Unit 10: Current Electricity & Magnetism

Current electricity and magnetism share 10–12 % of JEE-Main and are highly correlated topics. The chapter brings together circuit analysis (Kirchhoff/Wheatstone/potentiometer) and continuum field problems (Biot–Savart and Ampère's law). Expect:

• JEE-Main: One Kirchhoff problem, one Biot–Savart (loop or finite wire), one Lorentz/cyclotron numerical, one moving-conductor or instrument question.
• JEE-Advanced: Multi-loop network with a non-trivial Wheatstone bridge; a problem combining magnetic force with circular motion or projectile motion; an experimental potentiometer setup.

Three skeletons unify it all:

1. Ohm: $V = IR$, microscopic $\vec J = \sigma \vec E$ with $\sigma = ne^2 \tau / m$.
2. Biot–Savart: $d\vec B = (\mu_0 / 4\pi)\, I\, d\vec\ell \times \hat r / r^2$.
3. Lorentz: $\vec F = q(\vec E + \vec v \times \vec B)$.

Concept Map

CURRENT ELECTRICITY
│
├── Drift velocity:  v_d = eEτ/m
├── Microscopic Ohm:  J = σE; σ = ne²τ/m
├── Resistance:  R = ρL/A
├── Combination
│     ├── Series: R = ΣR
│     └── Parallel: 1/R = Σ(1/R)
├── Cells
│     ├── EMF ε, internal r
│     ├── Series: ε_eq = Σε, r_eq = Σr
│     └── Parallel: 1/r_eq = Σ(1/r)
├── Kirchhoff's laws (KCL + KVL) → mesh / loop
├── Wheatstone bridge → P/Q = R/S
├── Potentiometer → measures EMF without drawing current
├── Power, Max-power transfer (R_L = R_int)
└── RC discharge & charging

MAGNETISM
│
├── Biot–Savart:  dB = μ₀ I dℓ × r̂ / (4π r²)
├── Standard fields
│     ├── Straight wire (finite): B = μ₀I(sin θ₁+sin θ₂)/(4πd)
│     ├── Circular loop axis: B = μ₀IR²/(2(R²+x²)^(3/2))
│     ├── Solenoid axis: B = μ₀nI
│     └── Toroid: B = μ₀NI/(2πr)
├── Ampère:  ∮B·dℓ = μ₀ I_enc
├── Lorentz:  F = qv×B
│     ├── Circular: r = mv/(qB), T = 2πm/(qB)
│     ├── Helical motion
│     └── Velocity selector E ⊥ B
├── Cyclotron — derivation
├── Force on wire:  F = IL × B
├── Two parallel wires:  F/L = μ₀I₁I₂/(2πd)
├── Torque on loop:  τ = m × B,  m = NIA
├── Galvanometer → ammeter (shunt) / voltmeter (series R)
└── Materials: dia / para / ferro, hysteresis

Topic 1: Current, Drift Velocity and Ohm's Law

Sub-topic A: Definitions

Current: $I = dQ/dt$. Unit: ampere (A) = C/s. Current is a scalar but has a direction (sign).

Current density: $\vec J$ with $I = \int \vec J\cdot d\vec A$. Microscopic relation:

where $n$ is free electron density, $\vec v_d$ the drift velocity.

Sub-topic B: Drift Velocity Derivation

A field $\vec E$ inside a conductor accelerates electrons, but collisions randomize their velocity every $\tau$ (mean free time):

(negative sign for electrons is absorbed in the direction of current — current opposite to electron drift.)

Mobility: $\mu = e\tau/m$, so $v_d = \mu E$.

Sub-topic C: Ohm's Law

Macroscopic: $V = IR$, where $R$ is the resistance.

Microscopic: $\vec J = \sigma \vec E$. Combine with $v_d = eE\tau/m$:

So conductivity

For a uniform conductor of length $L$, area $A$:

Sub-topic D: Temperature Dependence

For metals, $\rho$ increases with $T$ (collisions more frequent at higher $T$):

For semiconductors, $n$ increases with $T$ much faster than $\tau$ decreases — net $\rho$ falls with $T$ (negative $\alpha$).

Worked Problem 1

A copper wire of cross section $1\,\text{mm}^2$ carries $1$ A current. If $n = 8.5 \times 10^{28}$/m³, find the drift velocity.

Solution. $v_d = I/(neA) = 1/(8.5\times 10^{28}\cdot 1.6\times 10^{-19}\cdot 10^{-6}) \approx 7.35\times 10^{-5}$ m/s. Tiny — that's why steady-state response is fast (field acts on all electrons), but bulk fluid flow is slow.

Topic 2: Circuit Elements and Combinations

Sub-topic A: Resistor Combinations

Series: same $I$ through each, $V$ adds:

Parallel: same $V$ across each, $I$ adds:

Sub-topic B: Cells, EMF, Internal Resistance

A real cell has EMF $\varepsilon$ and internal resistance $r$. Terminal voltage when drawing current $I$:

(when current is drawn out). When charging, $V = \varepsilon + Ir$.

Cells in series (same $I$, EMFs add or subtract depending on orientation):

Cells in parallel (identical orientation, terminals connected):

Sub-topic C: Power Considerations

Power dissipated in a resistor: $P = I^2 R = V^2/R = IV$.

Max power transfer theorem: For source EMF $\varepsilon$ and internal $r$ driving load $R_L$:

Maximum at $dP_L/dR_L = 0 \Rightarrow R_L = r$. Maximum power $P_{\max} = \varepsilon^2/(4r)$.

(Efficiency at max-power-transfer is only 50%. For practical power delivery, $R_L \gg r$ gives lower power but high efficiency.)

Topic 3: Kirchhoff's Laws and Circuit Analysis

Sub-topic A: Two Laws

Kirchhoff's Current Law (KCL): At every junction $\sum I_{\text{in}} = \sum I_{\text{out}}$. Conservation of charge.

Kirchhoff's Voltage Law (KVL): Around any closed loop $\sum V = 0$. Conservation of energy.

Sub-topic B: Mesh and Loop Analysis Procedure

1. Assign current $I_i$ to each independent loop, with a chosen direction.
2. Apply KVL around each loop, tracking signs:
   • Going across a resistor in the direction of current: $-IR$.
   • Going from $-$ to $+$ terminal of EMF: $+\varepsilon$ (else $-\varepsilon$).
3. Solve the resulting linear system.

Worked Problem 2

In the figure: a $12$ V battery in series with $R_1 = 2\,\Omega$ feeds a parallel combination of $R_2 = 4\,\Omega$ and $R_3 = 4\,\Omega$. Find current from battery, and current through each parallel branch.

Solution. $R_{23} = 2\,\Omega$ (parallel of 4 & 4). Total $R = 4\,\Omega$. $I = 12/4 = 3$ A. Voltage across parallel = $3\times 2 = 6$ V. Each branch carries $6/4 = 1.5$ A.

Sub-topic C: Wheatstone Bridge

Four resistors $P, Q, R, S$ form a bridge with a galvanometer across the bridge diagonal:

• $P$ and $Q$ in series in one branch
• $R$ and $S$ in series in the other
• Galvanometer between the midpoints

Balanced condition (zero current through galvanometer):

Proof. At balance, the potentials at the two midpoints are equal. With current $I_1$ in the $PQ$ branch and $I_2$ in the $RS$ branch, $I_1 P = I_2 R$ and $I_1 Q = I_2 S$. Dividing gives the result.

The condition is independent of source EMF/internal resistance — robust experimental tool.

Sub-topic D: Meter Bridge

A practical version: a uniform resistance wire of $1$ m length forms two arms of the bridge by sliding a contact at distance $\ell$ from one end. If the unknown is $X$ and the standard is $S$:

Sub-topic E: Potentiometer

A long uniform wire with a current driven through it from a stable source. The potential drops linearly along the wire, providing a continuously variable reference voltage.

Used to measure EMF of an unknown cell without drawing current from it (the unknown cell is balanced against the potential drop, no current ⇒ no internal-resistance error).

If $\ell_1$ balances cell of EMF $\varepsilon_1$ and $\ell_2$ balances $\varepsilon_2$:

Also used to measure internal resistance: balance the cell with switch open ($\ell_1$) and closed through a known $R$ ($\ell_2$):

Topic 4: RC Circuit

Sub-topic A: Charging Derivation

A capacitor $C$ in series with $R$ and EMF $\varepsilon$:

Solving with $q(0) = 0$:

Time constant $\tau = RC$. Quantities reach $63\%$ at $\tau$, $95\%$ at $3\tau$.

Sub-topic B: Discharging

Battery removed, capacitor short-circuited through $R$:

Worked Problem 3

A capacitor of $10\,\mu$F is charged through a $1\,\text{M}\Omega$ resistor by a $100$ V battery. Find time for charge to reach $50\,\mu$C.

Solution. $q_\infty = CV = 1000\,\mu$C. $\tau = RC = 10$ s. $50 = 1000(1 - e^{-t/10})$ ⇒ $e^{-t/10} = 0.95$ ⇒ $t = -10\ln 0.95 \approx 0.513$ s.

Topic 5: Magnetic Field — Biot-Savart Law

Sub-topic A: Statement

A current element $I\,d\vec\ell$ produces an infinitesimal field at $\vec r$:

$\mu_0 = 4\pi\times 10^{-7}$ T·m/A (exact).

Right-hand rule: thumb along $I\,d\vec\ell$, fingers curl in the direction of $\vec B$.

Sub-topic B: Standard Fields by Biot–Savart

1. Straight wire of length, point at perpendicular distance $d$, with angles $\theta_1, \theta_2$ from perpendicular to ends:

For infinite wire ($\theta_1 = \theta_2 = \pi/2$):

For semi-infinite wire (extending from $d$-perpendicular point to infinity, $\theta_1 = 0, \theta_2 = \pi/2$):

2. Circular arc of radius $R$ subtending angle $\phi$ at the centre:

Full loop ($\phi = 2\pi$):

3. Loop on axis:

For $x = 0$: $B = \mu_0 I/(2R)$ (above). For $x \gg R$: $B \to \mu_0 I R^2/(2x^3) = \mu_0 m/(2\pi x^3)$ where $m = I\pi R^2$ is the magnetic moment — the "magnetic dipole" formula.

4. Solenoid (long, on axis):

where $n =$ turns per unit length. At the end of a long solenoid, $B = \mu_0 n I/2$.

5. Toroid:

where $N$ is total turns, $r$ is the distance from toroid axis.

6. Field of a moving point charge (small but useful):

This is the source of the Biot–Savart law in disguise.

Sub-topic C: Ampère's Law

The magnetic equivalent of Gauss's law. Useful for:

• Infinite straight wire: Amperian circle ⇒ $B(2\pi r) = \mu_0 I$.
• Infinite solenoid: Amperian rectangle ⇒ $B L = \mu_0 (nL) I$ ⇒ $B = \mu_0 n I$.
• Toroid: $B(2\pi r) = \mu_0 N I$.

Worked Problem 4

A square loop of side $a$ carries current $I$. Find $B$ at the centre.

Solution. Each side is a finite wire. From the centre, the perpendicular distance is $a/2$ and the ends subtend $\theta_1 = \theta_2 = 45^\circ$ from the perpendicular. So $B_{\text{side}} = \frac{\mu_0 I}{4\pi(a/2)}(2\sin 45^\circ) = \frac{\mu_0 I \sqrt 2}{2\pi a}$

Four sides, all contributing the same direction: $B_{\text{total}} = 4\cdot \frac{\mu_0 I\sqrt 2}{2\pi a} = \frac{2\sqrt 2 \mu_0 I}{\pi a}$

Worked Problem 5

Two long parallel wires carrying currents $I_1 = 10$ A and $I_2 = 20$ A in the same direction are $0.5$ m apart. Find the force per unit length on each wire.

Solution. $B_1$ at location of wire 2: $\mu_0 I_1 / (2\pi d) = 2\pi \times 10^{-7}\cdot 10/0.5 = 4\times 10^{-6}$ T.

$F/L = I_2 B_1 = 20\cdot 4\times 10^{-6} = 8\times 10^{-5}$ N/m, attractive (parallel currents attract).

Topic 6: Lorentz Force and Motion of Charged Particles

Sub-topic A: Lorentz Force

The magnetic part is always perpendicular to $\vec v$, hence does no work — magnetic force only changes direction of motion, not kinetic energy.

Sub-topic B: Motion in Uniform $\vec B$

For $\vec v \perp \vec B$: circular motion. Centripetal force $= q v B$:

Note: $T$ is independent of $v$ and $r$ — the cyclotron period.

If $\vec v$ has a component parallel to $\vec B$ (say $v_\parallel$), that component is unaffected (no force):

• Perpendicular component → circle of radius $r = m v_\perp/(qB)$.
• Parallel component → straight-line drift.
• Combined: helix with pitch $p = v_\parallel T = 2\pi m v_\parallel /(qB)$.

Sub-topic C: Velocity Selector

Crossed $\vec E$ and $\vec B$ (perpendicular). Net force on a particle moving with velocity $\vec v$ perpendicular to both:

(if $E$ and $v\times B$ are antiparallel). Force is zero when $v = E/B$.

Particles with this exact velocity pass through undeflected — used to filter velocities.

Sub-topic D: Cyclotron — Derivation

A cyclotron uses an alternating voltage applied between two D-shaped electrodes (dees) immersed in a perpendicular $\vec B$. A charged particle is accelerated as it crosses the gap between the dees; inside each dee, $\vec B$ makes it move in a circle.

Since the cyclotron period $T = 2\pi m/(qB)$ is independent of $v$, the same alternating frequency

works for all radii — the particle is in resonance with the alternating field.

Maximum kinetic energy: at maximum radius $R$ (limited by dee size),

Limits of cyclotron: relativistic increase of $m$ at high $v$ destroys resonance — need synchrocyclotron or synchrotron for high energies. Also, $T$ depends on $q/m$, so neutral particles cannot be accelerated.

Worked Problem 6

A proton ($m = 1.67 \times 10^{-27}$ kg, $q = 1.6\times 10^{-19}$ C) moves in a circle of radius $20$ cm in a $0.5$ T field. Find its kinetic energy in eV.

Solution. $v = qBR/m = 1.6\times 10^{-19}\cdot 0.5 \cdot 0.2/1.67\times 10^{-27} = 9.58\times 10^6$ m/s.

$K = \tfrac{1}{2}m v^2 = \tfrac{1}{2}\cdot 1.67\times 10^{-27}\cdot 9.18\times 10^{13} = 7.67\times 10^{-14}$ J $= 4.79\times 10^5$ eV $\approx 480$ keV.

Topic 7: Forces on Currents and Torque on Loops

Sub-topic A: Force on a Current-Carrying Wire

For a small element $d\vec\ell$ in field $\vec B$:

For a straight wire of length $L$ in uniform $\vec B$:

For any closed loop in uniform $\vec B$: $\vec F = 0$ (forces around a closed loop integrate to zero) — but there is a net torque.

Sub-topic B: Force Between Parallel Wires

Two infinite parallel wires carrying currents $I_1, I_2$ at separation $d$:

Field from wire 1 at wire 2: $B_1 = \mu_0 I_1/(2\pi d)$.

Force per unit length on wire 2:

Parallel currents attract; antiparallel repel. (This defined the SI ampere historically.)

Sub-topic C: Torque on a Current Loop

A rectangular loop of sides $a$ and $b$ carries current $I$ in field $\vec B$. With normal at angle $\theta$ to $\vec B$:

where $\vec m = NI \vec A$ is the magnetic moment (vector, by right-hand rule from current to normal).

Vector form:

Potential energy: $U = -\vec m\cdot\vec B$.

This is exactly the dipole-in-field analogue from electrostatics.

Sub-topic D: Galvanometer → Ammeter / Voltmeter

A galvanometer has a coil of resistance $G$ and gives full-scale deflection at current $i_g$ (typically a few mA).

Converting to ammeter (measuring large $I$): connect a small shunt resistance $S$ in parallel:

Converting to voltmeter (measuring large $V$): connect a large series resistance $R_s$:

An ideal ammeter has zero resistance; ideal voltmeter has infinite resistance.

Worked Problem 7

A galvanometer of $G = 50\,\Omega$ gives full-scale deflection at $i_g = 2$ mA. Convert it to (a) ammeter reading up to 5 A; (b) voltmeter reading up to 100 V.

Solution. (a) $S = i_g G/(I - i_g) = 0.002\cdot 50/4.998 \approx 0.02\,\Omega$ in parallel. (b) $R_s = V/i_g - G = 100/0.002 - 50 = 50000 - 50 = 49950\,\Omega$ in series.

Topic 8: Earth's Magnetism (Brief)

The Earth has a magnetic field with three local parameters:

• Magnetic declination $\delta$: angle between geographic and magnetic north.
• Dip (inclination) $\theta$: angle of $\vec B$ with horizontal.
• Horizontal component $B_H$.

Total field $B = B_H/\cos\theta$, vertical $B_V = B_H\tan\theta$.

For a magnet of moment $M$ at distance $d$ on axis, balance with $B_H$:

This setup (tangent galvanometer) historically measured $B_H$.

Topic 9: Magnetic Materials

Sub-topic A: Magnetisation, $\vec M$, $\vec H$

When matter is placed in $\vec B$, the atoms acquire magnetic moments. Magnetisation $\vec M$ = magnetic moment per unit volume. Define magnetic intensity:

In linear materials: $\vec M = \chi_m \vec H$, and

$\mu_r = 1 + \chi_m$ is relative permeability.

Sub-topic B: Classification

Diamagnetism: present in all materials (Lenz's-law-like response of orbital electrons); usually masked.

Paramagnetism: unpaired electron spins partially align with $\vec B$. Above the Curie temperature, ferromagnets become paramagnetic.

Ferromagnetism: spontaneous alignment of spins below Curie temperature, leading to domains. Hysteresis curve: $B$ vs $H$ loop shows remnant magnetisation $B_r$ and coercivity $H_c$. Area of loop = energy dissipated per cycle.

Worked Problem 8

A solenoid has $1000$ turns/m and carries $2$ A. It is filled with material of $\chi_m = 1500$. Find the magnetic field inside.

Solution. $H = nI = 2000$ A/m. $\mu_r = 1501$. $B = \mu_0\mu_r H = 4\pi\times 10^{-7}\times 1501\times 2000 \approx 3.77$ T. (Soft iron core.)

Problem-Solving Heuristics

1. For DC circuits, always reduce by series/parallel first. Only invoke Kirchhoff for irreducible meshes.
2. Wheatstone shortcut: if the bridge is balanced, the middle arm carries no current — replace with an open or short, doesn't matter.
3. Symmetric circuits: identify symmetric nodes (equipotential) — collapse them into one node.
4. Power = $I^2R$ but also $V^2/R$ — pick whichever quantity is constant.
5. For RC, time constant $RC$ tells you the half-life-ish scale; after $\sim 5\tau$ steady state.
6. For Biot–Savart of a wire, use the $\sin\theta$ formula with angles measured from the perpendicular foot.
7. For an axis point of a loop, $B = \mu_0 I R^2 / (2(R^2+x^2)^{3/2})$ — drop $x$ for centre, drop $R^2$ in denominator for far field.
8. Solenoid: $B = \mu_0 n I$ inside, zero outside (for ideal infinite).
9. Cyclotron period $T = 2\pi m / qB$ is independent of velocity.
10. For helical motion, decompose $\vec v$ into parallel + perpendicular components and treat each.
11. Force on a closed loop in uniform $B$ is zero — but the torque $\vec m\times\vec B$ is not.
12. Galvanometer conversion: ammeter → low shunt, voltmeter → high series.
13. Magnetic flux through a coil $\Phi = NBA\cos\theta$ — feeds straight into Faraday in the next unit.

Common Traps & Mistakes

• Conventional current vs electron drift are opposite. Always use conventional (positive-to-negative outside the cell).
• Ohm's law fails for diodes, gas discharge tubes, etc — they're "non-Ohmic". But the JEE rarely probes this beyond the I-V curve question.
• Internal resistance is not separate from the rest of the circuit — it's in series with the load.
• Wheatstone needs zero current through galvanometer, not zero potential difference end-to-end.
• Potentiometer is for null-method only, no current flowing through the unknown — that's the whole point.
• Magnetic force does no work, but the work-energy theorem still applies if there's also an electric field present.
• Direction of $\vec v\times\vec B$: use right-hand rule, then flip for negative charge.
• Force on a moving charge requires both $v$ and $B$, neither alone produces force.
• Solenoid field is uniform inside but suddenly drops outside — for a finite solenoid, the field decreases gradually.
• Bias on parallel wires: same current direction = attractive (counter-intuitive — currents are like things, but they attract).
• Curie temperature drops $\chi_m$ from ferro down to typical para values, but doesn't make it negative.

Quick Revision Card

• Drift: $v_d = eE\tau/m$; $\sigma = ne^2\tau/m$, $\rho = 1/\sigma$.
• Resistance: $R = \rho L/A$. Temp: $\rho = \rho_0(1 + \alpha\Delta T)$.
• Series: $R = \sum R$. Parallel: $1/R = \sum 1/R$.
• Cell: $V = \varepsilon - Ir$. Max power: $R_L = r$.
• Kirchhoff: KCL (junction), KVL (loop).
• Wheatstone: $P/Q = R/S$ at balance.
• Potentiometer: $\varepsilon \propto \ell$. Internal $r = R(\ell_1 - \ell_2)/\ell_2$.
• RC: $q = q_0(1 - e^{-t/RC})$, $\tau = RC$.
• Biot-Savart: $dB = (\mu_0/4\pi)I\,d\ell\sin\theta/r^2$.
• Straight wire: $B = \mu_0 I/(2\pi d)$.
• Loop centre: $B = \mu_0 I/(2R)$. Axis: $\mu_0 I R^2/(2(R^2+x^2)^{3/2})$.
• Solenoid: $B = \mu_0 n I$. Toroid: $B = \mu_0 N I/(2\pi r)$.
• Ampère: $\oint B\cdot d\ell = \mu_0 I_{\text{enc}}$.
• Lorentz: $\vec F = q\vec v\times \vec B$, no work.
• Circular: $r = mv/(qB)$, $T = 2\pi m/(qB)$.
• Force on wire: $\vec F = I\vec L\times \vec B$. Parallel wires: $F/L = \mu_0 I_1 I_2/(2\pi d)$.
• Loop torque: $\vec\tau = \vec m\times\vec B$, $\vec m = NIA$.
• Ammeter: shunt $S = i_g G/(I - i_g)$. Voltmeter: $R_s = V/i_g - G$.

Formula Sheet