Area $A = (0.1 \text{ m})^2 = 0.01 \text{ m}^2$. Since the loop is perpendicular to the field, angle $\theta = 0^\circ$ (between normal and field).
$\Phi_B = B A \cos(0^\circ) = 0.5 \times 0.01 \times 1 = \mathbf{0.005 \text{ Wb}}$.
Area $A = \pi (0.05)^2 = 0.0025\pi \text{ m}^2$. Angle $\theta = 60^\circ$.
$\Phi_B = B A \cos(60^\circ) = 0.2 \times (0.0025\pi) \times 0.5 = \mathbf{0.00025\pi \text{ Wb}} \approx \mathbf{7.85 \times 10^{-4} \text{ Wb}}$.
1st Law: A changing magnetic flux linked with a coil induces an EMF in it.
2nd Law: The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux. $e = -N\frac{d\Phi}{dt}$.
$|e| = \left| \frac{\Phi_{final} - \Phi_{initial}}{\Delta t} \right| = \left| \frac{2 - 10}{0.1} \right| = \left| \frac{-8}{0.1} \right| = \mathbf{80 \text{ V}}$.
$e = -N \frac{\Delta\Phi}{\Delta t} = -100 \times \frac{0 - 5 \times 10^{-3}}{0.05} = 100 \times \frac{0.005}{0.05} = 100 \times 0.1 = \mathbf{10 \text{ V}}$.
Pushing the magnet changes the magnetic flux linked with the coil. According to Faraday's law, this changing flux induces an EMF, which drives a momentary induced current through the galvanometer.
$|e| = \left| \frac{d\Phi}{dt} \right| = \frac{d}{dt}(3t^2 + 4t + 9) = 6t + 4$.
At $t = 2$ s, $|e| = 6(2) + 4 = 12 + 4 = \mathbf{16 \text{ V}}$.
The SI unit is **Weber (Wb)**. It is a **scalar** quantity.
If the speed is doubled, the time taken for the flux change ($\Delta t$) is halved. Since $e \propto 1/\Delta t$, the induced EMF is **doubled**.
If the loop is parallel to the field, the area vector (normal) is strictly perpendicular to the field lines ($\theta = 90^\circ$). $\Phi = BA\cos(90^\circ) = \mathbf{0 \text{ Wb}}$.
The polarity of the induced EMF is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
The induced current opposes the motion of the magnet (due to magnetic repulsion/attraction). Mechanical work must be done against this opposing force to move the magnet. This mechanical work is converted into the electrical energy of the induced current.
To oppose the approaching North pole, the face of the loop must act as a North pole. Therefore, the current will flow in an **anti-clockwise** direction.
As the North pole approaches from above, the ring acts as a North pole to repel it. Thus, the current as viewed from above is **anti-clockwise**.
**Less than $g$**. The induced current creates a magnetic field that opposes the downward motion of the magnet, providing an upward repelling force, hence reducing its net downward acceleration.
The induced EMF depends only on the rate of change of flux ($d\Phi/dt$), not on the material. Therefore, the induced EMF will be **the same** in both loops.
Since $I = E/R$, the loop with lower resistance will have a higher current. Therefore, the induced current will be greater in the **copper** loop.
**Yes**. As the wire falls vertically, it cuts the horizontal component of the Earth's magnetic field, inducing a motional EMF across its ends.
Pulling the loop out decreases the flux. By Lenz's law, the induced current creates a field to oppose this decrease, resulting in an attractive magnetic force pulling the loop back in. Mechanical work is needed to overcome this inward force.
**Yes**. Resistance $R = \rho L / A_{wire}$. A larger wire cross-sectional area lowers the resistance. Since $I = E/R$, a lower resistance results in a higher induced current for the same induced EMF.
Consider a rod of length $l$ moving with velocity $v$ in a perpendicular magnetic field $B$. The area swept in time $dt$ is $dA = l v dt$. Change in flux $d\Phi = B dA = B l v dt$. From Faraday's law, $|E| = d\Phi/dt = B l v$.
$E = Blv = 0.2 \times 0.5 \times 4 = \mathbf{0.4 \text{ V}}$.
Velocity $v = 360 \text{ km/h} = 360 \times \frac{5}{18} = 100 \text{ m/s}$. The wings cut the vertical component $B_v$.
$E = B_v l v = (4 \times 10^{-5}) \times 40 \times 100 = \mathbf{0.16 \text{ V}}$.
$E = \frac{1}{2} B \omega L^2$
Angular velocity $\omega = 120 \text{ rev/min} = \frac{120 \times 2\pi}{60} = 4\pi \text{ rad/s}$.
$E = \frac{1}{2} B \omega l^2 = \frac{1}{2} \times 0.5 \times 4\pi \times (0.8)^2 = 0.64\pi \text{ V} \approx \mathbf{2.01 \text{ V}}$.
**Fleming’s Right-Hand Rule**. (Thumb points to motion, forefinger to magnetic field, middle finger gives direction of induced current).
Induced current $I = \frac{Blv}{R}$. Magnetic force $F_m = IlB = \frac{B^2 l^2 v}{R}$. Mechanical power to balance this force $P = F_m v = \mathbf{\frac{B^2 l^2 v^2}{R}}$.
Since $E = Blv$, the motional EMF is directly proportional to velocity. If velocity is halved, the EMF is **halved**.
**Yes**. The horizontal axle of the train moves North-South, slicing through the **vertical component** ($B_v$) of the Earth's magnetic field, inducing an EMF.
It is fundamentally caused by the **magnetic Lorentz force**. The free electrons in the conductor move with velocity $v$, experiencing a magnetic force $F = q(\vec{v} \times \vec{B})$ that pushes them to one end, creating a potential difference.
Eddy currents are circulating loops of electrical current induced within bulk pieces of conductors. They are produced whenever the bulk conductor is subjected to a changing magnetic flux.
1. They cause severe power loss in the form of heat ($I^2R$ heating).
2. The excessive heating can physically damage the insulation of the windings in the machine.
By using **laminated cores**. The core is made of thin metallic sheets coated with insulating varnish. This severely restricts the available area for the eddy currents to circulate, heavily reducing their magnitude.
Strong electromagnets are activated near the metallic rails. The relative motion induces massive eddy currents in the rails, which, by Lenz's law, create an opposing magnetic field that provides smooth, frictionless braking.
High-frequency alternating currents are passed through a coil surrounding the metal to be melted. This induces very strong eddy currents inside the metal, producing immense Joule heat that melts the metal quickly.
The slots break up the continuous surface area of the metal plate. This increases the electrical resistance for the circulating loops and severely restricts the formation of large eddy currents, preventing unwanted excessive damping when not needed.
The **solid copper cube** will come to rest faster. It allows larger eddy currents to flow compared to the laminated block, resulting in a stronger opposing magnetic force (higher electromagnetic damping).
**Yes**. They always circulate in a direction such that their own magnetic field strongly opposes the change in magnetic flux that created them.
**Diathermy** (deep heat therapy in medicine) or **Analog Speedometers/Energy meters** in older vehicles and homes.
Since $I = E/R$, an increase in the electrical resistance ($R$) of the bulk metal will proportionally **decrease** the magnitude of the circulating eddy currents.
Self-inductance is the property of a coil by virtue of which it opposes any change in the strength of current flowing through it by inducing an opposing EMF. SI Unit: **Henry (H)**.
$|e| = L \frac{\Delta I}{\Delta t} \implies 200 = L \left( \frac{5 - 0}{0.1} \right) \implies 200 = L (50) \implies L = \frac{200}{50} = \mathbf{4 \text{ H}}$.
$L = \frac{\mu_0 N^2 A}{l}$
Area $A = \pi (0.02)^2 = 0.0004\pi \text{ m}^2$. Length $l = 0.5 \text{ m}$. $N = 500$.
$L = \frac{4\pi \times 10^{-7} \times (500)^2 \times 0.0004\pi}{0.5} \approx \mathbf{7.9 \times 10^{-4} \text{ H}}$ (or $0.79 \text{ mH}$).
Mutual inductance is the property of a pair of coils due to which an EMF is induced in one coil when the current in the neighboring coil changes.
Magnitude of induced EMF $e_2 = M \frac{dI_1}{dt} = 1.5 \times 4 = \mathbf{6 \text{ V}}$.
$U = \frac{1}{2} L I^2 = \frac{1}{2} \times (50 \times 10^{-3}) \times (2)^2 = 25 \times 10^{-3} \times 4 = 100 \times 10^{-3} = \mathbf{0.1 \text{ Joules}}$.
It depends on their number of turns ($N_1, N_2$), common cross-sectional area, length, and the magnetic permeability of the core material filling them.
It **increases** the self-inductance enormously because soft iron has a very high relative magnetic permeability ($\mu_r$), which multiplies the internal magnetic flux.
Just as mechanical mass (inertia) opposes any change in the state of motion of a body, self-inductance directly opposes any change in the flow of electrical current in a circuit.
It works strictly on the principle of **Electromagnetic Induction**. When a coil rotates in a magnetic field, the magnetic flux linked with it continuously changes, inducing an alternating EMF.
$e = NAB\omega \sin(\omega t)$. Where $N$ is number of turns, $A$ is area, $B$ is magnetic field, and $\omega$ is the uniform angular velocity.
Frequency $f = 50 \text{ Hz} \implies \omega = 2\pi f = 100\pi \text{ rad/s}$.
Peak EMF $E_0 = NAB\omega = 100 \times 0.05 \times 0.2 \times 100\pi = 1 \times 100\pi \approx \mathbf{314 \text{ V}}$.
They provide a continuous, sliding electrical connection between the rotating armature coil and the stationary external circuit, allowing the alternating current to be tapped off without twisting wires.
It is maximum when the plane of the coil is exactly **parallel** to the magnetic field. (At this point, the rate of change of flux is highest, even though the flux itself is zero).
Since $E_0 = NAB\omega$, the peak EMF is directly proportional to $\omega$. If the angular speed is doubled, the peak EMF is exactly **doubled**.
It is a mathematically pure **Sine wave** (or Cosine wave).
A **Split-ring commutator** (two separate half-rings).
$\omega = 2\pi(50) = 100\pi$. The equation is: $e(t) = 314 \sin(100\pi t)$.
**Yes**. The positive half-cycle exactly mirrors and cancels out the negative half-cycle over one full rotation ($T = 2\pi/\omega$), making the integral/average mathematically zero.
