Magnetism — Concepts, Formulas & Solved Numericals

Magnetism — The Force That Runs Half Your Physics Paper

Magnetism shows up everywhere in Class 12 — from the force on a current-carrying conductor to the working of a cyclotron. Students who understand this chapter well typically pick up 8-12 marks in JEE Main without much struggle, because the pattern of questions is highly predictable.

The central idea is simple: moving charges create magnetic fields, and magnetic fields exert forces on moving charges. That’s it. Everything else — Biot-Savart, Ampere’s Law, the cyclotron — is just this idea applied in different geometries.

We’ll work through this chapter the way it actually appears in exams: starting from field calculations, moving to force and motion, then torque and magnetic moments. By the end, you’ll have a clean mental map and every formula you need.

Key Terms & Definitions

Magnetic Field (B): The region around a magnet or current-carrying conductor where a magnetic force acts on other magnets or moving charges. SI unit is Tesla (T). 1 T = 1 Wb/m².

Magnetic Flux (Φ): The total number of magnetic field lines passing through a surface. $\Phi = B \cdot A \cdot \cos\theta$ . Unit: Weber (Wb).

Permeability of Free Space (μ₀): A constant that appears in all magnetism formulas. $\mu_0 = 4\pi \times 10^{-7}$ T·m/A.

Magnetic Dipole Moment (m): For a current loop, $m = NIA$ . Direction: perpendicular to the loop plane, given by the right-hand rule. Unit: A·m².

Lorentz Force: The combined electric and magnetic force on a charge: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$ .

Ampere: Defined in terms of the force between two parallel current-carrying wires — this definition appears in CBSE theoretical questions regularly.

Core Concepts and Methods

1. Biot-Savart Law — Finding B Due to a Current Element

The Biot-Savart law gives the magnetic field $d\vec{B}$ due to a small current element $Id\vec{l}$ :

d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \, d\vec{l} \times \hat{r}}{r^2}

where $r$ is the distance from the current element to the point, and $\hat{r}$ is the unit vector from element to field point.

Why the cross product? The field is perpendicular to both the current direction and the line joining the element to the point. This is what makes it different from Coulomb’s law, where the force is along the line joining charges.

Important results you must know cold:

Configuration	Magnetic Field
Infinite straight wire	$B = \dfrac{\mu_0 I}{2\pi r}$
Circular loop at centre	$B = \dfrac{\mu_0 I}{2R}$
Circular loop at axial point (distance $x$ )	$B = \dfrac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$
Solenoid (inside, infinite)	$B = \mu_0 n I$
Toroid (inside)	$B = \dfrac{\mu_0 N I}{2\pi r}$

Here $n$ = number of turns per unit length, $N$ = total turns, $r$ = radius of toroid.

For the circular loop axial field, when $x \gg R$ , the formula simplifies to $B \approx \dfrac{\mu_0 I R^2}{2x^3}$ , which looks like a magnetic dipole field. This connection between a current loop and a magnetic dipole is asked frequently in JEE.

2. Ampere’s Circuital Law

When the current distribution has symmetry, Biot-Savart becomes messy. Ampere’s Law is the cleaner tool:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}

The line integral of $\vec{B}$ around any closed loop equals $\mu_0$ times the net current enclosed by that loop.

How to apply it — 3 steps:

Identify the symmetry (cylindrical → choose a circular Amperian loop)
Choose the Amperian loop so that $B$ is constant and parallel to $d\vec{l}$
Solve: $B \cdot 2\pi r = \mu_0 I_{\text{enc}}$

This directly gives the $B = \dfrac{\mu_0 I}{2\pi r}$ result for an infinite wire.

3. Force on a Charged Particle in a Magnetic Field

\vec{F} = q(\vec{v} \times \vec{B})

Magnitude: $F = qvB\sin\theta$

Key insight: The magnetic force is always perpendicular to $\vec{v}$ . This means it does no work on the particle — it changes direction but not speed. This is a favourite conceptual question in CBSE boards.

Circular motion in a uniform magnetic field:

When $\vec{v} \perp \vec{B}$ , the particle moves in a circle. Equating magnetic force to centripetal force:

qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}

The time period $T = \dfrac{2\pi m}{qB}$ is independent of speed — this is the principle behind the cyclotron.

JEE Main pattern: Questions on charged particle motion often combine electric and magnetic fields. If a particle moves undeflected through crossed $E$ and $B$ fields, then $qE = qvB$ , giving the velocity selector condition $v = E/B$ . This appeared in JEE Main 2023 Session 1.

4. Force on a Current-Carrying Conductor

\vec{F} = I(\vec{L} \times \vec{B})

For a straight conductor of length $L$ : $F = BIL\sin\theta$

Force between two parallel wires: Two wires carrying currents $I_1$ and $I_2$ separated by distance $d$ :

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Same-direction currents attract; opposite-direction currents repel. Remember this with the logic: same-direction currents create fields that squeeze toward each other.

5. Torque on a Current Loop — The Galvanometer Principle

A rectangular current loop in a uniform magnetic field $B$ experiences a torque:

\vec{\tau} = \vec{m} \times \vec{B}

Magnitude: $\tau = NIAB\sin\theta$

where $\theta$ is the angle between $\vec{m}$ (magnetic moment) and $\vec{B}$ .

The loop reaches equilibrium when $\vec{m} \parallel \vec{B}$ (i.e., $\theta = 0$ ). Maximum torque occurs at $\theta = 90°$ .

The moving coil galvanometer uses this torque against a restoring spring. At equilibrium: $NIAB = k\phi$ , giving $\phi = \dfrac{NIAB}{k}$ , where $\phi$ is the deflection.

6. Magnetism in Matter

Material	Behaviour	$\chi_m$	Example
Diamagnetic	Weakly repelled	Small, negative	Bismuth, Copper
Paramagnetic	Weakly attracted	Small, positive	Aluminium, Oxygen
Ferromagnetic	Strongly attracted	Large, positive	Iron, Cobalt

Curie’s Law for paramagnetics: $M = C \cdot \dfrac{B}{T}$ , where $C$ is the Curie constant. Above the Curie temperature, ferromagnetics become paramagnetic.

Diamagnetism is a quantum effect — classical physics cannot explain it. Paramagnetism comes from permanent magnetic moments aligning with $B$ . Ferromagnetism arises from magnetic domains. CBSE asks the distinction between these every few years.

Solved Examples

Example 1 — Easy (CBSE Level)

A straight wire carries a current of 5 A. Find the magnetic field at a perpendicular distance of 10 cm from the wire.

Using $B = \dfrac{\mu_0 I}{2\pi r}$ :

B = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.10} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.10}

$B = \frac{2 \times 10^{-6}}{0.10} = 1 \times 10^{-5} \text{ T} = 10 \text{ }\mu\text{T}$

Example 2 — Medium (JEE Main Level)

A proton moving with velocity $2 \times 10^6$ m/s enters a uniform magnetic field $B = 0.1$ T perpendicular to its motion. Find the radius of its circular path. (Mass of proton = $1.67 \times 10^{-27}$ kg, charge = $1.6 \times 10^{-19}$ C)

r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 0.1}

r = \frac{3.34 \times 10^{-21}}{1.6 \times 10^{-20}} = 0.209 \text{ m} \approx 20.9 \text{ cm}

The time period: $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.1} \approx 6.56 \times 10^{-7}$ s

Note that $T$ is independent of velocity — this is key to how a cyclotron accelerates particles without changing frequency.

Example 3 — Hard (JEE Advanced Level)

A circular loop of radius $R = 0.1$ m carries current $I = 2$ A. Find the magnetic field at a point on its axis at distance $x = 0.1$ m from the centre. Also find the position where the axial field equals the field at the centre.

Axial field:

B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} = \frac{4\pi \times 10^{-7} \times 2 \times (0.1)^2}{2((0.1)^2 + (0.1)^2)^{3/2}}

(R^2 + x^2)^{3/2} = (0.01 + 0.01)^{3/2} = (0.02)^{3/2} = 2^{3/2} \times 10^{-3} = 2.83 \times 10^{-3}

B = \frac{4\pi \times 10^{-7} \times 2 \times 0.01}{2 \times 2.83 \times 10^{-3}} = \frac{8\pi \times 10^{-9}}{5.66 \times 10^{-3}} \approx 4.44 \times 10^{-6} \text{ T}

When does axial field equal centre field? At centre, $B_0 = \dfrac{\mu_0 I}{2R}$ . Setting axial field equal:

\frac{R^2}{(R^2 + x^2)^{3/2}} = \frac{1}{R} \implies (R^2 + x^2)^{3/2} = R^3

This gives $x = 0$ — the only solution. So the axial field is always less than the centre field, with maximum at $x = 0$ . A follow-up question often asks: at what $x$ does the axial field fall to half? That gives $(R^2 + x^2)^{3/2} = 2R^3$ , so $x = R(2^{2/3} - 1)^{1/2} \approx 0.766R$ .

Exam-Specific Tips

CBSE Class 12 (Board Exam)

The board exam typically has:

1 mark: State Biot-Savart law or Ampere’s law
2 marks: Derive field due to straight wire or circular loop
3 marks: Numerical on force, torque, or galvanometer sensitivity
5 marks: Complete derivation (moving coil galvanometer, force between parallel wires)

Scoring tip: The 5-mark derivation almost always asks for moving coil galvanometer OR cyclotron. Learn both derivations with diagrams. Labelled diagrams alone fetch 1-2 marks even if the derivation has errors.

JEE Main

Magnetism carries roughly 3-4 questions per session. High-weightage areas:

Magnetic field due to symmetric current distributions (Biot-Savart + Ampere)
Charged particle motion in combined $E$ and $B$ fields
Force on current-carrying conductors in non-trivial geometries

The velocity selector ( $v = E/B$ ) and the cyclotron frequency formula ( $\nu = qB/2\pi m$ ) appear almost every year in some form.

JEE Main 2024 had a question on a rectangular current loop in a non-uniform field, where you had to find the net force on the loop. The net force is zero in a uniform field — but in non-uniform fields, it’s non-zero. Know this distinction.

Common Mistakes to Avoid

Mistake 1 — Wrong direction for $\vec{F} = q\vec{v} \times \vec{B}$

Students apply the right-hand rule for positive charge and forget to reverse for negative charge (electrons). If $q$ is negative, the force direction flips. Always write the sign of $q$ explicitly before applying the cross product.

Mistake 2 — Confusing $n$ (turns/length) with $N$ (total turns)

In the solenoid formula $B = \mu_0 nI$ , $n$ is turns per unit length. If the problem gives total turns $N$ and length $L$ , then $n = N/L$ . Writing $B = \mu_0 NI$ is a very common error that costs full marks.

Mistake 3 — Forgetting that magnetic force does no work

“The speed of a charged particle in a magnetic field remains constant” is always true (for pure $B$ fields). If a question asks about kinetic energy change in a magnetic field alone — the answer is zero. Students sometimes calculate work using $F \cdot d$ and get a non-zero answer by mistake.

Mistake 4 — Applying Biot-Savart to infinite wire problems instead of Ampere’s Law

Both give the same answer, but Biot-Savart requires a messy integral for an infinite wire. Use Ampere’s Law when there’s sufficient symmetry (infinite wire, solenoid, toroid). Save Biot-Savart for finite wires and circular loops.

Mistake 5 — Torque on a loop: confusing angle convention

$\tau = NIAB\sin\theta$ where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$ , NOT the angle between the plane of the loop and $B$ . These two angles are complementary. When the plane is parallel to $B$ , $\theta = 90°$ and torque is maximum — not zero. Students mix these up every year.

Practice Questions

Q1. A long straight wire carries a current of 10 A. An electron moves parallel to the wire at a distance of 2 cm with velocity $3 \times 10^7$ m/s. Find the force on the electron if it moves in the same direction as the current.

First find $B$ due to wire at distance 2 cm:

B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.02} = 10^{-4} \text{ T}

Force on electron: $F = qvB\sin 90° = 1.6 \times 10^{-19} \times 3 \times 10^7 \times 10^{-4}$

F = 4.8 \times 10^{-16} \text{ N}

Direction: The wire’s $B$ at the electron’s location points into the page (using right-hand rule for wire current going right). Electron moves right with $v$ . $\vec{F} = q(\vec{v} \times \vec{B}) = (-e)(v\hat{x} \times (-B\hat{z})) = evB(-\hat{y})$ — the force is away from the wire (repulsive). This is expected: the electron moves opposite to conventional current, so it’s like an antiparallel current, which repels.

Q2. A circular coil of 50 turns, radius 4 cm carries a current of 2 A. Find the magnetic field at (a) the centre and (b) a point on the axis 3 cm from the centre.

$R = 0.04$ m, $N = 50$ , $I = 2$ A

(a) At centre:

B = \frac{\mu_0 N I}{2R} = \frac{4\pi \times 10^{-7} \times 50 \times 2}{2 \times 0.04} = \frac{4\pi \times 10^{-5}}{0.08} = 5\pi \times 10^{-4} \approx 1.57 \times 10^{-3} \text{ T}

(b) At axial point, $x = 0.03$ m:

R^2 + x^2 = 0.0016 + 0.0009 = 0.0025

(R^2 + x^2)^{3/2} = (0.0025)^{3/2} = (2.5 \times 10^{-3})^{3/2} = 1.25 \times 10^{-4}

B = \frac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}} = \frac{4\pi \times 10^{-7} \times 50 \times 2 \times 0.0016}{2 \times 1.25 \times 10^{-4}}

= \frac{4\pi \times 10^{-7} \times 0.16}{2.5 \times 10^{-4}} = \frac{6.4\pi \times 10^{-8}}{2.5 \times 10^{-4}} \approx 8.04 \times 10^{-4} \text{ T}

Q3. A solenoid of length 0.5 m has 1000 turns and carries a current of 5 A. Find the magnetic field inside. If an iron core with relative permeability 200 is inserted, what is the new field?

$n = 1000/0.5 = 2000$ turns/m

Without core: $B = \mu_0 n I = 4\pi \times 10^{-7} \times 2000 \times 5 = 4\pi \times 10^{-3} \approx 0.01257$ T

With iron core: $B = \mu_r \mu_0 n I = 200 \times 0.01257 \approx 2.51$ T

This factor-of-200 amplification is why iron cores are used in electromagnets and transformers.

Q4. Two parallel wires 20 cm apart carry currents of 6 A and 4 A in opposite directions. Find the force per unit length between them and state whether it’s attractive or repulsive.

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 6 \times 4}{2\pi \times 0.20} = \frac{24 \times 4 \times 10^{-7}}{0.40} = 2.4 \times 10^{-5} \text{ N/m}

Since currents are in opposite directions, the force is repulsive.

Q5. A rectangular coil of dimensions 8 cm × 5 cm with 100 turns carries a current of 3 A in a uniform magnetic field of 0.5 T. The plane of the coil makes 30° with the field. Find the torque.

When the plane makes 30° with $\vec{B}$ , the magnetic moment $\vec{m}$ makes 60° with $\vec{B}$ (since $\vec{m}$ is perpendicular to the plane).

$\tau = NIAB\sin\theta$ where $\theta = 60°$

$A = 0.08 \times 0.05 = 4 \times 10^{-3}$ m²

\tau = 100 \times 3 \times 4 \times 10^{-3} \times 0.5 \times \sin 60° = 0.6 \times \frac{\sqrt{3}}{2} = 0.3\sqrt{3} \approx 0.52 \text{ N·m}

Common trap: If the question says “plane makes 30° with field”, don’t use sin(30°). The angle in the torque formula is between $\vec{m}$ and $\vec{B}$ , which is 90° − 30° = 60°.

Q6. In a moving coil galvanometer, the coil has 50 turns, area 4 cm², in a radial field of 0.1 T. Spring constant is $2 \times 10^{-8}$ N·m/degree. Find the current sensitivity (deflection per unit current).

At equilibrium: $NIAB = k\phi$

Current sensitivity = $\dfrac{\phi}{I} = \dfrac{NAB}{k} = \dfrac{50 \times 4 \times 10^{-4} \times 0.1}{2 \times 10^{-8}}$

= \frac{2 \times 10^{-3}}{2 \times 10^{-8}} = 10^5 \text{ degree/A} = 10^5 \text{ div/A}

This is also expressed as $10^5$ divisions per ampere, or equivalently $0.1$ μA/division — a highly sensitive instrument.

Q7. A proton and an alpha particle enter a uniform magnetic field perpendicularly with the same kinetic energy. Find the ratio of their radii.

Kinetic energy $K = \frac{1}{2}mv^2$ , so $v = \sqrt{2K/m}$ and $p = mv = \sqrt{2mK}$

Radius $r = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2mK}}{qB}$

For proton: $m_p = m$ , $q_p = e$ , so $r_p = \dfrac{\sqrt{2mK}}{eB}$

For alpha: $m_\alpha = 4m$ , $q_\alpha = 2e$ , so $r_\alpha = \dfrac{\sqrt{8mK}}{2eB} = \dfrac{2\sqrt{2mK}}{2eB} = \dfrac{\sqrt{2mK}}{eB}$

\frac{r_p}{r_\alpha} = 1

The radii are equal! This surprises most students. It’s because doubling the charge and quadrupling the mass cancel out exactly at equal kinetic energies.

Q8. A current of 2 A flows in a semicircular arc of radius 5 cm. Find the magnetic field at the centre of the circle.

A full circle gives $B = \dfrac{\mu_0 I}{2R}$ . A semicircle gives half this, plus we need to check contributions from the straight portions.

The straight wire segments pass through the centre — a point lying on the line of the wire contributes zero field (since $d\vec{l} \times \hat{r} = 0$ ).

So the field comes entirely from the semicircular arc:

B = \frac{1}{2} \cdot \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R} = \frac{4\pi \times 10^{-7} \times 2}{4 \times 0.05} = \frac{8\pi \times 10^{-7}}{0.2} = 4\pi \times 10^{-6} \approx 1.26 \times 10^{-5} \text{ T}

FAQs

Why does a magnetic field do no work on a charged particle?

Work = $\vec{F} \cdot d\vec{r}$ . The magnetic force $\vec{F} = q\vec{v} \times \vec{B}$ is always perpendicular to $\vec{v}$ , which is the direction of $d\vec{r}$ . So the dot product is always zero. Speed stays constant; only direction changes.

What is the difference between magnetic flux density and magnetic field intensity?

$\vec{B}$ (magnetic flux density, in Tesla) is the actual field experienced by a particle. $\vec{H}$ (magnetic field intensity, in A/m) is related by $\vec{B} = \mu \vec{H}$ . In vacuum, $\mu = \mu_0$ . Inside a material, $\mu = \mu_r \mu_0$ . For board exams, the two are used somewhat interchangeably, but for JEE conceptual questions, know the distinction.

Why do same-direction currents attract each other?

Wire 1 creates a magnetic field at Wire 2’s location. By the right-hand rule, this field points toward Wire 1 (for parallel currents going upward, $B$ due to Wire 1 at Wire 2 points to the left). The force on Wire 2 carrying current upward in this leftward $B$ is $F = IL \times B$ — pointing left, i.e., toward Wire 1. Hence attraction.

What is the cyclotron frequency and why is it velocity-independent?

The time period $T = \dfrac{2\pi m}{qB}$ depends only on mass and charge, not on speed. As the particle is accelerated, its speed increases, but the radius also increases proportionally ( $r = mv/qB$ ), keeping the period constant. This is why the alternating voltage frequency in a cyclotron doesn’t need to change as the particle speeds up.

What happens to a magnetic dipole in a non-uniform field?

In a uniform field, a magnetic dipole experiences torque but zero net force. In a non-uniform field, it experiences both torque and a net translational force. This is why paramagnetic/ferromagnetic materials are attracted toward regions of stronger field.

What is magnetic susceptibility and what does its sign tell us?

$\chi_m = M/H$ where $M$ is magnetisation. If $\chi_m < 0$ : diamagnetic (repelled). If $\chi_m > 0$ and small: paramagnetic. If $\chi_m \gg 1$ : ferromagnetic. The sign directly tells you the direction of induced magnetisation relative to the applied field.

Can a uniform magnetic field confine a charged particle indefinitely?

Yes — if the velocity is purely perpendicular to $\vec{B}$ , the particle moves in a perfect circle and stays in that plane forever (ignoring radiation). If there’s a component parallel to $\vec{B}$ , the path becomes a helix. This helical motion along field lines is used in magnetic confinement in tokamak reactors.

Is the magnetic force between two moving charges the same as the force between two current-carrying wires?

Conceptually yes — both arise from $q\vec{v} \times \vec{B}$ . But for individual charges, we use the Lorentz force directly. For macroscopic wires, we first find $B$ due to one wire and then calculate $F = IL \times B$ on the other. The macroscopic formula is just a statistical sum over all the individual charge interactions.