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JEE Main chapters · Class 12 · Physics

Magnetic Effects of Current and Magnetism: JEE Main Physics formulas and notes

Magnetic Effects of Current and Magnetism gives about 3 questions in JEE Mains every year, mostly short numericals and statements. The ones that come most are the field of wires, loops and solenoids, the radius and the helix of a charged particle, the torque and energy of a loop or a magnet, the galvanometer and its conversions, and the elements of the earth’s field. Learn the formulas of the fields and the motion with their directions, and use about 8 hours over 5 days.

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Formulas with solved examples

Biot–Savart law and fields of simple currents

Biot–Savart law

How to apply. Divide the conductor into elements, use the symmetry to see which components cancel, and add the rest. The direction comes from the right-hand rule: thumb along the current, fingers curl along B.

Example. A proton moves at 10⁶ m/s. At 1 mm from it, on a line perpendicular to its velocity, B = 10⁻⁷ × 1.6 × 10⁻¹⁹ × 10⁶ ÷ 10⁻⁶ = 1.6 × 10⁻¹⁴ T.

Long straight wire

How to apply. Use B = 2 × 10⁻⁷ I/d with d in metre. A wire that stops at the foot of the perpendicular gives exactly half the infinite value.

Example. I = 10 A, d = 5 cm: B = 2 × 10⁻⁷ × 10 ÷ 0.05 = 4 × 10⁻⁵ T.

Finite straight wire

How to apply. Draw the perpendicular from the point to the wire. Measure both angles from this perpendicular, then add the two sines. If the point is beyond an end, subtract.

Example. Wire of length 20 cm, I = 10 A, point on the bisector at 10 cm: α = β = 45°, B = 10⁻⁷ × 10 ÷ 0.1 × 2 × 0.707 = 1.41 × 10⁻⁵ T.

Centre of a circular loop and of an arc

How to apply. Count the arcs as a fraction of the full circle: B = (θ/2π) × μ₀I/2R. Add the arcs as vectors with their directions.

Example. Semicircle of R = 20 cm, I = 5 A: B = 4π × 10⁻⁷ × 5 ÷ (4 × 0.2) = 7.85 × 10⁻⁶ T. A 90° arc with R = 10 cm and I = 4 A gives 6.28 × 10⁻⁶ T.

Axis of a circular loop

How to apply. Put x and R in the same unit and keep the 3/2 power. For x ≫ R the loop behaves like a bar magnet of moment NIA.

Example. R = 3 cm, x = 4 cm, N = 1, I = 5 A: B = 4π × 10⁻⁷ × 5 × 9 × 10⁻⁴ ÷ (2 × (25 × 10⁻⁴)^3/2) = 2.26 × 10⁻⁵ T (the value at the centre is 1.05 × 10⁻⁴ T).

Square and polygon loops

How to apply. Find the field of one side with the finite-wire formula (both angles π/n), then multiply by the number of sides.

Example. Square of side 20 cm, I = 5 A: B = 2√2 × 4π × 10⁻⁷ × 5 ÷ (π × 0.2) = 2.83 × 10⁻⁵ T.

Rotating charged ring or disc

How to apply. Turn the revolving charge into a current I = Q/T and use the loop formulas. For a disc add thin rings.

Example. Ring of R = 10 cm, Q = 3 μC, ω = 100 rad/s: B = 10⁻⁷ × 3 × 10⁻⁶ × 100 ÷ 0.1 = 3 × 10⁻¹⁰ T.

Two fields at one point

How to apply. Find the direction of each field with the right-hand rule first. Add magnitudes only when the directions are the same or opposite.

Example. Two long wires 20 cm apart carry 10 A in opposite directions. Midway the fields add: 2 × 2 × 10⁻⁷ × 10 ÷ 0.1 = 4 × 10⁻⁵ T. With equal currents in the same direction the field midway is zero.

Ampere’s law, solenoid and toroid

Ampere’s circuital law

How to apply. Choose a closed path on which B is constant (or zero) along each piece, write B × length = μ₀I_enc, and solve for B.

Example. A path that encloses 5 A one way and 2 A the other way has ∮B·dl = μ₀ × 3 = 3.77 × 10⁻⁶ T m.

Cylindrical conductor

How to apply. For a solid cylinder only the current inside the radius r counts: I_enc = I r²/a². Outside, treat it as a wire on the axis.

Example. I = 10 A, a = 2 cm, r = 1 cm: B = 2 × 10⁻⁷ × 10 × 0.01 ÷ (0.02)² = 5 × 10⁻⁵ T, which is half the value at the surface.

Solenoid

How to apply. Use n = turns per metre (N/L). Convert turns per cm to turns per metre first (×100). The field at the end is half the middle value.

Example. n = 20 turns per cm = 2000 per m, I = 2 A: B = 4π × 10⁻⁷ × 2000 × 2 = 5.03 × 10⁻³ T, and 2.5 × 10⁻³ T at an end.

Toroid

How to apply. Take a circle of radius r through the ring: B × 2πr = μ₀NI. The field is larger on the inner side of the ring.

Example. N = 500 turns, I = 2 A, mean radius 10 cm: B = 2 × 10⁻⁷ × 500 × 2 ÷ 0.1 = 2 × 10⁻³ T.

Solenoid problems and winding wire

How to apply. Write N or n in terms of the data first. Then put it into μ₀nI.

Example. A tube of radius 2 cm, length 20 cm is wound with 25.1 m of wire in one layer: N = 25.1 ÷ (2π × 0.02) = 200 turns, n = 1000 per metre. With I = 1 A, B = 1.26 × 10⁻³ T.

Force on moving charges and on wires

Force on a moving charge

How to apply. Find the component of v across B. The force is qvB for that component, perpendicular to both. Never write that the magnetic field changes the energy.

Example. A proton at 2 × 10⁶ m/s perpendicular to B = 0.5 T feels F = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.5 = 1.6 × 10⁻¹³ N.

Force on a current-carrying wire

How to apply. For a bent wire in a uniform field, replace it by the straight chord joining its ends. A semicircle of radius R has L_eff = 2R.

Example. I = 5 A, L = 0.4 m, B = 0.3 T at 30°: F = 5 × 0.4 × 0.3 × 0.5 = 0.3 N. A semicircular wire of R = 10 cm with the same I and B = 0.4 T (field in the plane of the loop, at right angles to the chord): F = 2 × 0.1 × 5 × 0.4 = 0.4 N.

Force between parallel wires

How to apply. Take the field of one wire at the position of the other and then the force on a length ℓ of it. Check the direction by the rule: parallel currents attract.

Example. Two wires 10 cm apart, both 5 A in the same direction: F/ℓ = 2 × 10⁻⁷ × 25 ÷ 0.1 = 5 × 10⁻⁵ N/m, attraction.

Definition of the ampere

How to apply. Put I₁ = I₂ = 1 A and d = 1 m in the formula for parallel wires.

Example. F/ℓ = 2 × 10⁻⁷ × 1 × 1 ÷ 1 = 2 × 10⁻⁷ N/m.

Wire in the field of another wire or loop

How to apply. Calculate the force on each side separately (the two sides perpendicular to the wire cancel), then add the signed forces.

Example. Square loop of side 10 cm, I = 5 A, with a side parallel to a long wire carrying 10 A at 10 cm. The near side feels 2 × 10⁻⁷ × 10 × 5 ÷ 0.1 × 0.1 = 10⁻⁵ N, the far side (20 cm) feels 5 × 10⁻⁶ N the other way. Net force 5 × 10⁻⁶ N.

Motion in fields: circle, helix, cyclotron, selector

Circular motion in a magnetic field

How to apply. Pick the form of r that fits the data: v, p, K or V. Remember that T is the same for all speeds.

Example. An electron accelerated through 100 V enters B = 1 mT at right angles: v = 5.93 × 10⁶ m/s, r = 9.1 × 10⁻³¹ × 5.93 × 10⁶ ÷ (1.6 × 10⁻¹⁹ × 10⁻³) = 3.4 cm.

Comparing particles in the same field

How to apply. Write r in terms of the quantity that is common. Do not use r = mv/qB when the energy or the voltage is given: it hides the mass dependence.

Example. Proton and α particle enter the same B with the same K: r_p : r_α = (1/1) : (√4/2) = 1 : 1. With the same V it is 1 : √2.

Helical motion

How to apply. Split v along and across B. The radius uses v⊥, the pitch uses v∥ × T.

Example. Proton at 10⁶ m/s, θ = 60° to B = 0.1 T: r = 1.67 × 10⁻²⁷ × 10⁶ × 0.866 ÷ (1.6 × 10⁻¹⁹ × 0.1) = 9.0 cm and pitch = 2π × 1.67 × 10⁻²⁷ × 10⁶ × 0.5 ÷ (1.6 × 10⁻²⁰) = 32.8 cm.

Deflection in a region of width d

How to apply. Find r first. Compare d with r. Then sinθ = d/r gives the angle turned and t = (θ/2π)T.

Example. A proton of r = 10 cm in B = 0.1 T crosses a strip 5 cm wide: sinθ = 0.5, θ = 30°. The time is (π/6) × 1.67 × 10⁻²⁷ ÷ (1.6 × 10⁻¹⁹ × 0.1) = 5.5 × 10⁻⁸ s.

Velocity selector

How to apply. Equate the electric and the magnetic force. If E and B are given, v = E/B at once.

Example. E = 3000 V/m and B = 0.01 T pass particles of v = 3000 ÷ 0.01 = 3 × 10⁵ m/s.

Cyclotron

How to apply. Find f from B and the mass. For the energy, put R of the last orbit in K = q²B²R²/2m.

Example. Proton, B = 1 T: f = 1.6 × 10⁻¹⁹ ÷ (2π × 1.67 × 10⁻²⁷) = 15.2 MHz. With R = 0.5 m, K = (1.6 × 10⁻¹⁹)² × 1 × 0.25 ÷ (2 × 1.67 × 10⁻²⁷) = 1.9 × 10⁻¹² J = 12 MeV.

Cycloid and crossed fields

How to apply. Write the net force qE + qv × B. If it is zero the motion is straight; otherwise find which force is larger.

Example. E = 600 V/m, B = 0.2 T: the drift speed is 3000 m/s and the highest speed of a particle released from rest is 6000 m/s.

Torque on a loop and the galvanometer

Torque on a current loop and magnetic moment

How to apply. Find m first, then θ between the normal of the loop (not its plane) and B. If the plane is parallel to B, θ = 90° and τ is largest.

Example. 100 turns, area 0.02 m², I = 0.5 A, B = 0.2 T with the plane parallel to B: τ = 100 × 0.5 × 0.02 × 0.2 = 0.2 N m.

Dipole in a magnetic field: energy and work

How to apply. Find the angle between m and B in the start and end positions. The work done by an outside agent equals the change of U.

Example. m = 3 A m², B = 0.2 T, θ = 30°: τ = 0.30 N m and U = −0.52 J. Turning from 0° to 180° needs 2 × 3 × 0.2 = 1.2 J.

Magnetic moment of an orbiting electron

How to apply. Write I = e/T and A = πr². Then μ = IA = evr/2 = (e/2m)L.

Example. Hydrogen ground state (n = 1): μ = 1.6 × 10⁻¹⁹ × 6.63 × 10⁻³⁴ ÷ (4π × 9.1 × 10⁻³¹) = 9.28 × 10⁻²⁴ A m².

Moving coil galvanometer

How to apply. Write the balance of the magnetic torque and the spring torque. Make a galvanometer more sensitive by raising N, A, B and lowering k.

Example. A coil with N = 100, A = 2 × 10⁻³ m², B = 0.2 T and k = 4 × 10⁻⁶ N m/rad turns by θ/I = 100 × 2 × 10⁻³ × 0.2 ÷ (4 × 10⁻⁶) = 10⁴ rad/A.

Conversion to ammeter and voltmeter

How to apply. Ig and G are the full-scale current and the resistance of the galvanometer. The shunt carries the extra current I − Ig, with the same voltage IgG.

Example. G = 50 Ω, Ig = 2 mA. For 0 to 1 A: S = 0.002 × 50 ÷ 0.998 = 0.10 Ω. For 0 to 10 V: R = 10 ÷ 0.002 − 50 = 4950 Ω in series.

Tangent galvanometer

How to apply. Put the coil in the meridian so that its field is perpendicular to B_H. Then tanθ = B_coil/B_H.

Example. N = 50, R = 10 cm, B_H = 3 × 10⁻⁵ T: K = 2 × 0.1 × 3 × 10⁻⁵ ÷ (4π × 10⁻⁷ × 50) = 0.0955 A, so a deflection of 45° means a current of 0.0955 A.

Bar magnet as a magnetic dipole

Field of a short bar magnet

How to apply. Quote the point as axial or equatorial first. Use r from the centre of the magnet. The same formulas hold for a small current loop of moment NIA.

Example. m = 2 A m² at r = 0.2 m: axial B = 10⁻⁷ × 2 × 2 ÷ 0.008 = 5 × 10⁻⁵ T and equatorial B = 2.5 × 10⁻⁵ T.

Poles, moment and dividing a magnet

How to apply. Start with m = qm × 2l and apply the cut. The pole strength is set by the area and the magnetisation, so it does not change when the length is cut.

Example. A magnet of moment 4 A m² is cut into two equal pieces across its length: each piece has m = 2 A m². The same magnet cut along its length gives two pieces of 2 A m².

Torque and energy of a bar magnet

How to apply. Treat the magnet exactly as a loop: the moment is m, the angle is between m and B.

Example. A magnet of m = 0.5 A m² in B = 0.4 T at 60°: τ = 0.5 × 0.4 × 0.866 = 0.173 N m.

Oscillation of a magnet

How to apply. The restoring torque is mB sinθ ≈ mBθ, so the motion is simple harmonic with ω² = mB/I. Use B_H for a magnet in the earth’s field.

Example. m = 2 A m², I = 2 × 10⁻⁵ kg m², B = 0.1 T: T = 2π√(2 × 10⁻⁵ ÷ 0.2) = 0.0628 s. With B four times larger, T = 0.0314 s.

Vibration magnetometer and null points

How to apply. Use the same m and I; only B changes, so T² ∝ 1/B. For a null point set the magnet field equal and opposite to B_H and solve for r.

Example. A magnet makes 10 oscillations in 20 s in the field B₁ and in 40 s in field B₂ (T₂ = 2T₁): B₁ : B₂ = 4 : 1.

Earth’s magnetism

Elements of the earth’s field

How to apply. Draw the right triangle B_H, B_V, B. Any two of the elements give the others.

Example. B_H = 3 × 10⁻⁵ T and δ = 60°: B = 3 × 10⁻⁵ ÷ 0.5 = 6 × 10⁻⁵ T and B_V = 3 × 10⁻⁵ × tan 60° = 5.2 × 10⁻⁵ T.

Apparent dip

How to apply. The vertical component stays the same. Only the horizontal component along the new plane is B_H cos α.

Example. δ = 45° and α = 60°: tan δ′ = 1 ÷ 0.5 = 2, so δ′ = 63.4°.

Earth as a dipole and field in magnets

How to apply. Keep the naming straight: a compass needle’s N-pole points north because the earth’s magnetic pole in the north is a magnetic south pole.

Example. At the magnetic equator B_V = 0 and B = B_H. At a magnetic pole B_H = 0 and a dip needle stands vertical.

Magnetic materials and Gauss’s law

Gauss’s law for magnetism

How to apply. Use it to check a claim: if a question asks for the flux of B through a closed surface, the answer is zero. For an open surface use Φ = BA cosθ.

Example. Flux of a uniform field of 0.2 T through a loop of area 0.05 m² with the normal at 60° to B: Φ = 0.2 × 0.05 × 0.5 = 5 × 10⁻³ Wb. Through a closed box it is 0.

H, M, χ and μr

How to apply. H comes only from the free current (nI); M is the response of the material. Use B = μ₀(1 + χ)H for a quick answer.

Example. Solenoid n = 1000 per metre, I = 0.5 A, iron core μr = 500: B = 4π × 10⁻⁷ × 500 × 1000 × 0.5 = 0.314 T. A material with χ = 299 has μr = 300.

Dia, para and ferro magnetism

How to apply. Classify by the sign and size of χ. For temperature changes use χ ∝ 1/T for para and χ ∝ 1/(T − T_c) for ferro above T_c.

Example. χ of a paramagnetic salt is 3 × 10⁻³ at 300 K; at 400 K it is 3 × 10⁻³ × 300 ÷ 400 = 2.25 × 10⁻³.

Hysteresis and uses of materials

How to apply. Match the use with the loop: tall and thin for cores that are magnetised again and again, wide for permanent magnets.

Example. A core taken through 50 cycles per second with a loop area (B–H) of 400 J/m³ per cycle loses 400 × 50 = 2 × 10⁴ W per cubic metre.

Electromagnets, Curie point and magnetisation

How to apply. Use B = μ₀μr nI to compare an air-core and an iron-core solenoid. The ratio is μr.

Example. A solenoid with 2000 turns per metre carries 1 A. The field is 2.5 × 10⁻³ T in air and 0.5 T with a core of μr = 200, an increase of 200 times.

Most asked question types

How do I solve “Field of a straight wire, finite or infinite” questions in JEE Main?

A long or a finite wire carries a current. Find B at a point, or the distance at which B has a stated value. 1. Infinite wire: B = 2 × 10⁻⁷ I/d. 2. Finite wire: B = (μ₀I/4πd)(sinα + sinβ). 3. Use the right-hand rule for the direction. Example: 10 A, 5 cm: 4 × 10⁻⁵ T.

How do I solve “Centre of loops, arcs and combinations” questions in JEE Main?

A wire is bent into a circle, semicircle or arcs joined by straight pieces. Find B at the centre. 1. Arc: B = μ₀Iθ/4πR. 2. Straight pieces on a line through the centre give zero. 3. Add the arcs as vectors. Example: Semicircle R = 20 cm, 5 A: 7.85 × 10⁻⁶ T.

How do I solve “Field on the axis of a loop or a coil” questions in JEE Main?

A coil of N turns carries a current. Find B at a distance x on its axis, or the ratio to the value at the centre. 1. B = μ₀NIR²/2(R² + x²)^3/2. 2. Ratio to centre: [R²/(R² + x²)]^3/2. 3. Far away B = (μ₀/4π)(2m/x³). Example: R = 3 cm, x = 4 cm: 0.216 of the centre value.

How do I solve “Ampere’s law for cylinders and sheets” questions in JEE Main?

A thick wire or a conducting tube has a current. Find B inside and outside, or where B is the same. 1. Choose a circular path of radius r. 2. I_enc = I r²/a² inside a solid cylinder. 3. Outside B = μ₀I/2πr. Example: 10 A, a = 2 cm, r = 1 cm: 5 × 10⁻⁵ T.

How do I solve “Solenoid and toroid” questions in JEE Main?

The turns, the length and the current of a solenoid or toroid are given. Find B inside, at the end or in terms of the radius. 1. n = N/L in turns per metre. 2. Solenoid B = μ₀nI, end half of it. 3. Toroid B = μ₀NI/2πr. Example: 2000 per m, 2 A: 5.03 × 10⁻³ T.

How do I solve “Force on a charge in a magnetic field” questions in JEE Main?

A charge enters a magnetic field with a given velocity. Find the force, its direction or the acceleration. 1. Take the component of v across B. 2. F = qvB sinθ. 3. Direction by v × B, reversed for a negative charge. Example: Proton, 2 × 10⁶ m/s, 0.5 T: 1.6 × 10⁻¹³ N.

How do I solve “Force on a straight or bent wire” questions in JEE Main?

A wire of given length and shape carries a current in a uniform field. Find the force on it. 1. F = BIL sinθ for a straight piece. 2. Bent wire: use the chord between the ends. 3. Closed loop: net force zero. Example: 5 A, 0.4 m, 0.3 T, 30°: 0.3 N.

How do I solve “Force between parallel wires and the ampere” questions in JEE Main?

Two long wires carry given currents at a given distance. Find the force per metre or the place where the net field or force is zero. 1. F/ℓ = 2 × 10⁻⁷ I₁I₂/d. 2. Same direction: attract. 3. Zero field lies between the wires only if the currents are in the same direction. Example: 5 A and 5 A at 10 cm: 5 × 10⁻⁵ N/m.

How do I solve “Radius and period of circular motion” questions in JEE Main?

A particle enters a field at right angles with a given speed, K or V. Find r, T or f, or compare two particles. 1. r = mv/qB = √(2mK)/qB. 2. T = 2πm/qB. 3. Same K: r ∝ √m/q; same V: r ∝ √(m/q). Example: Electron, 100 V, 1 mT: 3.4 cm.

How do I solve “Helical motion” questions in JEE Main?

The velocity makes an angle with B. Find the radius, the pitch or the time for one turn. 1. v∥ = v cosθ, v⊥ = v sinθ. 2. r = mv sinθ/qB. 3. Pitch = v∥ × T. Example: Proton, 10⁶ m/s, 60°, 0.1 T: r = 9.0 cm, pitch = 32.8 cm.

How do I solve “Velocity selector and crossed fields” questions in JEE Main?

E and B are crossed. Find the speed of the undeflected charge, or the path of one with another speed. 1. qE = qvB. 2. v = E/B. 3. Larger force decides the side of bending. Example: 3000 V/m, 0.01 T: 3 × 10⁵ m/s.

How do I solve “Cyclotron frequency and energy” questions in JEE Main?

The field and the radius of a cyclotron are given. Find the frequency, the final speed or the energy. 1. f = qB/2πm. 2. v = qBR/m. 3. K = q²B²R²/2m. Example: Proton, 1 T, 0.5 m: 15.2 MHz, 12 MeV.

How do I solve “Torque and magnetic moment of a loop” questions in JEE Main?

A coil is placed in a field at a given angle. Find the moment, the torque, the energy or the work to turn it. 1. m = NIA along the normal. 2. τ = mB sinθ. 3. U = −mB cosθ. Example: 3 A m², 0.2 T, 30°: 0.30 N m.

How do I solve “Magnetic moment of a revolving charge” questions in JEE Main?

An electron or a charged ring goes round. Find the current, the moment or the ratio of the moment to the angular momentum. 1. I = q/T. 2. μ = IA = (q/2m)L. 3. Bohr model: L = nh/2π. Example: Hydrogen n = 1: 9.27 × 10⁻²⁴ A m².

How do I solve “Galvanometer sensitivity” questions in JEE Main?

A coil of given N, A, B and k is given. Find the deflection per unit current or voltage, or the effect of changing N. 1. NIAB = kθ. 2. θ/I = NAB/k. 3. θ/V = NAB/kR. Example: N = 100, 2 × 10⁻³ m², 0.2 T, 4 × 10⁻⁶: 10⁴ rad/A.

How do I solve “Ammeter and voltmeter from a galvanometer” questions in JEE Main?

G and Ig are given. Find the shunt for a range of current or the series resistance for a range of voltage. 1. S = IgG/(I − Ig). 2. R = V/Ig − G. 3. Combined resistance GS/(G + S) or G + R. Example: 50 Ω, 2 mA: S = 0.10 Ω, R = 4950 Ω.

How do I solve “Field of a bar magnet or a small loop” questions in JEE Main?

A short magnet has a moment m. Find B at a point on the axis or on the equatorial line, or the ratio of the two. 1. Axial (μ₀/4π)2m/r³. 2. Equatorial (μ₀/4π)m/r³. 3. General angle: √(1 + 3cos²θ). Example: 2 A m², 0.2 m: 5 × 10⁻⁵ T and 2.5 × 10⁻⁵ T.

How do I solve “Time period of a magnet in a field” questions in JEE Main?

A magnet vibrates in a field. Find T, compare two fields or two magnets, or find B_H from the oscillations. 1. T = 2π√(I/mB). 2. B ∝ 1/T². 3. Add I and m for a combination. Example: 2 A m², 2 × 10⁻⁵ kg m², 0.1 T: 0.0628 s.

How do I solve “Earth’s magnetic elements” questions in JEE Main?

Two of B, B_H, B_V and the dip are given. Find the others. 1. B_H = B cosδ. 2. B_V = B sinδ. 3. tanδ = B_V/B_H. Example: B_H = 3 × 10⁻⁵ T, 60°: B = 6 × 10⁻⁵ T.

How do I solve “Apparent dip and tangent galvanometer” questions in JEE Main?

A dip needle is in a plane at an angle to the meridian, or a coil is in the meridian. Find the apparent dip or the current. 1. tanδ′ = tanδ/cosα. 2. B_coil = B_H tanθ. 3. I = K tanθ with K = 2RB_H/μ₀N. Example: 45°, α = 60°: 63.4°.

How do I solve “Dia, para and ferromagnetic materials” questions in JEE Main?

Values of χ, μr or the behaviour in a field or with temperature are given. Classify the material. 1. Dia: χ < 0, para: χ > 0 small, ferro: large χ. 2. Para χ ∝ 1/T. 3. Ferro above T_c becomes para. Example: 3 × 10⁻³ at 300 K gives 2.25 × 10⁻³ at 400 K.

How do I solve “Hysteresis, electromagnets and permanent magnets” questions in JEE Main?

The loop of a material is described. Say which is suitable for a core or a permanent magnet, or find the energy lost. 1. Retentivity and coercivity from the loop. 2. Loop area = energy loss per cycle per volume. 3. Soft iron for cores, steel for magnets. Example: 400 J/m³ per cycle, 50 Hz: 2 × 10⁴ W/m³.

Common myths and the truth

Mistakes to avoid

Simulations for Magnetic Effects of Current and Magnetism

More: Class 12 Physics simulations · all JEE Main chapters.