Physics: Class 12 (Book 1) — Chapter 6: Electromagnetic Induction

The discovery of electromagnetic induction was based on three seminal experiments:

• Experiment 6.1 (Coil and Magnet): A coil $C_1$ is connected to a sensitive galvanometer $G$. When the North pole of a bar magnet is pushed towards the coil, the galvanometer pointer deflects, indicating the induction of an electric current. The deflection lasts only as long as the magnet is in motion. Pulling the magnet away reverses the deflection. Moving the South pole produces opposite deflections. Faster motion yields larger deflections. Moving the coil while keeping the magnet stationary produces identical effects, proving that relative motion is the key driver.
• Experiment 6.2 (Coil and Coil with Relative Motion): The bar magnet is replaced by a second coil $C_2$ connected to a battery. The steady current in $C_2$ produces a steady magnetic field. When $C_2$ is moved towards or away from $C_1$, a current is induced in $C_1$, indicated by momentary galvanometer deflections.
• Experiment 6.3 (Coil and Coil with Stationary Setup): Coils $C_1$ and $C_2$ are both held stationary. $C_2$ is connected to a battery through a tapping key $K$. When $K$ is pressed, a momentary deflection is observed in $C_1$. When $K$ is held continuously, there is no deflection. When $K$ is released, a momentary deflection in the opposite direction occurs. Inserting an iron rod along the axis dramatically increases the deflection.

For a plane area $A$ placed in a uniform magnetic field $\mathbf{B}$, the magnetic flux $\Phi_B$ is defined as:

$$\Phi_B = \mathbf{B} \cdot \mathbf{A} = BA \cos \theta$$

where $\theta$ is the angle between the magnetic field vector $\mathbf{B}$ and the area vector $\mathbf{A}$. For non-uniform fields or curved surfaces, the flux is summed over all infinitesimal area elements $d\mathbf{A}_i$:

$$\Phi_B = \sum_{\text{all}} \mathbf{B}_i \cdot d\mathbf{A}_i$$

• SI Unit: weber ($\text{Wb}$) or tesla meter squared ($\text{T m}^2$).
• Physical Nature: Scalar quantity.

Faraday's law states that:

The magnitude of the induced electromotive force (emf) in a circuit is equal to the time rate of change of magnetic flux through the circuit.

$$\epsilon = -\frac{d\Phi_B}{dt}$$

For a closely wound coil consisting of $N$ turns, the total induced emf is given by:

$$\epsilon = -N \frac{d\Phi_B}{dt}$$

The negative sign indicates the direction of the induced emf (and current), which is determined by Lenz's Law.

Mathematical Problem Statement (Example 6.2): A square loop of side 10 cm and resistance 0.5 Ω is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s at a steady rate. Determine the magnitudes of induced emf and current during this time-interval.

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Step-by-Step Algorithmic Solution Trace:

1. Calculate the area of the square loop: $A = (0.10\text{ m})^2 = 10^{-2}\text{ m}^2$.
2. Identify the angle $\theta$ between the area vector (pointing North or South) and the magnetic field (pointing North-East): $\theta = 45^\circ$.
3. Calculate the initial magnetic flux $\Phi_{B\text{ (initial)}}$: $$\Phi_{B\text{ (initial)}} = BA \cos \theta = 0.10 \times 10^{-2} \times \cos 45^\circ = \frac{10^{-3}}{\sqrt{2}}\text{ Wb} \approx 0.707 \times 10^{-3}\text{ Wb}$$
4. The final flux is $\Phi_{B\text{ (final)}} = 0$.
5. Calculate the magnitude of the induced emf using Faraday's Law: $$\epsilon = \frac{|\Delta \Phi_B|}{\Delta t} = \frac{0.707 \times 10^{-3}\text{ Wb}}{0.70\text{ s}} \approx 1.0 \times 10^{-3}\text{ V} = 1.0\text{ mV}$$
6. Calculate the induced current using Ohm's Law: $$I = \frac{\epsilon}{R} = \frac{1.0 \times 10^{-3}\text{ V}}{0.5\ \Omega} = 2.0 \times 10^{-3}\text{ A} = 2.0\text{ mA}$$

Final Evaluation Output: Induced emf = 1.0 mV , Induced current = 2.0 mA

Mathematical Problem Statement (Example 6.3): A circular coil of radius 10 cm , 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through $180^\circ$ in 0.25 s . Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is $3.0 \times 10^{-5}\text{ T}$.

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Step-by-Step Algorithmic Solution Trace:

1. Calculate the area of the circular coil: $A = \pi r^2 = \pi \times (0.1\text{ m})^2 = \pi \times 10^{-2}\text{ m}^2$.
2. Calculate the initial flux: $\Phi_{B\text{ (initial)}} = BA \cos 0^\circ = 3.0 \times 10^{-5} \times \pi \times 10^{-2} = 3\pi \times 10^{-7}\text{ Wb}$.
3. Calculate the final flux after $180^\circ$ rotation: $\Phi_{B\text{ (final)}} = BA \cos 180^\circ = -3\pi \times 10^{-7}\text{ Wb}$.
4. Calculate the change in flux: $\Delta \Phi_B = \Phi_{B\text{ (final)}} - \Phi_{B\text{ (initial)}} = -6\pi \times 10^{-7}\text{ Wb}$.
5. Calculate the induced emf for $N = 500$ turns: $$\epsilon = N \frac{|\Delta \Phi_B|}{\Delta t} = 500 \times \frac{6\pi \times 10^{-7}}{0.25} = 2000 \times 6\pi \times 10^{-7} \approx 3.8 \times 10^{-3}\text{ V}$$
6. Calculate the induced current: $$I = \frac{\epsilon}{R} = \frac{3.8 \times 10^{-3}\text{ V}}{2\ \Omega} = 1.9 \times 10^{-3}\text{ A}$$

Final Evaluation Output: Induced emf = 3.8 mV , Induced current = 1.9 mA

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.

This law explains the negative sign in Faraday's equation: $\epsilon = -\frac{d\Phi_B}{dt}$.

Lenz's law is a direct consequence of the law of conservation of energy:

• When the North pole of a magnet is pushed towards a coil, the induced current flows in a counter-clockwise direction (viewed from the magnet side), creating a North pole that repels the incoming magnet.
• To keep moving the magnet, an external agent must perform mechanical work to overcome this repulsive force.
• This mechanical work is converted into electrical energy, which is then dissipated as heat via Joule heating ($I^2 R$).
• If the induced current flowed in the opposite direction (attracting the magnet), a gentle push would cause the magnet to accelerate indefinitely, continuously increasing its kinetic energy and electrical energy without any external work, violating the law of conservation of energy.

Consider a rectangular conductor $PQRS$ in a uniform magnetic field $\mathbf{B}$ perpendicular to the plane of the system. Let the arm $PQ$ of length $l$ slide to the left with velocity $v$.

1. The area enclosed by the loop is $A = lx$, where $x$ is the width of the loop.
2. The magnetic flux enclosed is: $\Phi_B = Blx$.
3. Since $x$ changes with time, the rate of change of flux is: $$\epsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Blx) = -Bl\frac{dx}{dt}$$
4. Since the rod moves to the left, its speed is $v = -\frac{dx}{dt}$. Substituting this yields: $$\epsilon = Blv$$

The same expression can be derived by considering the Lorentz force acting on the free charge carriers (charge $q$) inside the moving conductor $PQ$:

1. As the rod moves with velocity $v$ in a magnetic field $B$, each charge experiences a magnetic force: $F = qvB$.
2. The work done in moving a charge from end $P$ to $Q$ is: $W = F \cdot l = qvBl$.
3. Since emf is defined as work done per unit charge: $$\epsilon = \frac{W}{q} = Blv$$

Mathematical Problem Statement (Example 6.6): A metallic rod of 1 m length is rotated with a frequency of 50 rev/s , with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m , about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

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Step-by-Step Algorithmic Solution Trace:

1. Identify the given parameters: length $R = 1\text{ m}$, frequency $\nu = 50\text{ Hz}$, magnetic field $B = 1\text{ T}$.
2. Calculate the angular velocity $\omega$: $$\omega = 2\pi \nu = 2\pi \times 50 = 100\pi\text{ rad/s}$$
3. The linear velocity of an infinitesimal element $dr$ at distance $r$ from the center is $v = \omega r$.
4. Integrate the motional emf $d\epsilon = Bv\,dr$ from $r = 0$ to $r = R$: $$\epsilon = \int_0^R B(\omega r)\,dr = B\omega \left[ \frac{r^2}{2} \right]_0^R = \frac{1}{2} B\omega R^2$$
5. Substitute the values: $$\epsilon = \frac{1}{2} \times 1\text{ T} \times 100\pi\text{ rad/s} \times (1\text{ m})^2 = 50\pi\text{ V} \approx 50 \times 3.14 = 157\text{ V}$$

Final Evaluation Output: Induced emf = 157 V

When two coils are placed close to each other, the magnetic flux linked with coil 1 ($\Phi_1$) is proportional to the current in coil 2 ($I_2$):

$$N_1\Phi_1 = M_{12} I_2$$

where $M_{12}$ is the mutual inductance of coil 1 with respect to coil 2. By reciprocity, $M_{12} = M_{21} = M$.

For two long coaxial solenoids of length $l$, inner radius $r_1$, and turn densities $n_1, n_2$:

$$M = \mu_0 n_1 n_2 \pi r_1^2 l$$

The induced emf in coil 1 due to a changing current in coil 2 is:

$$\epsilon_1 = -M \frac{dI_2}{dt}$$

When the current in a single isolated coil changes, the flux linked with the coil itself changes, inducing a back emf. The flux linkage is proportional to the current:

$$N\Phi_B = L I$$

where $L$ is the self-inductance of the coil. The induced back emf is:

$$\epsilon = -L \frac{dI}{dt}$$

For a long solenoid of cross-sectional area $A$, length $l$, and turn density $n$:

$$L = \mu_0 n^2 A l$$

If filled with a material of relative permeability $\mu_r$:

$$L = \mu_r \mu_0 n^2 A l$$

Work must be done against the back emf to establish a current $I$ in an inductor. This work is stored as magnetic potential energy:

$$W = \frac{1}{2} L I^2$$

The magnetic energy stored per unit volume (magnetic energy density $u_B$) inside a solenoid is:

$$u_B = \frac{B^2}{2\mu_0}$$

An armature coil of $N$ turns and area $A$ is rotated with a constant angular speed $\omega$ in a uniform magnetic field $\mathbf{B}$.

1. The angle between the magnetic field vector $\mathbf{B}$ and the area vector $\mathbf{A}$ at any instant $t$ is $\theta = \omega t$.
2. The magnetic flux linked with the coil at time $t$ is: $$\Phi_B = BA \cos(\omega t)$$
3. According to Faraday's Law, the induced emf is: $$\epsilon = -N \frac{d\Phi_B}{dt} = -N \frac{d}{dt}[BA \cos(\omega t)] = NBA\omega \sin(\omega t)$$
4. The maximum (peak) emf is $\epsilon_0 = NBA\omega$. Thus, the instantaneous emf is: $$\epsilon = \epsilon_0 \sin(\omega t)$$

The induced emf varies sinusoidally with time, producing an alternating current.

Mathematical Problem Statement (Example 6.10): Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m² . The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil?

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Step-by-Step Algorithmic Solution Trace:

1. Identify the given parameters: $N = 100$, $A = 0.10\text{ m}^2$, $B = 0.01\text{ T}$, frequency $\nu = 0.5\text{ Hz}$.
2. Calculate the angular frequency $\omega$: $$\omega = 2\pi \nu = 2\pi \times 0.5 = \pi\text{ rad/s} \approx 3.14\text{ rad/s}$$
3. Calculate the maximum voltage $\epsilon_0$: $$\epsilon_0 = NBA\omega = 100 \times 0.01 \times 0.10 \times 3.14 = 0.314\text{ V}$$

Final Evaluation Output: Maximum voltage = 0.314 V