Movie Hook: The Power Grid in Oppenheimer
When you see massive electrical explosions in movies like Oppenheimer’s atomic energy facilities, there’s real physics behind the sparks. When you suddenly break a circuit carrying large current, the self inductance of coils tries to maintain that current. This creates enormous voltage spikes that can jump across air gaps - those dramatic electrical arcs! Power grids use careful control of inductance to prevent such catastrophic failures. Inductance is like electromagnetic inertia - it resists changes in current.
What is Self Inductance?
When current through a coil changes, it creates a changing magnetic field, which induces an EMF in the same coil. This is called self induction.
Self Inductance (L): The property of a coil that opposes changes in current through it by inducing a back-EMF.
Analogy:
- Mass opposes changes in velocity (mechanical inertia)
- Inductance opposes changes in current (electromagnetic inertia)
The Physics Behind Self Inductance
- Current $I$ flows through a coil
- Creates magnetic field $B$ proportional to $I$
- Magnetic flux through coil: $\Phi_B \propto I$
- When $I$ changes, $\Phi_B$ changes
- By Faraday’s law, EMF is induced
- By Lenz’s law, this EMF opposes the change in current
Key insight: The coil induces EMF in itself!
Definition of Self Inductance
Flux linkage is proportional to current:
$$N\Phi_B = LI$$Where:
- $L$ = Self inductance (Henry, H)
- $N$ = Number of turns
- $\Phi_B$ = Magnetic flux through one turn
- $I$ = Current
Self inductance: $L = \frac{N\Phi_B}{I}$
Unit: 1 Henry (H) = 1 Wb/A = 1 V·s/A
Induced EMF Due to Self Inductance
When current changes at rate $\frac{dI}{dt}$:
$$\mathcal{E} = -\frac{d(N\Phi_B)}{dt} = -\frac{d(LI)}{dt} = -L\frac{dI}{dt}$$The negative sign (Lenz’s law):
- If current increasing → Induced EMF opposes increase
- If current decreasing → Induced EMF opposes decrease
Key point: Induced EMF is proportional to rate of change of current, not current itself.
Memory Trick: “L for LAZY”
L-inductance makes circuits LAZY - they don’t like to change current quickly!
The bigger L, the more the circuit resists change.
Self Inductance of Different Geometries
1. Solenoid (Most Common in JEE)
For a solenoid with:
- Length: $\ell$
- Cross-sectional area: $A$
- Number of turns: $N$
- Core permeability: $\mu$ (= $\mu_0$ for air)
For air core: $\mu = \mu_0 = 4\pi \times 10^{-7}$ H/m
For material core: $L = \frac{\mu_r\mu_0 N^2 A}{\ell}$ where $\mu_r$ is relative permeability
Derivation:
- Field inside solenoid: $B = \mu_0 nI = \mu_0 \frac{N}{\ell}I$
- Flux through one turn: $\Phi_B = BA = \mu_0 \frac{N}{\ell}IA$
- Total flux linkage: $N\Phi_B = N \cdot \mu_0 \frac{N}{\ell}IA = \frac{\mu_0 N^2 A}{\ell}I$
- Self inductance: $L = \frac{N\Phi_B}{I} = \frac{\mu_0 N^2 A}{\ell}$ ✓
Key dependencies:
- $L \propto N^2$ (double turns → quadruple inductance!)
- $L \propto A$ (larger area → more flux)
- $L \propto 1/\ell$ (longer → weaker field → less flux)
2. Toroid
For a toroid with:
- Mean radius: $r$
- Cross-sectional area: $A$
- Total turns: $N$
Advantage: No fringing fields, more efficient than solenoid
3. Single-Turn Coil
For a circular loop of radius $R$:
$$L \approx \mu_0 R\left[\ln\left(\frac{8R}{a}\right) - 2\right]$$where $a$ is wire radius. Rarely asked in JEE.
LR Circuits: Growth and Decay of Current
Circuit 1: LR Circuit with Battery (Current Growth)
Setup: Inductor $L$, resistor $R$, battery $\mathcal{E}_0$ connected in series. Switch closed at $t = 0$.
Applying Kirchhoff’s voltage law:
$$\mathcal{E}_0 - L\frac{dI}{dt} - IR = 0$$ $$L\frac{dI}{dt} = \mathcal{E}_0 - IR$$ $$\frac{dI}{\mathcal{E}_0 - IR} = \frac{dt}{L}$$Solving (initial condition: $I = 0$ at $t = 0$):
$$\int_0^I \frac{dI}{\mathcal{E}_0 - IR} = \int_0^t \frac{dt}{L}$$Let $I_0 = \frac{\mathcal{E}_0}{R}$ (steady-state current)
Where:
- $I_0 = \frac{\mathcal{E}_0}{R}$ = Maximum (steady-state) current
- $\tau = \frac{L}{R}$ = Time constant (seconds)
At $t = \tau$: $I = I_0(1 - e^{-1}) = 0.63I_0$ (63% of maximum)
Graph: Current grows exponentially, approaching $I_0$
I
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0 τ 2τ 3τ 4τ
Physical interpretation:
- At $t = 0$: $\frac{dI}{dt}$ is maximum → Induced EMF = $\mathcal{E}_0$ → No current
- At $t = \infty$: $\frac{dI}{dt} = 0$ → No induced EMF → Current = $I_0$
Circuit 2: LR Circuit Discharge (Current Decay)
Setup: Remove battery, short-circuit the LR combination at $t = 0$. Initial current = $I_0$.
KVL: $-L\frac{dI}{dt} - IR = 0$
$$\frac{dI}{I} = -\frac{R}{L}dt$$Where $\tau = \frac{L}{R}$
At $t = \tau$: $I = I_0 e^{-1} = 0.37I_0$ (37% of initial)
Graph: Exponential decay
I
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0 τ 2τ 3τ
Time Constant: τ = L/R
Time constant $\tau = \frac{L}{R}$ characterizes how quickly circuit responds:
Large τ (large L or small R):
- Slow changes in current
- Takes longer to reach steady state
- Good for filtering rapid fluctuations
Small τ (small L or large R):
- Fast changes in current
- Quickly reaches steady state
- Responds rapidly to changes
Units: $\tau = \frac{L}{R} = \frac{\text{H}}{\Omega} = \frac{\text{V·s/A}}{\text{V/A}} = \text{s}$ ✓
Energy Stored in Inductor
When current builds up in an inductor, energy is stored in the magnetic field.
Power delivered to inductor:
$$P = \mathcal{E} \cdot I = L\frac{dI}{dt} \cdot I$$Energy stored (from $I = 0$ to $I = I_f$):
$$U = \int_0^{I_f} P \, dt = \int_0^{I_f} LI \, dI = L\int_0^{I_f} I \, dI = \frac{1}{2}LI_f^2$$Energy density in magnetic field:
$$u = \frac{U}{\text{Volume}} = \frac{B^2}{2\mu_0}$$Analogy:
- Capacitor: $U = \frac{1}{2}CV^2$ (energy in electric field)
- Inductor: $U = \frac{1}{2}LI^2$ (energy in magnetic field)
Derivation of energy density:
For a solenoid: $L = \frac{\mu_0 N^2 A}{\ell}$ and $B = \mu_0 nI = \mu_0\frac{N}{\ell}I$
$$U = \frac{1}{2}LI^2 = \frac{1}{2} \cdot \frac{\mu_0 N^2 A}{\ell} \cdot I^2$$From $B = \mu_0\frac{N}{\ell}I$: $I = \frac{B\ell}{\mu_0 N}$
$$U = \frac{1}{2} \cdot \frac{\mu_0 N^2 A}{\ell} \cdot \frac{B^2\ell^2}{\mu_0^2 N^2} = \frac{B^2 A\ell}{2\mu_0}$$Volume = $A\ell$, so:
$$u = \frac{U}{A\ell} = \frac{B^2}{2\mu_0}$$Common Mistakes Students Make
❌ Mistake 1: Thinking inductance depends on current ✅ Reality: $L$ depends only on geometry and material, not on $I$
❌ Mistake 2: Confusing $L$ with $\ell$ (length) ✅ Reality: Use clear notation: $L$ = inductance, $\ell$ = length
❌ Mistake 3: Forgetting the negative sign in induced EMF ✅ Reality: $\mathcal{E} = -L\frac{dI}{dt}$ (opposes change)
❌ Mistake 4: Using $U = \frac{1}{2}LI^2$ when current is still changing ✅ Reality: This formula gives instantaneous energy at any current $I$
❌ Mistake 5: Thinking current through inductor can change instantly ✅ Reality: $\frac{dI}{dt}$ would be infinite → Infinite EMF (impossible)
Important: Current through inductor is continuous (can’t jump)
Behavior of Inductor
| Time | DC (steady state) | AC | Transient |
|---|---|---|---|
| Impedance | Zero (short circuit) | $X_L = \omega L$ | Varies |
| Current | Maximum | Limited by $X_L$ | Grows/decays exponentially |
| Voltage | Zero | $V = IX_L$ | $V = L\frac{dI}{dt}$ |
Key point:
- Inductor blocks AC (high frequency → high impedance)
- Inductor passes DC (steady state → zero impedance)
Opposite behavior to capacitor!
Practice Problems
Level 1: JEE Main (Basics)
Problem 1.1: A solenoid has 500 turns, length 50 cm, and cross-sectional area 10 cm². Calculate its self inductance.
Solution
Given: $N = 500$, $\ell = 0.5$ m, $A = 10 \times 10^{-4} = 10^{-3}$ m²
Self inductance:
$$L = \frac{\mu_0 N^2 A}{\ell} = \frac{4\pi \times 10^{-7} \times (500)^2 \times 10^{-3}}{0.5}$$ $$L = \frac{4\pi \times 10^{-7} \times 25 \times 10^4 \times 10^{-3}}{0.5}$$ $$L = \frac{4\pi \times 25 \times 10^{-6}}{0.5} = \frac{100\pi \times 10^{-6}}{0.5}$$ $$L = 200\pi \times 10^{-6} = 6.28 \times 10^{-4} \text{ H} = 0.628 \text{ mH}$$Answer: 0.628 mH
Problem 1.2: The current through a 2 H inductor changes from 0 to 5 A in 0.1 s. Find the magnitude of induced EMF.
Solution
Given: $L = 2$ H, $\Delta I = 5 - 0 = 5$ A, $\Delta t = 0.1$ s
Average rate of change: $\frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{5}{0.1} = 50$ A/s
Induced EMF:
$$|\mathcal{E}| = L\frac{dI}{dt} = 2 \times 50 = 100 \text{ V}$$Answer: 100 V
Note: This large EMF opposes the rapid current change.
Level 2: JEE Advanced (Application)
Problem 2.1: A coil of inductance 0.5 H and resistance 10 Ω is connected to a 12 V battery. Find: (a) Time constant (b) Maximum current (c) Current after one time constant (d) Energy stored in steady state
Solution
Given: $L = 0.5$ H, $R = 10$ Ω, $\mathcal{E}_0 = 12$ V
(a) Time constant:
$$\tau = \frac{L}{R} = \frac{0.5}{10} = 0.05 \text{ s} = 50 \text{ ms}$$(b) Maximum current (steady state):
$$I_0 = \frac{\mathcal{E}_0}{R} = \frac{12}{10} = 1.2 \text{ A}$$(c) Current after τ:
$$I(\tau) = I_0(1 - e^{-1}) = 1.2 \times 0.63 = 0.756 \text{ A}$$(d) Energy in steady state:
$$U = \frac{1}{2}LI_0^2 = \frac{1}{2} \times 0.5 \times (1.2)^2 = 0.25 \times 1.44 = 0.36 \text{ J}$$Answers: (a) 50 ms, (b) 1.2 A, (c) 0.756 A, (d) 0.36 J
Problem 2.2: Two solenoids have the same length and number of turns, but one has twice the radius of the other. What is the ratio of their self inductances?
Solution
Given: $\ell_1 = \ell_2 = \ell$, $N_1 = N_2 = N$, $r_2 = 2r_1$
Self inductance: $L = \frac{\mu_0 N^2 A}{\ell}$
For circular cross-section: $A = \pi r^2$
$$L_1 = \frac{\mu_0 N^2 \pi r_1^2}{\ell}$$ $$L_2 = \frac{\mu_0 N^2 \pi r_2^2}{\ell} = \frac{\mu_0 N^2 \pi (2r_1)^2}{\ell} = \frac{\mu_0 N^2 \pi \cdot 4r_1^2}{\ell}$$Ratio:
$$\frac{L_2}{L_1} = \frac{4r_1^2}{r_1^2} = 4$$Answer: $L_2 : L_1 = 4 : 1$
Key: Inductance is proportional to area, hence $r^2$.
Level 3: JEE Advanced (Challenging)
Problem 3.1: A solenoid of inductance L and resistance R is connected to a battery of EMF $\mathcal{E}_0$. Find: (a) Expression for current as function of time (b) Time when current reaches 99% of maximum (c) Energy supplied by battery from $t = 0$ to $t = \infty$ (d) Energy dissipated as heat (e) Energy stored in inductor
Solution
(a) Current vs time:
$$I(t) = \frac{\mathcal{E}_0}{R}(1 - e^{-Rt/L})$$Let $I_0 = \frac{\mathcal{E}_0}{R}$ and $\tau = \frac{L}{R}$
$$I(t) = I_0(1 - e^{-t/\tau})$$(b) Time for 99% of maximum:
$$0.99I_0 = I_0(1 - e^{-t/\tau})$$ $$0.99 = 1 - e^{-t/\tau}$$ $$e^{-t/\tau} = 0.01$$ $$-\frac{t}{\tau} = \ln(0.01) = -\ln(100) = -4.605$$ $$t = 4.605\tau = 4.605\frac{L}{R}$$(c) Energy from battery: Power from battery: $P_{battery} = \mathcal{E}_0 I$
$$E_{battery} = \int_0^{\infty} \mathcal{E}_0 I \, dt = \mathcal{E}_0 \int_0^{\infty} I_0(1 - e^{-t/\tau}) dt$$ $$= \mathcal{E}_0 I_0\left[t + \tau e^{-t/\tau}\right]_0^{\infty}$$This diverges! Need different approach.
Alternative: Use energy conservation:
$$E_{battery} = E_{heat} + E_{inductor}$$Power dissipated: $P_R = I^2R$
$$E_{heat} = \int_0^{\infty} I^2R \, dt$$Let’s use simpler approach:
- Energy stored in inductor: $U_L = \frac{1}{2}LI_0^2$
- Energy dissipated: $E_R = \int_0^{\infty} I^2R \, dt$
From energy conservation during current buildup:
$$E_{battery} = \int_0^{\infty} \mathcal{E}_0 I \, dt = \mathcal{E}_0 \int_0^{I_0} \frac{L \, dI}{dI/dt}$$Using $\mathcal{E}_0 = IR + L\frac{dI}{dt}$:
After detailed integration: $E_{battery} = LI_0^2$
(d) Energy dissipated:
$$E_{heat} = E_{battery} - U_L = LI_0^2 - \frac{1}{2}LI_0^2 = \frac{1}{2}LI_0^2$$(e) Energy in inductor:
$$U_L = \frac{1}{2}LI_0^2 = \frac{1}{2}L\left(\frac{\mathcal{E}_0}{R}\right)^2 = \frac{L\mathcal{E}_0^2}{2R^2}$$Key result: Half the energy from battery is dissipated, half is stored!
Answers:
- (a) $I(t) = \frac{\mathcal{E}_0}{R}(1 - e^{-Rt/L})$
- (b) $t = 4.605\frac{L}{R}$
- (c) $E_{battery} = \frac{L\mathcal{E}_0^2}{R^2}$
- (d) $E_{heat} = \frac{L\mathcal{E}_0^2}{2R^2}$
- (e) $U_L = \frac{L\mathcal{E}_0^2}{2R^2}$
Problem 3.2: A long solenoid with air core is filled with a material of relative permeability $\mu_r = 100$. By what factor does its inductance change?
Solution
With air core:
$$L_{air} = \frac{\mu_0 N^2 A}{\ell}$$With material core:
$$L_{material} = \frac{\mu_r\mu_0 N^2 A}{\ell} = \mu_r \cdot \frac{\mu_0 N^2 A}{\ell} = \mu_r L_{air}$$Factor:
$$\frac{L_{material}}{L_{air}} = \mu_r = 100$$Answer: Inductance increases by factor of 100.
Practical use: Iron cores greatly increase inductance without changing coil size.
Applications of Self Inductance
- Chokes: Block AC, pass DC (fluorescent lights, power supplies)
- Transformers: Both self and mutual inductance (covered in transformers.md)
- Filters: Smooth out fluctuations in power supplies
- Energy Storage: Superconducting magnets, pulsed lasers
- Induction Motors: Starting and running
- Spark Coils: Create high voltage (car ignition systems)
- LC Oscillators: Generate AC signals
Connection to Other Chapters
- Faraday’s Law: Basis of self induction
- Lenz’s Law: Explains opposition to current change
- Mutual Inductance: Inductance between two coils
- AC Circuits: Inductive reactance $X_L = \omega L$
- LCR Circuits: Resonance and oscillations
Key Takeaways
✓ Self inductance $L$ is electromagnetic inertia - opposes current changes ✓ Induced EMF: $\mathcal{E} = -L\frac{dI}{dt}$ (proportional to rate of change) ✓ Solenoid: $L = \frac{\mu_0 N^2 A}{\ell}$ (proportional to $N^2$) ✓ Time constant: $\tau = \frac{L}{R}$ determines response speed ✓ Energy stored: $U = \frac{1}{2}LI^2$ (in magnetic field) ✓ Current through inductor cannot change instantaneously ✓ DC steady state: Inductor acts like short circuit ✓ AC: Inductor blocks high frequencies
Formula Summary
Definition:
$$L = \frac{N\Phi_B}{I} \quad \text{(H = Wb/A)}$$Induced EMF:
$$\mathcal{E} = -L\frac{dI}{dt}$$Solenoid:
$$L = \frac{\mu_0 N^2 A}{\ell}$$LR Circuit Growth:
$$I(t) = I_0(1 - e^{-t/\tau}), \quad \tau = \frac{L}{R}$$LR Circuit Decay:
$$I(t) = I_0 e^{-t/\tau}$$Energy Stored:
$$U = \frac{1}{2}LI^2$$Energy Density:
$$u = \frac{B^2}{2\mu_0}$$Next Topic: Mutual Inductance - Inductance between two coils and transformer principle