Magnetic Effects of Current and Magnetism

Two long parallel wires carry current in the same direction. Do they attract or repel?

Most say repel. Like charges repel, so like currents should too.

They attract. Antiparallel currents are the ones that repel — the exact reverse of the electrostatic case, and this force is what defines the ampere.

The analogy misleads because there is no magnetic charge to be "like" in the first place. Three facts organise the chapter:

• No magnetic monopoles. Every magnetic field comes from moving charge. Field lines are closed loops, flux through any closed surface is zero, and a broken magnet gives two whole magnets.
• The magnetic force is always perpendicular to velocity, so it does no work. A magnetic field changes direction but never speed. Circular and helical paths follow from that one line.
• A current loop is a magnetic dipole, $m=NIA$. Every electric-dipole result transfers with $\vec{p}\to \vec{m}$ — torque, energy, even the factor of two between axial and equatorial.

Scope note. The cyclotron was removed from JEE Main in the 2023 revision and stays out for 2026, though it remains examinable in Advanced. Section 6 flags it.

1. Gauss's Law for Magnetism

${\Phi }_B=\oint \vec{B}\cdot d\vec{A}=0$

Always zero — in sharp contrast to the electric case, where the closed-surface flux equals $q_{enc}/{\epsilon }_0$.

There is no enclosed magnetic charge, ever. Every line entering a closed surface must leave again, because lines have nowhere to terminate. That is why magnetic field lines close on themselves while electric field lines start and stop on charges.

2. The Biot-Savart Law

$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2},{\mu }_0=4\pi \times 10^{-7}{\text{T m A}}^{-1}$

Trap. A current element produces no field along its own direction — the cross product vanishes. So a straight wire produces nothing at points on its own line, and a radial segment contributes nothing at the centre of an arc. Spotting this turns integration problems into ten-second problems.

Direction by the right-hand rule: thumb along the current, curled fingers give the sense in which $\vec{B}$ circles the wire.

3. Standard Field Results

The arc result is the workhorse: most exam configurations are arcs plus radial straights, and only the arcs need computing. A full loop is $\theta =2\pi$, recovering ${\mu }_{0}I/2R$.

A long solenoid's interior field depends only on turns per unit length — not on the total turns, not on the radius. Outside, it is nearly zero.

Illustration 1

Two concentric coplanar loops of radii $R$ and $2R$ carry equal currents $I$ in opposite senses. Find $\vec{B}$ at the common centre.

$B=\frac{{\mu }_{0}I}{2R}-\frac{{\mu }_{0}I}{2(2R)}\Rightarrow B=\frac{{\mu }_{0}I}{4R}$

directed as the smaller loop dictates. The inner loop always wins, because $B\propto 1/R$ at the centre — which is why a compact coil beats a large one at the same current.

Illustration 2

A wire carrying current $I$ comes in from far away along a radius, bends into a three-quarter circle of radius $R$, and leaves along another radius. Find $\vec{B}$ at the centre.

Both straight portions point directly at the centre, so $d\vec{l}\times \hat{r}=0$ everywhere along them and neither contributes anything at all. Only the arc counts, with $\theta =3\pi /2$:

$B=\frac{{\mu }_{0}I\theta }{4\pi R}=\frac{{\mu }_{0}I}{4\pi R}\cdot \frac{3\pi }{2}\Rightarrow B=\frac{3{\mu }_{0}I}{8R}$

The check worth running: an arc of angle $\theta$ should give the fraction $\theta /2\pi$ of the full-loop value ${\mu }_{0}I/2R$, and three-quarters of ${\mu }_{0}I/2R$ is indeed $3{\mu }_{0}I/8R$. Nothing else in the configuration matters at the centre.

4. Ampere's Circuital Law

$\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{enc}$

Ampere's law stands to Biot-Savart exactly as Gauss's law stands to Coulomb. No new physics — a repackaging so that symmetry does the integration, with the same limitation: it is always true, but useful only where a path exists on which $B$ is constant and either along or across $d\vec{l}$ everywhere.

Only current threading the loop counts. A wire lying outside changes $\vec{B}$ at every point of the path while contributing nothing to the integral.

Deriving the solenoid field

Take a rectangle with one long side inside the solenoid, the parallel side outside, and short sides across the axis.

• Outside side: $B\approx 0$, contributes nothing.
• Two short sides: $\vec{B}\perp d\vec{l}$, contribute nothing.
• Inner side: contributes $BL$, and encloses $nLI$.

$BL={\mu }_{0}nLI\Rightarrow B={\mu }_{0}nI$

One line. Summing Biot-Savart over every turn would be a substantial integral.

Illustration 3

A straight wire of radius $a$ carries current $I$ spread uniformly over its cross-section. Find $B$ inside and outside.

Outside, the whole current is enclosed. Inside, only the fraction within radius $r$:

$I_{enc}=I\frac{r^2}{a^2}\Rightarrow B(2\pi r)=\frac{{\mu }_{0}I{r}^2}{a^2}\Rightarrow B_{in}=\frac{{\mu }_{0}Ir}{2\pi a^2}$

The same shape as $E$ inside a uniformly charged solid sphere, and for the same reason: enclosed source grows faster than the geometric factor until you reach the boundary.

Magnetic flux

$\Phi =\int \vec{B}\cdot d\vec{A}[\text{weber}],1{\text{Wb m}}^{-2}=1\text{T}$

Through an open surface such as a loop, flux is generally not zero — unlike the closed-surface case of Section 1. This is the quantity the next chapter is built on, because induction depends on flux changing, not on the field itself.

5. Force on a Moving Charge

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

The magnetic term is zero when the charge is at rest, and zero when it moves parallel to $\vec{B}$. It is maximum when $\vec{v}\perp \vec{B}$.

Trap. The magnetic force never does work. Perpendicular to $\vec{v}$ at every instant, its dot product with displacement is always zero, so speed and kinetic energy are constant. Any question implying a magnetic field speeds a particle up is testing exactly this.

A velocity selector uses crossed $\vec{E}$ and $\vec{B}$ so the two forces oppose. Only particles with $v=E/B$ pass undeflected — independently of charge and mass, which is the whole point of the device.

Illustration 4

Crossed fields have $E=3\times 10^5$ V m⁻¹ and $B=0.6$ T. Find the selected speed, and say what happens to slower particles.

$qE=qvB\Rightarrow v=\frac{E}{B}=\frac{3\times 10^5}{0.6}=5\times 10^5{\text{m s}}^{-1}$

For a slower particle $qvB<qE$, so the electric force wins and it deflects toward the plates. Faster ones deflect the other way. Both charge and mass cancel out.

6. Motion of a Charged Particle in a Magnetic Field

With $\vec{v}\perp \vec{B}$, the magnetic force supplies the centripetal force:

$\frac{m{v}^2}{r}=qvB\Rightarrow r=\frac{mv}{qB}=\frac{p}{qB},T=\frac{2\pi m}{qB}$

The period is independent of speed and radius. A faster particle sweeps a proportionally larger circle in exactly the same time — the operating principle of the cyclotron, where a fixed-frequency voltage keeps accelerating particles through ever-larger orbits.

Trap. Watch whether a comparison is at equal speed or equal energy. At equal speed, $r=mv/qB\propto m/q$. At equal kinetic energy, $r=\sqrt{2mK}/qB\propto \sqrt{m}/q$ — a different ratio entirely.

If $\vec{v}$ has a component along $\vec{B}$, the path is a helix. The parallel component feels no force and coasts; the perpendicular component circles. The two are completely independent.

$\text{pitch}=v_{\parallel }T=v\cos \theta \cdot \frac{2\pi m}{qB}$

Illustration 5

Ions pass undeflected through crossed fields $E=2\times 10^4$ V m⁻¹ and $B_1=0.1$ T, then enter a second region of field $B_2=0.2$ T perpendicular to their motion. Singly charged ions strike a detector 0.5 m from the entry slit. Find the ion's mass.

The selector fixes the speed, and does so independently of mass:

$v=\frac{E}{B_1}=\frac{2\times 10^4}{0.1}=2\times 10^5{\text{m s}}^{-1}$

A semicircle in the second region lands the ion at $2r$ from the slit, so $r=0.25$ m:

$m=\frac{qr{B}_2}{v}=\frac{(1.6\times 10^{-19})(0.25)(0.2)}{2\times 10^5}\Rightarrow m=4.0\times 10^{-26}\text{kg}$

About 24 atomic mass units, which is magnesium. Two stages, each doing exactly one job: the crossed fields strip out the speed spread, and then the pure magnetic field sorts what is left by mass alone.

Illustration 6

A particle of charge $q$, mass $m$, speed $v$ enters a field region of width $d$ perpendicular to $\vec{B}$. Find the angle through which it is deflected on leaving.

Inside the region it turns along a circular arc of radius $r=mv/qB$. Geometry of the chord gives

$\sin \alpha =\frac{d}{r}=\frac{qBd}{mv}$

The particle escapes only while $d<r$. Once $d\ge r$ it turns right round and comes back out of the face it entered — which is precisely how a magnetic mirror confines plasma.

7. Force on a Conductor and Between Parallel Wires

$\vec{F}=I\vec{L}\times \vec{B},F=BIL\sin \theta$

A closed loop in a uniform field feels zero net force, because $\sum I\vec{L}$ around a closed path is zero. It does feel a torque — Section 8.

$\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$

Parallel currents attract; antiparallel currents repel. This force defines the ampere: one ampere in two infinite parallel wires one metre apart gives $2\times 10^{-7}$ N per metre.

Illustration 7

A wire of arbitrary crooked shape carries current $I$ between two points $A$ and $B$ in a uniform field $\vec{B}$. Find the net force.

Add the elements: $\vec{F}=I\left(\sum d\vec{l}\right)\times \vec{B}$, and $\sum d\vec{l}$ from $A$ to $B$ is just the straight vector $\overrightarrow{AB}$, whatever route the wire took.

$\vec{F}=I\overrightarrow{AB}\times \vec{B}$

So a semicircular wire of radius $R$ feels the same force as a straight wire of length $2R$ joining its ends. Only the endpoints matter — and a closed loop, whose endpoints coincide, feels nothing.

Illustration 8

A long straight wire carries $I_1$. A rectangular loop of sides $a$ parallel to the wire and $b$ perpendicular to it carries $I_2$, its near side at distance $d$. Find the net force on the loop.

The field of the wire is non-uniform, $B={\mu }_0{I}_1/2\pi r$, so the zero-net-force rule for uniform fields does not apply here.

The two sides perpendicular to the wire feel equal and opposite forces that cancel by symmetry. The two parallel sides do not, because they sit in different fields:

$F=\frac{{\mu }_0{I}_1{I}_{2}a}{2\pi }\left(\frac{1}{d}-\frac{1}{d+b}\right)=\frac{{\mu }_0{I}_1{I}_{2}ab}{2\pi d(d+b)}$

The near side dominates, so the loop is pulled toward the wire whenever its near-side current runs parallel to $I_1$. This is the magnetic counterpart of a charged rod attracting neutral paper: a dipole in a non-uniform field always feels a net force.

8. Current Loop as a Magnetic Dipole

$\vec{m}=NI\vec{A},\vec{\tau }=\vec{m}\times \vec{B},U=-\vec{m}\cdot \vec{B}$

Exactly the electric-dipole expressions with $\vec{p}\to \vec{m}$. Nothing new to learn; the whole framework carries over.

Stable when $\vec{m}$ is aligned with $\vec{B}$, unstable when antiparallel. Torque is maximum when the loop's plane contains the field — where the energy happens to be zero.

Illustration 9

A wire of fixed length $L$ carrying current $I$ is wound into $n$ circular turns. How does the magnetic moment depend on $n$?

$2\pi rn=L\Rightarrow r=\frac{L}{2\pi n},m=nI(\pi r^2)=nI\pi \frac{L^2}{4{\pi }^2{n}^2}$

$m=\frac{I{L}^2}{4\pi n}$

The moment falls as $1/n$. Winding more turns from the same wire loses more in area than it gains in turns, so a single big loop has the largest moment — the opposite of what most people guess.

9. Moving Coil Galvanometer

A coil in a radial field, restrained by a spring. The magnetic torque $NIAB$ balances the restoring torque $k\varphi$, so deflection is proportional to current.

$\text{current sensitivity}=\frac{NAB}{k},\text{voltage sensitivity}=\frac{NAB}{kR}$

Trap. More turns does not reliably improve voltage sensitivity. Raising $N$ also lengthens the wire, raising $R$ roughly in proportion — and the two effects cancel. The two sensitivities cannot be improved independently by winding.

Both conversions push the instrument toward its ideal form. Ask what the ideal looks like and the connection follows without memorising.

Illustration 10

A galvanometer of resistance 60 Ω gives full-scale deflection at 1 mA. Convert it into an ammeter reading to 5 A, and separately into a voltmeter reading to 10 V.

For the ammeter the excess current must bypass the coil, so the shunt goes in parallel:

$S=\frac{I_{g}G}{I-I_g}=\frac{(10^{-3})(60)}{5-10^{-3}}=0.012\Omega$

For the voltmeter the coil must be starved of current, so a large resistance goes in series:

$R=\frac{V}{I_g}-G=\frac{10}{10^{-3}}-60=9940\Omega$

Twelve milliohms against ten kilohms — a factor of a million between two conversions of the same instrument. That contrast is the sanity check. An ammeter must be near-zero resistance so it does not throttle the branch it sits in, and a voltmeter near-infinite so it draws nothing from the element it straddles.

10. Bar Magnet as an Equivalent Solenoid

A bar magnet and a current-carrying solenoid produce identical external field patterns. That is the clearest evidence that magnetism originates in circulating current — inside a magnet those currents are electron orbital and spin motions, not currents in wires, but the field is of the same kind.

Axial is twice equatorial, both falling as $1/r^3$ — identical to the electric dipole, for identical geometric reasons.

Illustration 11

A short bar magnet of moment 0.4 A m² lies with its axis along the north-south line. Find the field it produces 20 cm from its centre, first end-on and then broadside.

$B_{axial}=\frac{{\mu }_0}{4\pi }\frac{2m}{r^3}=(10^{-7})\frac{2(0.4)}{(0.2{)}^3}=10^{-5}\text{T}$

$B_{eq}=\frac{{\mu }_0}{4\pi }\frac{m}{r^3}=5\times 10^{-6}\text{T}$

Both are within a factor of a few of the Earth's horizontal field, roughly $3\times 10^{-5}$ T, which is exactly why a magnet this size visibly disturbs a compass placed near it. Two things are worth carrying away: the factor of two between the positions, and the steepness of $1/r^3$ — halving the distance multiplies both fields by eight.

11. Magnetic Materials

Diamagnetism is universal but usually swamped. Every material shows a weak induced opposition to an applied field; it is only visible when nothing stronger hides it.

Paramagnetism weakens with heating — thermal agitation randomises the atomic moments. That is Curie's law, $\chi \propto 1/T$.

Ferromagnetism depends on domains, regions where enormous numbers of moments align spontaneously. An applied field grows the favourable domains at the expense of the rest. Above the Curie temperature — 1043 K for iron — thermal energy destroys the alignment and the material becomes merely paramagnetic.

Illustration 12

A rod of magnetic susceptibility $\chi =599$ sits in a magnetising field of 1200 A m⁻¹. Find its relative permeability, its magnetisation, and the field $B$ inside it.

${\mu }_r=1+\chi =600,M=\chi H=(599)(1200)=7.19\times 10^5{\text{A m}}^{-1}$

$B={\mu }_0(H+M)={\mu }_0{\mu }_{r}H=(4\pi \times 10^{-7})(600)(1200)\Rightarrow B\approx 0.90\text{T}$

A susceptibility in the hundreds places this firmly among the ferromagnets. The two expressions for $B$ are one statement written twice, since $H+M=H(1+\chi )={\mu }_{r}H$, so whichever form the given data suits will do.

Hysteresis is the lagging of magnetisation behind the applied field, and the width of the loop decides the application. Soft iron loses little per cycle, so it suits electromagnets and transformer cores where the field reverses fifty times a second. Steel retains its magnetisation, so it suits permanent magnets.

Summary

• No magnetic monopoles: closed-surface flux is exactly zero, and field lines are closed loops.
• Every magnetic field comes from moving charge; a bar magnet is a solenoid whose currents are atomic.
• Biot-Savart is ${\mu }_{0}Id\vec{l}\times \hat{r}/4\pi r^2$ — perpendicular to the element, so radial segments contribute nothing at the centre.
• Arc at the centre: ${\mu }_{0}I\theta /4\pi R$. Most configurations reduce to arcs alone.
• Ampere's law is to Biot-Savart what Gauss is to Coulomb; the solenoid's ${\mu }_{0}nI$ falls out in one line.
• Inside a thick wire $B\propto r$; outside $B\propto 1/r$, peaking at the surface.
• The magnetic force never does work — direction changes, speed never does.
• $r=mv/qB$, $T=2\pi m/qB$: the period is independent of speed and radius.
• Equal speed gives $r\propto m/q$; equal energy gives $r\propto \sqrt{m}/q$. Different questions, different ratios.
• Helix: resolve first. The parallel component coasts, the perpendicular circles, pitch $=v_{\parallel }T$.
• Velocity selector passes $v=E/B$, whatever the charge and mass.
• A crooked wire in a uniform field feels $I\overrightarrow{AB}\times \vec{B}$ — only its endpoints matter, so a closed loop feels nothing.
• Parallel currents attract, unlike like charges — and this force defines the ampere.
• $\vec{m}=NI\vec{A}$, $\vec{\tau }=\vec{m}\times \vec{B}$, $U=-\vec{m}\cdot \vec{B}$: the electric dipole results with $\vec{p}\to \vec{m}$.
• Axial field of a magnetic dipole is twice equatorial, both as $1/r^3$.
• More galvanometer turns raises current sensitivity but not necessarily voltage sensitivity, since $R$ rises too.
• Shunt in parallel makes an ammeter; large series resistance makes a voltmeter.
• Diamagnetic: weakly repelled, temperature-independent, universal. Paramagnetic: $\chi \propto 1/T$. Ferromagnetic: domains, paramagnetic above the Curie point.
• Loop area is the energy lost per cycle: soft iron for transformers, steel for permanent magnets.