{avg}} = \frac{e_{\text{rms}}^2}{R}$.
Rearranging for resistance $R$,we get $R = \frac{e_{\text{rms}}^2}{P_{\text{avg}}} = \frac{(NBA\omega)^2}{2 \times P_{\text{avg}}}$.
Substituting the given values: $N = 40$,$B = 0.05 \ T$,$A = 0.01 \ m^2$,$\omega = 50 \ rad \ s^{-1}$,and $P_{\text{avg}} = 25 \times 10^{-3} \ W$.
$R = \frac{(40 \times 0.05 \times 0.01 \times 50)^2}{2 \times 25 \times 10^{-3}} = \frac{(1)^2}{50 \times 10^{-3}} = \frac{1}{0.05} = 20 \ \Omega$.

(B) The maximum induced electromotive force $(emf)$ in a rotating coil is given by the formula:
$e_{\max} = N B A \omega$
Where:
$N = 50$ (number of turns)
$B = 2 \times 10^{-2} \,T$ (magnetic field strength)
$A = 200 \,cm^2 = 200 \times 10^{-4} \,m^2 = 2 \times 10^{-2} \,m^2$ (area of the coil)
$\omega = 40 \,rad/s$ (angular speed)
Substituting these values into the formula:
$e_{\max} = 50 \times (2 \times 10^{-2}) \times (2 \times 10^{-2}) \times 40$
$e_{\max} = 50 \times 4 \times 10^{-4} \times 40$
$e_{\max} = 200 \times 40 \times 10^{-4}$
$e_{\max} = 8000 \times 10^{-4} = 0.8 \,V$
Therefore,the maximum induced $emf$ is $0.8 \,V$.

(A) Given:
Number of turns in the circular coil $(N) = 100$
Area $(A) = 2 \times 10^{-2} \,m^2$
Magnetic field $(B) = 0.01 \,T$
Frequency of rotation $(f) = 50 \,Hz$
The maximum induced electromotive force (emf) in a rotating coil is given by the formula:
$e_{max} = N B A \omega$
Since angular frequency $\omega = 2 \pi f$, we have:
$e_{max} = N B A (2 \pi f)$
Substituting the given values:
$e_{max} = 100 \times 0.01 \times (2 \times 10^{-2}) \times 2 \times 3.14159 \times 50$
$e_{max} = 1 \times 0.02 \times 314.159$
$e_{max} = 0.02 \times 314.159 = 6.283 \,V$
Thus, the maximum emf produced is approximately $6.28 \,V$.