Faraday's and Lenz's Law Questions in English

(B) Given,$B = 2t + 4t^{2}$.
At $t = 0 \ s$,$B_{1} = 2(0) + 4(0)^{2} = 0 \ T$.
At $t = 2 \ s$,$B_{2} = 2(2) + 4(2)^{2} = 4 + 16 = 20 \ T$.
The magnetic flux is $\phi = B \cdot A = B \cdot \pi r^{2}$.
The change in magnetic flux is $\Delta \phi = \phi_{2} - \phi_{1} = \pi r^{2} (B_{2} - B_{1})$.
Substituting the values,$\Delta \phi = \pi r^{2} (20 - 0) = 20 \pi r^{2} \ Wb$.
The induced charge $Q$ is given by $Q = \frac{\Delta \phi}{R}$.
Therefore,$Q = \frac{20 \pi r^{2}}{R} \ C$.

(D) According to the right-hand thumb rule,if the current in the wire flows upward,the magnetic field lines in the region of the coil are directed into the plane of the paper.
If the current in the wire flows downward,the magnetic field lines in the region of the coil are directed out of the plane of the paper.
For a clockwise induced current in the loop,the induced magnetic field must be directed into the plane of the paper (by the right-hand grip rule).
According to Lenz's law,the induced current opposes the change in magnetic flux.
Case $1$: If the current is flowing upward and increasing,the magnetic field into the plane increases,so the induced current will be counter-clockwise to oppose it. This does not match.
Case $2$: If the current is flowing upward and decreasing,the magnetic field into the plane decreases,so the induced current will be clockwise to support it.
Case $3$: If the current is flowing downward and increasing,the magnetic field out of the plane increases,so the induced current will be counter-clockwise to oppose it.
Case $4$: If the current is flowing downward and decreasing,the magnetic field out of the plane decreases,so the induced current will be clockwise to support it.
Since the current must be time-dependent to induce a current,and both decreasing upward current and decreasing downward current can produce a clockwise induced current,the most appropriate general description is that the current is time-dependent.

(A) Given the magnetic flux $\phi = 4t^2 + 6t + 9 \text{ Wb}$.
According to Faraday's law of electromagnetic induction, the induced emf $\varepsilon$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Taking the magnitude, $\varepsilon = \left| \frac{d\phi}{dt} \right|$.
Differentiating $\phi$ with respect to $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(4t^2 + 6t + 9) = 8t + 6$.
At $t = 2 \text{ s}$, the induced emf is:
$\varepsilon = 8(2) + 6 = 16 + 6 = 22 \text{ V}$.

(B) Given: Magnetic flux $\phi = 5t^2 - 4t + 1 \text{ Wb}$ and resistance $R = 10 \Omega$.
According to Faraday's law of induction, the induced electromotive force $(EMF)$ is $e = -\frac{d\phi}{dt}$.
Differentiating $\phi$ with respect to $t$: $\frac{d\phi}{dt} = \frac{d}{dt}(5t^2 - 4t + 1) = 10t - 4$.
The induced current $I$ is given by $I = \frac{|e|}{R} = \frac{|-d\phi/dt|}{R} = \frac{|-(10t - 4)|}{10} = \frac{|4 - 10t|}{10}$.
At $t = 0.2 \text{ s}$, the current is $I = \frac{|4 - 10(0.2)|}{10} = \frac{|4 - 2|}{10} = \frac{2}{10} = 0.2 \text{ A}$.

(B, D) According to Faraday's law of electromagnetic induction, an emf is induced in a loop when the magnetic flux $\Phi = B \cdot A \cos \theta$ linked with the loop changes with time.
$1$. If the loop is moved along the direction of $B$, the magnetic field $B$, area $A$, and the angle $\theta$ remain constant. Thus, the flux remains constant, and no emf is induced.
$2$. If the loop is squeezed to a smaller area, the area $A$ changes with time. Therefore, the magnetic flux $\Phi$ changes, and an emf is induced.
$3$. If the loop is rotated about its axis, the orientation of the area vector relative to the magnetic field remains constant $(\theta = 0^\circ)$. Thus, the flux remains constant, and no emf is induced.
$4$. If the loop is rotated about one of its diameters, the angle $\theta$ between the magnetic field $B$ and the area vector changes with time. Therefore, the magnetic flux $\Phi$ changes, and an emf is induced.
Since both $(b)$ and $(d)$ result in a change in magnetic flux, both conditions will induce an emf.

(B) The magnetic field $B$ at a distance $x$ from an infinitely long wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2 \pi x}$.
As the current $I$ in the wire is suddenly halved, the magnetic flux $\phi$ linked with the square loop decreases.
According to Lenz's law, the induced current in the loop will flow in such a direction as to oppose this decrease in magnetic flux. This means the induced current will create a magnetic field in the same direction as the original magnetic field.
For this to happen, the induced current in the loop must flow in the clockwise direction.
Now, consider the forces on the sides of the loop:
$1$. The side of the loop closer to the wire (let's call it $CD$) carries current in the same direction as the main wire. Thus, it experiences an attractive force $F_{CD}$ towards the wire.
$2$. The side of the loop farther from the wire (let's call it $AB$) carries current in the opposite direction to the main wire. Thus, it experiences a repulsive force $F_{AB}$ away from the wire.
Since the magnetic field is stronger closer to the wire, the attractive force on the closer side is greater than the repulsive force on the farther side $(F_{CD} > F_{AB})$.
Therefore, the net force on the loop is directed towards the wire, and the loop will move towards the wire.

(A) The magnetic field inside the long solenoid is given by $B = \mu_0 n I$,where $n = 500 \ turns/cm = 50000 \ turns/m$.
Given $I = 10 \sin(\omega t)$,so $B = \mu_0 n (10 \sin(\omega t))$.
The magnetic flux through the loop of radius $r = 1 \ cm = 0.01 \ m$ is $\phi = B \cdot A = \mu_0 n (10 \sin(\omega t)) \cdot (\pi r^2)$.
The induced $EMF$ is $\varepsilon = -\frac{d\phi}{dt} = -\mu_0 n \pi r^2 (10 \omega \cos(\omega t))$.
The magnitude of induced current is $i = \frac{|\varepsilon|}{R} = \frac{\mu_0 n \pi r^2 (10 \omega \cos(\omega t))}{R}$.
The peak current is $i_0 = \frac{\mu_0 n \pi r^2 (10 \omega)}{R}$.
Substituting values: $\mu_0 = 4\pi \times 10^{-7} \ T\cdot m/A$,$n = 5 \times 10^4 \ m^{-1}$,$r = 10^{-2} \ m$,$\omega = 10^3 \ rad/s$,$R = 10 \ \Omega$.
$i_0 = \frac{(4\pi \times 10^{-7}) \times (5 \times 10^4) \times \pi \times (10^{-2})^2 \times 10 \times 10^3}{10} = 20 \pi^2 \times 10^{-6} \ A$.
$i_0 = 20 \times (9.8696) \times 10^{-6} \ A \approx 197.39 \ \mu A$.
The r.m.s. current is $i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{197.39}{\sqrt{2}} \ \mu A$.
Thus,$\alpha \approx 197$.

(D) The radius of the circular loop is $r = 7 \ cm = 0.07 \ m$. The area of the circular loop is $A_1 = \pi r^2 = \pi (0.07)^2 = 0.0049 \pi \ m^2$.
The circumference of the loop is $C = 2 \pi r = 2 \pi (0.07) = 0.14 \pi \ m$.
When converted to a square loop,the perimeter remains the same. Let the side of the square be $a$. Then $4a = 0.14 \pi$,so $a = 0.035 \pi \ m$.
The area of the square loop is $A_2 = a^2 = (0.035 \pi)^2 = 0.001225 \pi^2 \ m^2$.
The change in magnetic flux is $\Delta \phi = B(A_1 - A_2) = 0.2 \times (0.0049 \pi - 0.001225 \pi^2)$.
Using $\pi \approx 3.14159$,$A_1 \approx 0.01539 \ m^2$ and $A_2 \approx 0.01208 \ m^2$.
$\Delta \phi = 0.2 \times (0.01539 - 0.01208) = 0.2 \times 0.00331 = 0.000662 \ Wb$.
The induced $EMF$ is $\epsilon = \frac{|\Delta \phi|}{\Delta t} = \frac{0.000662}{0.5} = 0.001324 \ V$.
Converting to $mV$,$\epsilon = 1.324 \ mV \approx 1.32 \ mV$.

(B) The magnetic field at the center of $C_2$ due to $C_1$ (anti-clockwise) is directed towards the right,and the magnetic field due to $C_3$ (clockwise) is directed towards the left. Since the coils are identical and $C_2$ is midway,the net magnetic field at $C_2$ is zero.
If $C_1$ moves towards $C_2$,the magnetic field from $C_1$ at $C_2$ increases (directed right). To oppose this,$C_2$ induces a current to create a magnetic field towards the left (clockwise).
If $C_3$ moves away from $C_2$,the magnetic field from $C_3$ at $C_2$ decreases (directed left). To oppose this decrease,$C_2$ induces a current to create a magnetic field towards the left (clockwise).
Thus,if $C_1$ moves towards $C_2$ and $C_3$ moves away from $C_2$,the net flux change induces a clockwise current in $C_2$.

According to Lenz's law, the induced current creates a magnetic field that opposes the change in magnetic flux.
In figure (a), the magnetic field is directed into the page $(\times)$ and is increasing. To oppose this increase, the induced current must create a magnetic field directed out of the page $(\bullet)$. According to the right-hand rule, an outward magnetic field corresponds to an anticlockwise current.
In figure (b), the magnetic field is directed out of the page $(\bullet)$ and is decreasing. To oppose this decrease, the induced current must create a magnetic field into the page $(\times)$. According to the right-hand rule, this corresponds to a clockwise current.
Therefore, the direction of current in ring (a) is anticlockwise and in ring (b) is clockwise.