Axiom Physics | Class 12 - Faraday's & Lenz's Laws

⚡ 01. The Spark (Analogy)

Socho ek aisi spring ko jo badlav (change) se sakht nafrat karti hai: jab aap use andar push karte hain toh woh aapko bahar dhakelti hai, aur jab aap use bahar khinchte hain toh woh aapko wapas kheenchti hai. Lenz's Law electrical circuits ka bilkul wahi universal inertia hai! Jab kisi coil ke paas magnet ka North pole aage badhta hai, toh coil turant apne samne North pole bana leti hai taaki use repel kar sake. Jab magnet door jaata hai, toh coil turant South pole bankar use attract karne lagti hai. Yeh virodh (opposition) koi accidental cheez nahi hai; agar yeh virodh na hota, toh bina koi mechanical work kiye free electricity create ho jaati, jo Conservation of Energy ke khilaaf hoti!

📐 02. The Core Law & Formulae

1. Magnetic Flux ($\Phi_B$)

Dot product of magnetic field and area vector element:

$$\Phi_B = \int \vec{B}\cdot d\vec{A} = B A \cos\theta$$

SI Unit: $\text{Weber (Wb)} = \text{Tesla}\cdot\text{m}^2 = \text{V}\cdot\text{s}$. Scalar quantity (can be positive, negative, or zero).

2. Faraday-Lenz Law of Induction

Induced EMF equals the negative time derivative of magnetic flux:

$$\mathcal{E} = -N \frac{d\Phi_B}{dt}, \quad I = \frac{\mathcal{E}}{R} = -\frac{N}{R}\frac{d\Phi_B}{dt}$$

The negative sign mathematically embodies Lenz's Law. Induced charge flow $\Delta q = \frac{\Delta \Phi_B}{R}$ is independent of time!

03. Interactive Visualizer: Faraday Induction & Lenz Opposition Engine

Magnet Position ($x$) -14.0 cm

Motion Velocity ($v = dx/dt$) Manual Drag Mode

Coil Turns ($N$) 100 Turns

Total Flux ($\Phi_B$):

2.45 mWb

Induced EMF ($\mathcal{E} = -d\Phi/dt$):

0.00 mV

Coil Face Polarity:

Neutral (No Motion)

Mechanical Interaction:

Zero Resistance

📝 04. Stepwise Derivations Blueprint

Fundamental Law

Mathematical Derivation of Faraday's Law & Induced Charge Independence

Fig 6.1A: North pole of bar magnet plunging towards coil; induced current creates a North face to resist the increase in flux.

1. Consider a closed loop of resistance $R$ linked with magnetic flux $\Phi_B = \vec{B}\cdot\vec{A}$.

2. By Faraday's second law of electromagnetic induction, the magnitude of induced EMF is proportional to the time rate of change of magnetic flux:

$$\mathcal{E} \propto \frac{d\Phi_B}{dt}$$

3. Incorporating Lenz's law, which states that the polarity of induced EMF always opposes the change in flux producing it:

$$\mathcal{E} = -N \frac{d\Phi_B}{dt}$$

4. By Ohm's law, the induced instantaneous current $I$ in a circuit of total resistance $R$ is:

$$I = \frac{\mathcal{E}}{R} = -\frac{N}{R} \frac{d\Phi_B}{dt}$$

5. The infinitesimal charge $dq$ that flows through the circuit in time interval $dt$:

$$dq = I \, dt = \left( -\frac{N}{R} \frac{d\Phi_B}{dt} \right) dt = -\frac{N}{R} d\Phi_B$$

6. Total induced charge $\Delta q$ when flux changes from $\Phi_1$ to $\Phi_2$:

$$\Delta q = \int dq = \frac{N}{R} |\Delta \Phi_B| = \frac{N (\Phi_1 - \Phi_2)}{R}$$

Vital Conclusion: While induced EMF ($\mathcal{E}$) and induced current ($I$) depend inversely on the time taken ($\Delta t$), the total charge passed ($\Delta q$) depends strictly on total change in flux ($\Delta \Phi$) and resistance ($R$), and is completely independent of time or rate of motion!

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Falling Magnet Through a Conducting Tube

*Board Question:* A bar magnet is dropped vertically through a long copper pipe. What is its acceleration?
• As it falls, increasing flux induces eddy currents in the pipe wall which exert an upward opposing magnetic force on the falling magnet.
• Hence, acceleration is strictly less than $g$ ($a < g$).
• Eventually, upward magnetic drag equals downward gravity ($mg$), and the magnet falls with a constant terminal velocity ($a = 0$)!

Trap 2: Magnet Dropped Through a Cut Ring

*Exam Trap:* What happens if the ring has a small cut (open circuit)?
• Change in magnetic flux still induces an EMF across the cut ends ($\mathcal{E} \ne 0$).
• However, because the circuit is broken, no continuous current can flow ($I = 0$).
• No opposing magnetic pole is developed; thus no retarding magnetic force acts on the magnet.
• The magnet falls freely with acceleration exactly equal to gravity: $a = g$.

Trap 3: Slow vs Fast Withdrawal Charge Transfer

*Conceptual Trap:* A magnet is pulled out of a coil rapidly in Case A, and very slowly in Case B. In which case is more charge transferred?
*Correct Answer:* EQUAL IN BOTH CASES! $$\Delta q = \frac{\Delta \Phi}{R}$$ Rapid withdrawal produces large $\mathcal{E}$ and large $I$ for a short time, while slow withdrawal produces small $I$ over a longer time. The total charge $\Delta q$ is identical.

Trap 4: Lenz's Law and Conservation of Energy

If induced current assisted the motion instead of opposing it, a gentle push would accelerate the magnet perpetually without consuming energy, violating the First Law of Thermodynamics. The mechanical work done by the external agent against the repulsive/attractive magnetic force is what transforms into electrical energy, and ultimately into Joule heat ($I^2Rt$).

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A square loop of side $10\text{ cm}$ and resistance $0.5\text{ }\Omega$ is placed vertically in the east-west plane. A uniform magnetic field of $0.10\text{ T}$ is set up across the plane in the north-east direction. The magnetic field is decreased to zero in $0.70\text{ s}$ at a steady rate. Determine the magnitudes of induced EMF and current during this time-interval.

Solution:

1. Area of the square loop $A = (0.10\text{ m})^2 = 1.0 \times 10^{-2}\text{ m}^2$.

2. The plane of the loop is in the east-west plane, so its normal points North. Field is in the North-East direction, making angle $\theta = 45^\circ$ with the area vector.

3. Initial magnetic flux $\Phi_1$:

$$\Phi_1 = B A \cos 45^\circ = (0.10) \times (1.0 \times 10^{-2}) \times \frac{1}{\sqrt{2}} = \frac{10^{-3}}{\sqrt{2}}\text{ Wb} \approx 0.707 \times 10^{-3}\text{ Wb}$$

4. Final magnetic flux $\Phi_2 = 0$. Change in flux $\Delta \Phi = \Phi_2 - \Phi_1 = -0.707 \times 10^{-3}\text{ Wb}$.

5. Induced EMF magnitude:

$$\mathcal{E} = \frac{|\Delta \Phi|}{\Delta t} = \frac{0.707 \times 10^{-3}\text{ Wb}}{0.70\text{ s}} \approx 1.01 \times 10^{-3}\text{ V} \approx 1.0\text{ mV}$$

6. Induced current:

$$I = \frac{\mathcal{E}}{R} = \frac{1.01 \times 10^{-3}\text{ V}}{0.5\text{ }\Omega} \approx 2.02 \times 10^{-3}\text{ A} \approx 2.0\text{ mA}$$

JEE/NEET Competitive Trap

Problem: The magnetic flux linked with a coil of resistance $10\text{ }\Omega$ varies with time according to the equation $\Phi_B = (6t^2 - 5t + 1)\text{ Wb}$. Find the magnitude of induced current at $t = 0.25\text{ s}$ and determine the total charge that passes through the coil between $t = 0$ and $t = 2\text{ s}$.

Solution:

1. Instantaneous induced EMF by differentiation:

$$\mathcal{E} = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(6t^2 - 5t + 1) = -(12t - 5)\text{ Volts}$$

2. At $t = 0.25\text{ s}$:

$$\mathcal{E}(0.25) = -(12 \times 0.25 - 5) = -(3 - 5) = +2\text{ Volts}$$ $$I = \frac{|\mathcal{E}|}{R} = \frac{2\text{ V}}{10\text{ }\Omega} = 0.2\text{ A}$$

3. Total charge transferred between $t = 0$ and $t = 2\text{ s}$:

$$\Phi_i(t=0) = 6(0)^2 - 5(0) + 1 = 1\text{ Wb}$$ $$\Phi_f(t=2) = 6(2)^2 - 5(2) + 1 = 24 - 10 + 1 = 15\text{ Wb}$$

$$\Delta q = \frac{|\Phi_f - \Phi_i|}{R} = \frac{|15 - 1|}{10} = \frac{14}{10} = 1.4\text{ Coulombs}$$

🗂️ 07. Quick Revision Flashcard Core Formulae

Magnetic Flux

$\Phi_B = BA\cos\theta$

Faraday's Law

$\mathcal{E} = -N\frac{d\Phi_B}{dt}$

Induced Charge

$\Delta q = \frac{\Delta\Phi_B}{R}$ (Time Indep.)

Lenz Origin

Conservation of Energy