Physics by Lamhi: Not Your Boring Physics

CLASS 12 · CHAPTER 4 · MAGNETISM

DETAILED NOTES

Moving Charges and Magnetism

The complete chapter, section by section: the force a magnetic field exerts on a moving charge and on a current-carrying wire, the circular motion that force produces, how currents themselves create magnetic fields (the Biot–Savart law, a current loop's axial field, and Ampere's circuital law), the solenoid, the force between parallel wires that defines the ampere, torque on a current loop, and how a moving-coil galvanometer is turned into a working ammeter or voltmeter.

4.1 Force on a Moving Charge

A charge qq moving with velocity v⃗\vec{v} through a magnetic field B⃗\vec{B} experiences the magnetic (Lorentz) force:

Magnetic force
F⃗=qv⃗×B⃗\vec{F} = q\vec{v}\times\vec{B}

Because this is a cross product, the force is always perpendicular to both v⃗\vec{v} and B⃗\vec{B}. Its magnitude is F=qvBsin⁡θF = qvB\sin\theta, where θ\theta is the angle between v⃗\vec{v} and B⃗\vec{B}: maximum when velocity is perpendicular to the field (θ=90∘\theta=90^\circ), and zero when the charge moves along the field (θ=0∘\theta=0^\circ), in which case it feels no magnetic force at all. The direction follows the right-hand rule: point the fingers along v⃗\vec{v}, curl them toward B⃗\vec{B}, and the thumb gives the direction of v⃗×B⃗\vec{v}\times\vec{B} (for a positive charge; reverse it for a negative one).

A force perpendicular to velocity does zero work on the charge, no matter how strong the field: magnetic forces can redirect a charged particle, but they can never speed it up or slow it down.

Worked example

Force on an electron entering at an angle

An electron moves at v = 2×10⁶ m/s through a field B = 0.5 T, with its velocity making a 30° angle to the field. Find the magnitude of the force on it.

F=qvBsin⁡θ=(1.6×10−19)(2×106)(0.5)sin⁡30∘=8×10−14 NF = qvB\sin\theta = (1.6\times10^{-19})(2\times10^{6})(0.5)\sin30^\circ = \textcolor{#e08a1e}{8\times10^{-14}\text{ N}}

4.2 Circular Motion & the Cyclotron

When a charge enters a uniform field with its velocity exactly perpendicular to B⃗\vec{B}, the magnetic force is always perpendicular to the velocity, which is exactly the condition for uniform circular motion: the force supplies the centripetal force, with a fixed magnitude and continuously changing direction.

Centripetal condition
qvB=mv2rqvB = \dfrac{mv^2}{r}

4.2.1 Radius

Orbit radius
r=mvqBr = \dfrac{mv}{qB}

directly proportional to speed: a faster particle traces a bigger circle.

4.2.2 Period and frequency

Time for one revolution is circumference over speed, T=2πr/vT=2\pi r/v. Substituting r=mv/(qB)r=mv/(qB) cancels vv out completely:

Period (independent of v)
T=2πmqBT = \dfrac{2\pi m}{qB}
Cyclotron frequency
f=1T=qB2πmf = \dfrac{1}{T} = \dfrac{qB}{2\pi m}

Both depend only on the charge-to-mass ratio and the field strength, never on the particle’s speed or the radius that speed produces. A proton moving twice as fast traces a circle twice as wide, but completes each lap in exactly the same time. See the interactive orbit for the live version.

This is the operating principle of the cyclotron, a particle accelerator that repeatedly kicks a charged particle to higher speeds using an alternating voltage held at one fixed frequency. Because the particle’s own revolution frequency never changes as it speeds up, that fixed external frequency stays in step with it, lap after lap, letting a single alternating voltage keep accelerating the particle indefinitely (this synchrony eventually fails at speeds approaching light speed, outside CBSE scope).

Worked example

Electron cyclotron motion

An electron (q = 1.6×10⁻¹⁹ C, m = 9.11×10⁻³¹ kg) moves perpendicular to a field B = 2×10⁻³ T at v = 4×10⁶ m/s. Find its orbit radius and period.

r=mvqB=(9.11×10−31)(4×106)(1.6×10−19)(2×10−3)≈11.39 mmr = \dfrac{mv}{qB} = \dfrac{(9.11\times10^{-31})(4\times10^{6})}{(1.6\times10^{-19})(2\times10^{-3})} \approx \textcolor{#e08a1e}{11.39\text{ mm}}T=2πmqB=2π(9.11×10−31)(1.6×10−19)(2×10−3)≈17.89 nsT = \dfrac{2\pi m}{qB} = \dfrac{2\pi(9.11\times10^{-31})}{(1.6\times10^{-19})(2\times10^{-3})} \approx \textcolor{#e08a1e}{17.89\text{ ns}}

Note that this T is different from the 35.77 ns in the interactive sim above, because the field here (2 mT) is different too. T depends on B, just never on v.

4.3 Force on a Current-Carrying Conductor

A current is just a stream of moving charges, so a current-carrying wire in a magnetic field feels a force too, the sum of the Lorentz forces on every charge drifting through it. For a straight wire of length LL carrying current II in a uniform field BB:

Force on a current-carrying wire
F=BILsin⁡θF = BIL\sin\theta

where θ\theta is the angle between the wire (current direction) and the field. The force is maximum when the wire is perpendicular to BB, and zero when the current flows along the field.

Worked example

Force on a straight current-carrying wire

A straight wire of length 0.5 m carries a current of 4 A, perpendicular to a uniform field of 0.25 T. Find the force on it.

F=BILsin⁡90∘=(0.25)(4)(0.5)=0.5 NF = BIL\sin90^\circ = (0.25)(4)(0.5) = \textcolor{#e08a1e}{0.5\text{ N}}

4.4 The Biot–Savart Law

Section 4.3 took a magnetic field as given and asked what force it exerts. This section asks the reverse question: where does the field itself come from? Every magnetic field traced back far enough comes from a moving charge or a current, and the Biot–Savart law tells us exactly how much field a small piece of current produces.

Take a tiny element of wire, length dl⃗d\vec{l}, carrying current II. At a point P, a displacement r⃗\vec{r} away from the element (with r^\hat{r} the unit vector along it), this element contributes a field:

Biot–Savart law
dB⃗=μ04πI dl⃗×r^r2d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\hat{r}}{r^2}

where μ0=4π×10−7 T m/A\mu_0 = 4\pi\times10^{-7}\text{ T m/A} is the permeability of free space, the magnetic counterpart of ε0\varepsilon_0 in Coulomb’s law. In magnitude, writing θ\theta for the angle between dl⃗d\vec{l} and r⃗\vec{r}:

dB=μ04πI dlsin⁡θr2dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}

The direction of dB⃗d\vec{B} is perpendicular to the plane containing dl⃗d\vec{l} and r⃗\vec{r} (the right-hand screw rule: curl the fingers from dl⃗d\vec{l} toward r⃗\vec{r}, and the thumb gives the direction of dl⃗×r^d\vec{l}\times\hat{r}). To get the actual field at P due to a real, finite wire, every element’s contribution has to be summed (in practice, integrated) over the whole length of the wire, using the same principle of superposition that governs the electric field of a continuous charge distribution.

4.4.1 Comparing it with Coulomb's law

The Biot–Savart law and Coulomb’s law share a family resemblance, both obey an inverse-square dependence on distance, and both let fields from separate sources simply add, but they differ in two genuinely important ways. First, the source of dB⃗d\vec{B} is a vector, I dl⃗I\,d\vec{l}, while the source of the electric field is a scalar, the charge qq. Second, the electric field points along the line joining source and field point, while the magnetic field points perpendicular to the plane containing that line and the current element, and it carries an angle dependence (sin⁡θ\sin\theta) that Coulomb’s law simply doesn’t have: a point lying exactly along the direction of the current element itself (θ=0\theta=0) sees zero contribution from it, no matter how close it is.

Worked example

Field from a single current element

A short element of wire, length 2 cm, carries a current of 15 A. Find the magnitude of the field it produces at a point 40 cm away, measured along a direction perpendicular to the element.

dB=μ04πI dlsin⁡θr2=(10−7)(15)(0.02)sin⁡90∘(0.4)2=1.875×10−7 TdB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2} = (10^{-7})\dfrac{(15)(0.02)\sin90^\circ}{(0.4)^2} = \textcolor{#e08a1e}{1.875\times10^{-7}\text{ T}}

Tiny, as a single element’s contribution always is: a measurable field only builds up once enough of these elements, all along a real wire, are added together.

4.5 Magnetic Field on the Axis of a Circular Current Loop

As a first real application of the Biot–Savart law, consider a circular loop of radius RR, carrying current II, and ask for the field at a point P on its axis, a distance xx from the centre.

Every element dl⃗d\vec{l} of the loop sits a distance r=x2+R2r=\sqrt{x^2+R^2} from P, and is perpendicular to that displacement, so each contributes dB=μ04πI dlx2+R2dB=\frac{\mu_0}{4\pi}\frac{I\,dl}{x^2+R^2}. That field has a component along the axis and a component perpendicular to it; but for every element, there is a diametrically opposite element whose perpendicular component points exactly the other way, so all the perpendicular components cancel in pairs, leaving only the axial component to survive the sum. That axial component is dBx=dBcos⁡ϕdB_x = dB\cos\phi, where cos⁡ϕ=R/x2+R2\cos\phi = R/\sqrt{x^2+R^2} (the angle between dB⃗d\vec{B} and the axis), so:

dBx=μ04πI dlx2+R2⋅Rx2+R2=μ0IR4π(x2+R2)3/2 dldB_x = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{x^2+R^2}\cdot\dfrac{R}{\sqrt{x^2+R^2}} = \dfrac{\mu_0 I R}{4\pi (x^2+R^2)^{3/2}}\,dl

Summing dldl all the way around the loop just gives its circumference, 2πR2\pi R:

Field on the axis of a circular loop
B=μ0IR22(x2+R2)3/2\textcolor{#e08a1e}{B} = \dfrac{\mu_0 I R^2}{2(x^2+R^2)^{3/2}}

Setting x=0x=0 gives the field at the centre of the loop itself, a clean special case worth remembering on its own:

Field at the centre of a loop
B0=μ0I2RB_0 = \dfrac{\mu_0 I}{2R}

Worked example

Field at the centre of a loop, and further out along its axis

A loop of radius 5 cm carries a current of 2 A. Find the field at its centre, and at a point on the axis one radius away (x = R = 5 cm) from the centre.

B0=μ0I2R=(4π×10−7)(2)2(0.05)≈2.51×10−5 TB_0 = \dfrac{\mu_0 I}{2R} = \dfrac{(4\pi\times10^{-7})(2)}{2(0.05)} \approx \textcolor{#e08a1e}{2.51\times10^{-5}\text{ T}}B(x=R)=μ0IR22(2R2)3/2=B022≈8.89×10−6 TB(x{=}R) = \dfrac{\mu_0 I R^2}{2(2R^2)^{3/2}} = \dfrac{B_0}{2\sqrt{2}} \approx \textcolor{#e08a1e}{8.89\times10^{-6}\text{ T}}

Moving out along the axis by just one radius already cuts the field to roughly 35% of its centre value, a reminder of how quickly a loop’s field falls away once you leave its centre.

4.6 Ampere’s Circuital Law

The Biot–Savart law is always true, but integrating it by brute force can be hard work. For situations with enough symmetry, Ampere’s circuital law gives the same physics in a far more convenient package, exactly the way Gauss’s law is Coulomb’s law repackaged for symmetric charge distributions.

Ampere's circuital law
∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enclosed}}

The integral runs around any closed loop (an “Amperian loop”), and IenclosedI_{\text{enclosed}} is the total current passing through the surface that loop bounds. In the common case where a loop can be chosen on which BB is tangential and has the same magnitude all the way around (or is zero, or purely perpendicular, over the rest of the loop), this collapses to simple algebra:

BL=μ0IeBL = \mu_0 I_e

with LL the length of the loop over which BB is tangential and IeI_e the current it encloses.

4.6.1 Field of an infinite straight wire

A long, straight wire carrying current II has an obvious symmetry: by the right-hand rule, its field forms circles centred on the wire, with the same magnitude at every point on a given circle. Choosing that circle, radius rr, as the Amperian loop, BB is tangential to it everywhere and constant in magnitude, so L=2πrL=2\pi r and Ie=II_e=I:

B(2πr)=μ0IB(2\pi r) = \mu_0 I
Field of a long straight wire
B=μ0I2πr\textcolor{#e08a1e}{B} = \dfrac{\mu_0 I}{2\pi r}

This is the same inverse-distance result (not inverse-square, notice) used without proof back in §4.8’s force between parallel wires, now derived properly from first principles rather than assumed.

Worked example

Field inside and outside a current-carrying wire of finite thickness

A long straight wire of radius a = 3 mm carries a current of 6 A, spread uniformly over its circular cross-section. Find the field at r = 2 mm (inside the wire) and at r = 9 mm (outside it).

Inside, the Amperian circle of radius r<ar<a only encloses the fraction r2/a2r^2/a^2 of the total current (since the current is spread uniformly over the area), so Ie=I(r2/a2)I_e = I(r^2/a^2):

B(2πr)=μ0Ir2a2  ⇒  B=μ0Ir2πa2=(4π×10−7)(6)(0.002)2π(0.003)2≈2.67×10−4 TB(2\pi r) = \mu_0 I\dfrac{r^2}{a^2} \;\Rightarrow\; B = \dfrac{\mu_0 I r}{2\pi a^2} = \dfrac{(4\pi\times10^{-7})(6)(0.002)}{2\pi(0.003)^2} \approx \textcolor{#e08a1e}{2.67\times10^{-4}\text{ T}}

Outside, the full current is enclosed, so the ordinary straight-wire formula applies directly:

B=μ0I2πr=(4π×10−7)(6)2π(0.009)≈1.33×10−4 TB = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi\times10^{-7})(6)}{2\pi(0.009)} \approx \textcolor{#e08a1e}{1.33\times10^{-4}\text{ T}}

Inside the wire the field climbs linearly with r (more current enclosed the further out you go); outside, it falls off as 1/r, the field peaking exactly at the wire’s surface, r = a.

4.7 Field of a Solenoid

A solenoid is a tightly wound helical coil. Deep inside a long solenoid, far from its ends, the magnetic field is uniform and directed along its axis, essentially zero just outside.

4.7.1 Deriving the field, using Ampere's law

Take a rectangular Amperian loop with one side, length hh, running along the axis inside the solenoid, the opposite side running outside it, and the two remaining sides crossing the windings. The outside segment contributes nothing (field there is taken as zero), and the two crossing segments contribute nothing either, since BB there runs along the axis, perpendicular to them. Only the inside segment survives, so L=hL=h. If the solenoid has nn turns per unit length, that segment encloses nhnh turns, each carrying current II, so Ie=I(nh)I_e = I(nh):

Bh=μ0I(nh)Bh = \mu_0 I(nh)
Field inside a long solenoid
B=μ0nI\textcolor{#e08a1e}{B} = \mu_0 nI

Unlike the field near a single wire, this field doesn’t fall off inside the solenoid, it stays essentially constant across the entire interior cross-section, independent of position, right up to the ends.

4.8 Force Between Parallel Wires & the Ampere

Each current-carrying wire creates its own magnetic field (now derived properly in §4.6.1), which exerts a force on any other current sitting in it. Two long, straight, parallel wires carrying currents I1I_1 and I2I_2, separated by a distance dd, exert a force on each other per unit length of:

Force per unit length between two parallel wires
FL=μ0I1I22πd\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}

Currents in the same direction attract; currents in opposite directions repel, the reverse of what happens between like and unlike electric charges.

This result is historically how the ampere itself was defined: one ampere is the constant current which, flowing in each of two infinitely long parallel wires held 1 m apart in vacuum, produces a force of exactly 2×10−7 N2\times10^{-7}\text{ N} on each metre of either wire.

Worked example

Force between two parallel wires

Two long parallel wires, 0.1 m apart, carry currents of 5 A and 10 A in the same direction. Find the force per unit length between them, and state whether it’s attractive or repulsive.

FL=μ0I1I22πd=(4π×10−7)(5)(10)2π(0.1)=1×10−4 N/m, attractive\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d} = \dfrac{(4\pi\times10^{-7})(5)(10)}{2\pi(0.1)} = \textcolor{#e08a1e}{1\times10^{-4}\text{ N/m, attractive}}

4.9 Torque on a Current Loop & Magnetic Dipole

A current loop of NN turns, each of area AA, carrying current II, placed in a field BB making angle θ\theta with the loop’s own magnetic moment, experiences a torque that tries to rotate it into alignment with the field:

Torque on a current loop
τ=NIABsin⁡θ\tau = NIAB\sin\theta

This has exactly the same mathematical form as the torque on any magnetic dipole, τ⃗=m⃗×B⃗\vec{\tau} = \vec{m}\times\vec{B}, since a current loop simply is a magnetic dipole, with dipole moment m=NIAm = NIA.

The torque vanishes only when m⃗\vec{m} is parallel or antiparallel to B⃗\vec{B}: these are the loop’s two equilibrium orientations. Parallel (θ=0\theta=0) is stable, any small rotation away from it is pushed straight back; antiparallel (θ=180∘\theta=180^\circ) is unstable, the restoring torque there actually grows the rotation instead of opposing it. This is the same stable/unstable pair, for exactly the same reason, as a bar magnet’s two alignments with an external field.

4.10 The Moving Coil Galvanometer

A moving-coil galvanometer puts §4.9’s torque to work as a measuring instrument. A coil of NN turns and area AA sits in a radial magnetic field, radial by design so that BB is always perpendicular to the coil’s plane, whatever angle the coil has rotated to, keeping sin⁡θ=1\sin\theta=1 throughout its swing. The magnetic torque NIABNIAB is balanced against a spring’s restoring torque, kϕk\phi, where kk is the spring’s torsional constant and ϕ\phi the steady deflection angle:

kϕ=NIABk\phi = NIAB
Galvanometer deflection
ϕ=(NABk)I\phi = \left(\dfrac{NAB}{k}\right)I

the deflection ends up directly proportional to the current, which is exactly what lets the instrument measure current in the first place. The quantity in brackets is called the current sensitivity, ϕ/I\phi/I, a fixed property of a given galvanometer.

4.10.1 Converting it into an ammeter

A bare galvanometer makes a poor ammeter for two reasons: it is far too sensitive (full-scale deflection at a current of a few microamps or milliamps), and its own coil resistance RGR_G is large enough to noticeably disturb the very current it’s trying to measure. The fix is a small shunt resistance rsr_s wired in parallel with the coil, so most of the current bypasses the galvanometer entirely. If the galvanometer deflects full-scale at current IgI_g, and the ammeter should read full-scale at some larger current II, then at that reading the galvanometer branch carries exactly IgI_g and the shunt carries the rest, I−IgI-I_g. Since the two branches are in parallel, they share the same voltage:

IgRG=(I−Ig) rsI_g R_G = (I-I_g)\,r_s
Shunt resistance
rs=IgRGI−Igr_s = \dfrac{I_g R_G}{I - I_g}

4.10.2 Converting it into a voltmeter

A voltmeter, by contrast, is connected in parallel with whatever it’s measuring, and so must draw as little current as possible to avoid disturbing the circuit. A large resistance RR wired in series with the galvanometer does this: to read full-scale at a chosen voltage VV, the combined series resistance must limit the current to exactly IgI_g:

V=Ig(RG+R)V = I_g(R_G+R)
Series resistance
R=VIg−RGR = \dfrac{V}{I_g} - R_G

The voltage sensitivity, ϕ/V\phi/V, then comes out to 1R(NABk)\frac{1}{R}\left(\frac{NAB}{k}\right). Note that doubling N doubles the current sensitivity, but since a coil with twice the turns also has roughly twice the resistance, the voltage sensitivity is left unchanged, which is why converting a galvanometer into a good ammeter and converting the same galvanometer into a good voltmeter are two genuinely different modifications, not the same one in reverse.

Worked example

Designing a shunt and a series resistor for the same galvanometer

A galvanometer has resistance RG = 50 Ω and deflects full-scale at Ig = 1 mA. Find (a) the shunt resistance needed to turn it into an ammeter reading 0–3 A, and (b) the series resistance needed to turn it into a voltmeter reading 0–15 V.

(a)rs=IgRGI−Ig=(0.001)(50)3−0.001≈0.0167 Ω(a)\quad r_s = \dfrac{I_g R_G}{I-I_g} = \dfrac{(0.001)(50)}{3-0.001} \approx \textcolor{#e08a1e}{0.0167\ \Omega}(b)R=VIg−RG=150.001−50=14,950 Ω(b)\quad R = \dfrac{V}{I_g}-R_G = \dfrac{15}{0.001}-50 = \textcolor{#e08a1e}{14{,}950\ \Omega}

A shunt a few hundredths of an ohm, against a series resistor of nearly 15 kΩ: the two conversions pull the same instrument in opposite directions, tiny added resistance for the ammeter, huge added resistance for the voltmeter.

— From the NCERT exercises

Two of the chapter’s own exercise questions, with original worked solutions, both exercising the field-generation formulas derived in §4.5 and §4.6.1 above.

NCERT Exercise 4.1

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Solution

Direct application of the centre-of-loop result from §4.5, generalised to NN turns by simply multiplying by NN (each turn contributes the same field, and they all add):

B=μ0NI2R=(4π×10−7)(100)(0.40)2(0.08)≈3.14×10−4 TB = \dfrac{\mu_0 NI}{2R} = \dfrac{(4\pi\times10^{-7})(100)(0.40)}{2(0.08)} \approx \textcolor{#e08a1e}{3.14\times10^{-4}\text{ T}}

NCERT Exercise 4.2

A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Solution

Direct application of the straight-wire result derived in §4.6.1:

B=μ0I2πr=(4π×10−7)(35)2π(0.20)=3.5×10−5 TB = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi\times10^{-7})(35)}{2\pi(0.20)} = \textcolor{#e08a1e}{3.5\times10^{-5}\text{ T}}
Want the interactive version instead? Try the cyclotron simulation → or take a practice paper →