Magnetism in One Shot for JEE 2026

Mastering Magnetism in One Shot for JEE 2026 is important for any aspirant, as magnetism is a fundamental force and a cornerstone of physics crucial for the exam. Here is a crisp overview of key concepts, starting with how moving charges and currents generate magnetic fields. It progresses to the forces these fields exert, the behavior of charged particles within them, and the principles of electromagnetic induction. Understanding magnetic properties of materials and devices like galvanometers and transformers is also essential for total exam readiness.

Magnetic Field Due to Straight Current-Carrying Wire

For a finite straight wire carrying current I, the magnetic field (B) at a point P is given by:

B = (μ₀I / 4πR) * (sin α + sin β)

Here, R is the perpendicular distance from point P to the wire, and α and β are angles from the perpendicular to the ends of the wire. The permeability of free space is μ₀ = 4π × 10⁻⁷ T·m/A.

Direction Rule for a Straight Wire

To find the direction of the magnetic field B:

1. Use your right hand.

2. Place your thumb on the wire, pointing in the direction of the current (I).

3. Point your fingers towards the point (P) where you want the field.

4. The direction your palm faces gives the direction of the magnetic field (B).
   (Memory Tip: Imagine you are a "magnetic field god"; your thumb is current, fingers point to the target, and your palm "shoots" the field.)

The magnetic field on the axis of the wire (along its length) is zero.

• Dot (●) symbol: Current coming outward from the plane.

• Cross (⊗) symbol: Current going inward into the plane.

Special Cases for Straight Wires

1. Infinitely Long Wire: When α and β approach 90°, the formula simplifies to:
   B = μ₀I / 2πR
   B is inversely proportional to R.

2. Semi-Infinite Wire: If point P is aligned with one end, one angle is 0° and the other is 90°.
   B = μ₀I / 4πR

3. Point Outside the End-Points: If the perpendicular falls on the extension of the wire, and both angles are on the same side:
   B = (μ₀I / 4πR) * |sin α - sin β|

Magnetic Field Due to a Circular Current-Carrying Coil

1. At the Center of the Coil

For a coil with N turns, radius R, and current I, the magnetic field at its center is:

B_center = μ₀NI / 2R

(Memory Tip: The straight wire formula has π, but the circular coil's center formula does not. Do not add π for circular coils.)

Direction Rule for a Circular Coil/Arc:

1. Curl your right-hand fingers in the direction of the current flow.

2. Your thumb will point in the direction of the magnetic field along the axis.

2. On the Axis of the Coil

At a distance x from the center along the axis, the magnetic field is:

B_axis = (μ₀NIR²) / [2 * (R² + x²)^(3/2)]

The magnetic field is maximum at the center (x=0) and continuously decreases with increasing x.

3. Magnetic Field due to a Circular Arc

For a circular arc subtending an angle θ (in radians) at the center:

B_arc = μ₀Iθ / 4πR

For N turns, multiply by N.

Solved Examples (Magnetic Field Calculations)

Example: Loop ABCD (April 2023)

A loop with semicircular segments (R₁, R₂) and straight segments carries current. To find the magnetic field at the center O:

1. Straight segments: Magnetic field is zero as O lies on their axis.

2. Semicircles: Calculate fields from each semicircle using B = μ₀I / 4R. Determine direction (inward/outward) using the right-hand rule.

3. Net field: Sum the vector fields. For example, if outer semicircle is outward and inner is inward, B_net = B_outer - B_inner.

Example: Square Loop with Current Division (JEE Mains)

A square loop has resistance R for path ABC and 2R for path ADC.

1. Current Division: Calculate currents I₁ (ABC) and I₂ (ADC) using inverse proportionality to resistance: I₁ = (2I/3), I₂ = (I/3).

2. Field from one side: For a square, perpendicular distance from center to side is a/2, and angles are 45°. B_wire = μ₀i√2 / 2πa.

3. Net Field: Each side of ABC path produces an inward field, each side of ADC path produces an outward field. Sum fields: B_net = 2 * B(I₁) - 2 * B(I₂).

Example: Infinite Wire Bent into a Loop (Jan 2024)

An infinite wire bent into a circular loop.

1. Two semi-infinite wires: Treated as one infinite straight wire at distance R. Field is μ₀I / 2πR.

2. Circular loop: Field is μ₀I / 2R.

3. Net Field: Combine fields considering their directions (e.g., one outward, one inward). B_net = (μ₀I / 2R) * (1/π - 1).

Magnetic Field of Current Configurations

1. Magnetic Field due to a Hollow Conducting Cylinder

• Outside (r ≥ R): Behaves like an infinite straight wire carrying total current I.
  B_out = (μ₀ * I) / (2πr)

• Inside (r < R): B_in = 0
  (Memory Tip: Outside, imagine current concentrated on the axis.)

2. Magnetic Field due to a Solid Conducting Cylinder

• Outside (r > R): Same as hollow cylinder.
  B_out = (μ₀ * I) / (2πr)

• Inside (r < R): Field increases linearly with r.
  B_in = (μ₀ * I * r) / (2πR²)

3. Magnetic Field of a Solenoid

For an infinitely long solenoid with n turns per unit length:

• Inside: B_in = μ₀ * n * I = μ₀ * (N / L) * I (strong, uniform, parallel to axis).

• Outside: B_out = 0

• North/South Poles: Right-hand rule (fingers curl with current, thumb points to North pole).

Energy Stored in a Solenoid:

• Magnetic Energy Density: u_B = B² / (2μ₀)

• Total Magnetic Energy: U_B = ½LI²
  If filled with a medium of relative permeability μ_r, B_in = μ₀ \* μ_r \* n \* I.

4. Magnetic Field of a Toroid

A solenoid bent into a circular shape (doughnut).

• Inside the core: B_in = μ₀ * n * I = (μ₀ * N * I) / (2πr) (where r is the average radius).

• Outside the core (inner or outer space): B = 0

Ampere's Circuital Law

∮ B ⋅ dl = μ₀ * I_enclosed

The line integral of the magnetic field around a closed loop equals μ₀ times the net current enclosed by the loop. The sign of I_enclosed is determined by a right-hand rule based on the loop's traversal direction.

Magnetic Force on a Moving Charge

The magnetic force F_m on a charge q moving with velocity v in a magnetic field B is:

F_m = q (v × B)

Magnitude: F_m = |q|vBsinθ, where θ is the angle between v and B.

Key Properties of the Magnetic Force

1. Direction: Always perpendicular to both v and B.

2. Work and Power: Work done and power delivered by magnetic force are always zero because F_m is perpendicular to v.

3. Effect on Motion: Cannot change kinetic energy or speed, only the direction of velocity. Acts as a centripetal force.

(Memory Tip for Direction (positive charge): Right-Hand Rule: Fingers along v, palm towards B, thumb points to F_m.)

Motion of a Charge in a Uniform Magnetic Field

1. Charge Released from Rest (v = 0): F_m = 0. The charge remains at rest.

2. Velocity Parallel or Anti-parallel to B (θ = 0° or 180°): F_m = 0. Particle moves in a straight line with constant velocity.

3. Velocity Perpendicular to B (v ⊥ B): Highly important. The particle moves in a circular path.

• Radius (r): r = mv / qB = p / qB = √(2m·KE) / qB = √(2mqV) / qB

• Time Period (T): T = 2πm / qB (independent of speed).

• Angular Speed (ω): ω = qB / m (independent of speed).

4. Velocity at an Angle θ to B: Very important. The motion is a helical path.

• Parallel Component (v cosθ): Causes linear motion along the field.

• Perpendicular Component (v sinθ): Causes circular motion.

• Time Period (T): T = 2πm / qB

• Pitch of the Helix: Pitch = (v cosθ) × T = (v cosθ) × (2πm / qB)

Solved Problems: Motion of a Charge in a Magnetic Field

Example: Return to Original Location (JEE 2025)

If a particle is fired perpendicular to B, returning to its original location means completing one circle. The question asks for the Time Period (T = 2πm / qB). (Crucial: Convert mass from μg to kg).

Example: Two Semicircles (JEE 2025)

Particle moves through two regions with different fields (B₁, B₂) forming semicircles of radii r₁ = mv / qB₁ and r₂ = mv / qB₂. The net displacement is 2r₁ - 2r₂, as they trace paths in opposite directions.

Example: Rod in Solenoid (JEE 2026)

A charge released from rest moves along the axis of a solenoid. The magnetic field is along the axis. Since velocity (due to gravity) is parallel to B, magnetic force is zero. Acceleration is g.

Example: Work Done in E and B Fields (Conceptual)

Work done by magnetic force is always zero because F_m is always perpendicular to velocity. Total work done is only by the electric field (W = q(E ⋅ S)).

Magnetic Force on a Current-Carrying Wire

For a straight wire of length L carrying current I in a uniform magnetic field B:

F = I(L × B)

Magnitude: F = ILB sinθ, where θ is between L (current direction) and B.

• F_max = ILB (when L ⊥ B)

• F = 0 (when L || B)

(Memory Tip for Direction: Right-Hand Rule: Fingers in current direction (I), palm towards magnetic field (B), thumb points to force (F). )

For an arbitrarily shaped wire in a uniform magnetic field, the force is F = I(L_equivalent × B), where L_equivalent is the straight vector connecting start and end points. The net force on a closed loop in a uniform magnetic field is always zero.

Force Between Two Parallel Wires

Two long, parallel wires carrying currents i₁ and i₂, separated by distance r.

• Currents in Same Direction: Wires attract.

• Currents in Opposite Directions: Wires repel.

• Magnitude of Force per Unit Length: F/L = μ₀ \* i₁ \* i₂ / (2πr)

• Total Force: F = (μ₀ \* i₁ \* i₂ \* L) / (2πr)

Torque on a Current-Carrying Coil

A current-carrying coil in a uniform magnetic field experiences a torque, but the net force is zero.

• Magnetic Moment (M): M = NIA (N=turns, I=current, A=area). Direction is found by the right-hand rule (fingers curl with current, thumb points to M).

• Torque (τ): τ = M × B. Magnitude: τ = MB sinθ (θ between M and B). (Memory Tip: τ = BINA sinθ).

• Potential Energy (U): U = -M ⋅ B = -MB cosθ

Stable and Unstable Equilibrium

Magnetic Moment of Uniformly Charged Rotating Objects

For a uniformly charged object (charge q, mass M, moment of inertia I) rotating with angular velocity ω, its magnetic moment m is:

m = (q / 2M) * Iω

This applies to point charges, rings (I=MR²), discs (I=MR²/2), etc.

Moving Coil Galvanometer (MCG)

An MCG works on the principle that magnetic torque on a current coil is balanced by a spring's restoring torque.

CΦ = BINA (where C = torsional constant, Φ = deflection angle).

• Current Sensitivity (CS): CS = Φ / i = BNA / C

• Voltage Sensitivity (VS): VS = Φ / V = BNA / CR

Gauss's Law for Magnetic Field

∮ B ⋅ dA = 0

The net magnetic flux through any closed surface is always zero. This implies that magnetic monopoles do not exist, and magnetic field lines form closed loops.

Bar Magnet

A bar magnet is the magnetic analogue of an electric dipole.

• Magnetic Moment (M): M = m \* L (m=pole strength, L=magnetic length from S to N pole).

• Magnetic Field of a Bar Magnet: Analogous to an electric dipole.

• Axial Line: B_axial = (μ₀/4π) \* (2M/r³)

• Equatorial Line: B_equatorial = (μ₀/4π) \* (M/r³) (opposite to M)

Bar Magnet in an External Magnetic Field

In a uniform external magnetic field B:

• Net Force: F_net = 0

• Torque: τ = M × B = MB sinθ

• Potential Energy: U = -M ⋅ B = -MB cosθ

• Small Oscillations: For small displacements from stable equilibrium, it performs SHM with Time Period T = 2π√(I / MB).

Magnetic Properties of Materials

Materials placed in an external magnetic field B₀ develop a net field B_net = μ_r * B₀ = (1 + χ) * B₀.

• μ_r: Relative Permeability.

• χ: Magnetic Susceptibility. (μ_r = 1 + χ) (Memory Tip: Think of it as profit (χ) on an investment (μ_r = 1 + profit)).

Electromagnetic Induction

1. Magnetic Flux (Φ)

Magnetic flux measures the total magnetic field lines passing through an area.

For N turns, area A, in uniform field B: Φ = NBA cos(θ) (θ between B and area vector A).

• Maximum Flux: θ = 0° (plane ⊥ B).

• Zero Flux: θ = 90° (plane || B).

• Unit: Weber (Wb) or T·m².

2. Faraday's Law of Induction

The magnitude of induced EMF (ε) is the rate of change of magnetic flux (Φ):

ε = -dΦ/dt

EMF is induced by changing:

1. Magnetic Field (B) (e.g., B(t)).

2. Area (A) (e.g., motional EMF).

3. Angle (θ) (e.g., rotating coil, generators).
   For a rotating coil: Φ(t) = NBA cos(ωt), ε(t) = NBAω sin(ωt).

3. Lenz's Law

The negative sign in Faraday's Law represents Lenz's Law: The induced EMF and current oppose the change in magnetic flux that produced them.

• Increasing Flux: Induced field opposes (B_induced opposite to B_external).

• Decreasing Flux: Induced field supports (B_induced same direction as B_external).
  (Memory Tip: "will neither let it live nor let it die" – it counteracts the change.)

4. Average EMF, Current, and Total Charge Flow

• Average EMF: ε_avg = -ΔΦ / Δt

• Average Current: I_avg = ε_avg / R = -(1/R) \* (ΔΦ / Δt)

• Total Charge Flow (purely resistive circuit): q = ΔΦ / R

Solved Problems (PYQs - EMI)

Example: Loop Shape Conversion

A circular loop (radius r) becomes a square (side a = πr/2). The induced EMF is ε = B \* |A_final - A_initial| / Δt. Calculate A_initial = πr² and A_final = a².

Example: Loop in Solenoid with Variable Current

If solenoid current I(t) varies, it creates a time-varying B(t) = μ₀nI(t). This induces EMF ε = -dΦ/dt = -d(μ₀nI(t)A)/dt in an inner loop. Calculate I_rms = ε_rms / R. (Irrelevant data like loop velocity should be ignored).

Example: Thermal Energy Dissipated

For a loop with B(t) = B₀ sin(ωt), find Φ(t) = B(t)A. Then ε(t) = -dΦ/dt = -B₀Aω cos(ωt). Calculate peak I₀ = ε₀/R, then I_rms = I₀/√2. Average power P_avg = I_rms²R. Energy dissipated over one period T = 2π/ω is Energy = P_avg \* T.

Motional EMF

EMF induced across a conductor of length L moving with velocity V through magnetic field B:

ε = vBL (if V, B, L are mutually perpendicular).

• Polarity (High-Potential End): Right-hand rule (Fingers → V, Palm → B, Thumb → Positive End).

• Zero EMF Condition: If any two of V, B, L are collinear.

• Angled Vectors: Resolve velocity or length into components perpendicular to the other two.

• Irregularly Shaped Wires: Use L_equivalent (straight vector from start to end points).

Rotational EMF

For a rod of length L rotating about one end with angular velocity ω in magnetic field B:

ε = ½ BωL²

Solved Problems on Motional & Rotational EMF

Example: Falling Wire

Wire falling under gravity: v = √(2gh). Induced EMF ε = vBL.

Example: Oscillating Pendulum

Metallic pendulum wire of length L oscillates. Max EMF occurs at the bottom of the swing (max ω). Find ω_max using energy conservation (mgh = ½mv²), then ε_max = ½ Bω_max L².

Example: Terminal Velocity of Sliding Rod

Rod sliding on rails, in B-field. Terminal velocity v_t when gravitational force (mg) equals magnetic braking force (F_m = ILB). Since I = (v_tBL)/R, then mg = (v_tB²L²)/R. Solve for v_t = mgR / (B²L²).

Induced Electric Field

A time-varying magnetic field (dB/dt ≠ 0) creates a non-conservative induced electric field (E) with closed field lines.

• Inside cylindrical region (r < R): E = (r/2) \* (dB/dt)

• Outside region (r > R): E = (R²/2r) \* (dB/dt)

Self-Inductance

Self-inductance (L) is a coil's property to oppose changes in current.

• Depends on geometry and medium (μ).

• Analogue to mass in mechanics.

• Stores magnetic energy: U = ½ LI².

Mutual Inductance of Two Coaxial Solenoids

For two long coaxial solenoids (n₁ and n₂ turns per unit length) with inner volume V:

M = μ₀ n₁ n₂ V

If a core material of relative permeability μ_r is used, M = μ_r μ₀ n₁ n₂ V.

Transformer

A transformer uses a primary (N₁) and secondary (N₂) coil to change AC voltage.

• Voltage Relation: V₂ / V₁ = N₂ / N₁ (Memory Tip: More turns, more voltage.)

• Efficiency (η): η = (P_out / P_in) \* 100 = (V₂I₂ / V₁I₁) \* 100

• Ideal Transformer: η = 100%, so P_out = P_in, meaning V₂I₂ = V₁I₁. Assume ideal if efficiency is not given.