Moving Charges and Magnetism: theory

CONCEPT

Biot-Savart Law & Field at the Centre of a Coil

In one line
Every small piece of a current-carrying wire makes a magnetic field around it. Biot-Savart law gives that field.

OERSTED

21.1.1 Current makes a magnetic field

A compass needle near a current-carrying wire deflects (Oersted).
A moving charge produces both an electric and a magnetic field.
A charge at rest produces only an electric field.
Reversing the current reverses the magnetic field.

BIOT-SAVART LAW

21.1.2 Field of a current element

Element dl, point P at position r

dB = μ₀4π · i dl sin θr²

dB = μ₀4π · i (dl × r)r³

μ₀ = 4π × 10⁻⁷ T m A⁻¹, so μ₀4π = 10⁻⁷ in SI.
i dl is the current element, a vector along the current.
dB is perpendicular to both dl and r.
θ = 0° or 180° (point on the line of the element): dB = 0.
θ = 90°: dB is maximum.
dB ∝ i, dB ∝ dl, dB ∝ sin θ and dB follows the inverse-square law.

Quantity	Value
SI unit of B	tesla (T) = Wb m⁻²
CGS unit	1 T = 10⁴ gauss
Nature	B is a vector

DIRECTION RULES

21.1.3 Which way does B point?

Right-hand thumb rule for a wire and for a coil

Right-hand thumb rule: thumb along the current, curled fingers give B.
Maxwell’s cork screw rule: the screw advances along the current, its rotation gives B.
Coil: anticlockwise current gives B out of the page, clockwise gives B into it.
Symbols: ⊙ out of the page, ⊗ into the page.

CENTRE OF A COIL

21.1.4 Circular coil of n turns

B = μ₀ni2R

Every element is at the same distance R.
Every element is at θ = 90° to r.
All the elements add up in one direction.
B ∝ n, B ∝ i and B ∝ 1 over R.

Field at the centre

REWINDING

21.1.5 Same wire, new number of turns

Same length of wire: n times the turns, 1 over n of the radius

Length is fixed: L = n × 2πr, so r = Rn.
New field at the centre:

B_n = n²B₁

From n₁ turns to n₂ turns: B₂B₁ = (n₂n₁)².
Same wire, same p.d.: resistance ∝ r, so i ∝ Vr and B ∝ Vr².

ROTATING CHARGE

21.1.6 A moving charge is a current

i = qf = qω2π

B = μ₀qf2R = μ₀qω4πR

A charged ring turning about its axis behaves as a coil of current i.
A point charge on a circular path behaves in the same way.

ORBITING ELECTRON

21.1.7 Field at the nucleus

i = ev2πr

B = μ₀4π · evr²

Bohr model: v ∝ 1n and r ∝ n².
So B ∝ n⁻⁵.

Electron in a circular orbit

General moving charge: B = μ₀4π · qv sin θr².

Practise PYQs on this topicBiot-Savart Law & Circular Coils · 115 questions with solutions

CONCEPT

Field on the Axis of a Coil

In one line
On the axis, the field of a coil points along the axis and falls off with distance from the centre.

AXIAL FIELD

21.2.1 Point at distance x from the centre

Coil of radius R, point P on its axis

B_a = μ₀niR²2(R² + x²)^3/2

Direction is along the axis: use the right-hand thumb rule on the coil.
At x = 0 it becomes the field at the centre.
The components perpendicular to the axis cancel in pairs.

RATIO TO THE CENTRE

21.2.2 The shortcut for most questions

B_cB_a = (1 + x²R²)^3/2

x	B_cB_a
R	2√2
√3 R	8
2√2 R	27

FAR AWAY

21.2.3 x much larger than R

B_a = μ₀niR²2x³ = μ₀4π · 2Mx³

M = niA is the magnetic moment of the coil (Concept 10).
Far from the coil, B ∝ 1x³.
The coil behaves as a magnetic dipole.

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B against x: inflection points

B is largest at the centre and falls on both sides

The curve changes its bending at x = ±R2.
Between these points the field changes almost linearly with x.
The two points are a distance R apart.

Helmholtz coils

Two equal coils, same current, separation R

Two identical coaxial coils carry equal currents in the same sense.
They are kept a distance R apart.
Field in the middle region is nearly uniform.

B = 8μ₀Ni5√5 R = 0.716 μ₀NiR

Practise PYQs on this topicBiot-Savart Law & Circular Coils · 115 questions with solutions

CONCEPT

Arcs & Concentric Coils

In one line
An arc gives the same fraction of the full-coil field as the fraction of the circle it covers. Fields of several coils add as vectors.

CIRCULAR ARC

21.3.1 Angle θ at the centre

B = μ₀4π · iθR

θ is in radian.
B ∝ θ.
The straight lead wires pointing to the centre give no field.

Arc of angle θ

Arc	B at the centre
θ	μ₀4π · iθR
(2π − θ)	μ₀4π · i(2π − θ)R
semicircle	μ₀i4R
quarter circle	μ₀i8R
34 of a circle	3μ₀i8R
full circle	μ₀i2R

CONCENTRIC COILS

21.3.2 Common centre

Two coils with a common centre

Arrangement	B at the centre
same plane, same sense	μ₀2(n₁i₁r₁ + n₂i₂r₂)
same plane, opposite sense	μ₀2(n₁i₁r₁ − n₂i₂r₂)
planes perpendicular	√(B₁² + B₂²)
planes at angle θ	√(B₁² + B₂² + 2B₁B₂ cos θ)

Same plane, opposite sense, B = 0 when i₁r₁ = i₂r₂.
Identical coils at right angles: B = √2 B₁.

DIVISION OF CURRENT

21.3.3 Current enters a uniform loop at two points

Current divides in the inverse ratio of the resistances of the two parts.
Each part gives a field ∝ i × length.
i × length is the same for both (same p.d.).
The fields oppose, so B = 0 at the centre.
True for any two points of a uniform circle, and for a square frame fed at a diagonal.

B = 0 at the centre

Practise PYQs on this topicBiot-Savart Law & Circular Coils · 115 questions with solutions

CONCEPT

Straight Wires

In one line
Around a long straight wire, B forms circles and falls as 1 over the distance.

FINITE WIRE

21.4.1 Angles at the point P

Wire, perpendicular distance r, angles φ₁ and φ₂

B = μ₀4π · ir (sin φ₁ + sin φ₂)

Both angles are measured from the perpendicular dropped on the wire.
If P is beyond the end of the wire, use (sin φ₁ − sin φ₂).

STANDARD CASES

21.4.2 Long wires

Position of the point relative to the wire

Wire	B at distance r
infinite	μ₀2π · ir
semi-infinite, foot at its end	μ₀4π · ir
point on the line of the wire	0

B does not depend on the wire’s thickness (outside it).
r halved ⇒ B doubles: B ∝ 1r.
Field lines are circles centred on the wire.

TWO PARALLEL WIRES

21.4.3 Vector addition

Top view: fields midway between two wires

Currents	B midway (separation d)
same direction	μ₀πd (i₁ − i₂)
opposite direction	μ₀πd (i₁ + i₂)

Equal currents, same direction: B = 0 midway.
Measuring both cases gives i₁i₂ from the ratio of the two fields.
Neutral point (same direction): i₁x = i₂d − x.
Point outside both wires, same direction: fields add.
P with its lines to the wires at 90°: B₁ ⊥ B₂, so B = √(B₁² + B₂²).

BENT WIRES

21.4.4 Break the wire into parts

Straight parts and arcs add with their own signs

Find B of each straight part and each arc separately.
Fix each direction: out of the page + , into the page −.
A part pointing at the point gives zero.

Figure	B at O
(a) semicircle, wires in line with O	μ₀i4r
(b) quarter arc + two semi-infinite	μ₀4π · ir(2 + π2)
(c) long wire with a loop	μ₀i2πr(π − 1)

WIRE AND LOOP

21.4.5 Field zero at the centre of a loop

μ₀I_c2R = μ₀I_w2πd ⇒ d = RI_wπI_c

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Regular polygon, side a, current I

B = n × μ₀4π · Id × 2 sin πn

d = a2 tan(πn) is the distance of each side from the centre.

Polygon	B at the centre
triangle	9μ₀I2πa
square	2√2 μ₀Iπa
hexagon	√3 μ₀Iπa

Point on the bisector of a finite wire

Wire of length L, point at distance r on its perpendicular bisector.

B = μ₀4π · 2Ir · L√(L² + 4r²)

Practise PYQs on this topicMagnetic Field of Straight Wires · 75 questions with solutions

CONCEPT

Ampere’s Law & Cylinders

In one line
Going once round any closed loop, B·dl adds up to μ₀ times the current enclosed.

AMPERE’S CIRCUITAL LAW

21.5.1 Closed loop in space

∮B·dl = μ₀ I_enc

Only enclosed current counts.
Currents outside do change B, but not the sum.
Opposite currents subtract.

I_enc = i₁ − i₂

Sign: curl the right-hand fingers along the loop; a current along the thumb is +.
It is easy to use when B is the same all along a symmetric loop.
Biot-Savart works for any wire; Ampere’s law needs symmetry.
Bundle of long wires: B = μ₀2π · I_netr.

SOLID CYLINDER

21.5.2 Uniform current, radius R

B rises up to the surface, then falls

Point	B
inside (r < R)	μ₀ir2πR²
surface (r = R)	μ₀i2πR
outside (r > R)	μ₀i2πr

On the axis B = 0.
B is largest at the surface.
B = half of the maximum at r = R2 and at r = 2R.

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Hollow cylinder, current on the surface

B = 0 inside, then a jump at r = R

Inside: no current is enclosed, so B = 0.

outside: B = μ₀i2πr

Thick hollow cylinder (radii a and b)

a < r < b: B = μ₀i2πr · r² − a²b² − a²

r < a: B = 0.
r > b: B = μ₀i2πr.

Coaxial cable

Equal and opposite currents in the two conductors.
Outside the cable: enclosed current is zero, so B = 0.
Between the conductors: B = μ₀i2πr.

Infinite current sheet

B = μ₀j2

j = current per unit width of the sheet.
B does not depend on the distance from the sheet.

Practise PYQs on this topicAmpere's Law, Cylinders, Solenoid & Toroid · 61 questions with solutions

CONCEPT

Solenoid & Toroid

In one line
Inside a long solenoid the field is uniform and depends only on the turns per unit length and the current.

SOLENOID

21.6.1 A long helical coil

Field inside is along the axis

n = NL · B = μ₀ni

n = number of turns per unit length.
B does not depend on the radius of the solenoid or of its wire.
Outside a long solenoid: B ≈ 0.
Several layers: count all the turns in N.
Field lines inside are straight and parallel to the axis.

Change	New B
n × a, i × b	ab B
n doubled, i to one-third	23 B
same N, length halved	2B

END AND FINITE LENGTH

21.6.2 Short solenoids

finite: B = μ₀ni2 (sin α + sin β)

at the end of a long solenoid: B = μ₀ni2

α and β are the angles the two ends make at the point.
End field is half the field at the centre.

CORE AND INTENSITY

21.6.3 Soft iron inside

with a core: B = μni = μ₀μ_rni

magnetic intensity H = Bμ₀ = ni

A core of higher permeability gives a stronger field.
Unit of H is A m⁻¹; H does not depend on the core.

TOROID

21.6.4 A solenoid bent into a ring

Field lies only inside the core

B = μ₀Ni2πr = μ₀ni

n = N2πr, with r the mean radius.
Field lines are circles with a common centre.
B = 0 outside and in the central hollow.
For fixed n, B does not depend on the radius.
Ampere’s law with a circular loop inside the core gives this at once.

Practise PYQs on this topicAmpere's Law, Cylinders, Solenoid & Toroid · 61 questions with solutions

CONCEPT

Moving Charge in a Magnetic Field

In one line
A magnetic field pushes a moving charge sideways. It changes the direction of the velocity, never its speed.

MAGNETIC FORCE

21.7.1 On a moving charge

Force is perpendicular to both v and B

F = q(v × B)

F = qvB sin θ

Case	Force
charge at rest (v = 0)	0
neutral particle (q = 0)	0
θ = 0° or 180°	0
θ = 90°	qvB (maximum)

Fleming’s left-hand rule: first finger B, middle finger current, thumb F.
For a positive charge, current is along v.
For an electron, current is opposite to v.

NO WORK

21.7.2 Speed stays the same

F ⊥ v at every instant, so power = 0.
Work done by the magnetic force = 0.
Speed and kinetic energy do not change.
Momentum changes, because its direction changes.

PATH

21.7.3 Depends on the angle between v and B

θ	Path
0° or 180°	straight line
90°	circle
any other	helix

CIRCULAR PATH

21.7.4 v perpendicular to B

qvB gives the centripetal force

qvB = mv²r

r = mvqB = pqB = √(2mK)qB = 1B √(2mVq)

T = 2πmqB · ν = qB2πm · ω = qBm

T does not depend on v or r.
Faster particle: larger circle, same time.
K = energy; V = accelerating potential difference.
K = q²B²r²2m.

COMPARING PARTICLES

21.7.5 Proton, deuteron, α-particle

Same	r ∝	p : d : α
v	mq	1 : 2 : 2
p	1q	2 : 2 : 1
K	√mq	1 : √2 : 1
V	√(mq)	1 : √2 : √2

Electron and proton with the same momentum: same radius, opposite sense.
Field region of width x: deviation θ = sin⁻¹(xr).

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Helical path (θ other than 0°, 90°, 180°)

Circle + steady drift along B

v sin θ makes the circle; v cos θ carries it along B.

r = mv sin θqB · T = 2πmqB

pitch p = v cos θ · T = 2πmv cos θqB

Number of pitches in length L: N = Lp.
Pitch = 2πr when θ = 45°.

Practise PYQs on this topicMoving Charge in a Magnetic Field · 134 questions with solutions

CONCEPT

Lorentz Force, Velocity Selector & Cyclotron

In one line
With both E and B present, the force is the sum of the two. Crossed fields let only one speed pass.

LORENTZ FORCE

21.8.1 E and B together

F = qE + q(v × B)

E, B and v along one line: no magnetic force.
Then the charge speeds up or slows down in a straight line.
An electron moving along E slows down.
Released from rest in parallel E and B: straight line.

VELOCITY SELECTOR

21.8.2 E, B and v mutually perpendicular

Electric and magnetic forces balance

qE = qvB ⇒ v = EB

The selected speed does not depend on m or q.
Other speeds are deflected out of the beam.

CYCLOTRON

21.8.3 Accelerator for positive ions

Energy is gained at every crossing of the gap

Two hollow dees D₁ and D₂ with an oscillator across the gap.
B is perpendicular to the dees.
Inside a dee: circular path, no speed change.
In the gap: electric field accelerates the ion.
Radius grows, but the time per half circle stays the same.

ν_osc = qB2πm (resonance)

K_max = q²B²r²2m · v_max = qBrm

ω is constant; v increases with r.
K_max does not depend on the oscillator frequency.
It cannot accelerate neutrons (no charge).
It cannot accelerate electrons: being light, their mass changes with speed and they fall out of step.

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Hall effect

Charges pile up on one edge of the strip

A current-carrying strip is placed in a perpendicular field.
Charges are pushed to one edge, which sets up a transverse field E_H.
Steady state: eE_H = evB.

V_H = vBd = Binet

d = width, t = thickness, n = charge carriers per unit volume.
Hall coefficient R_H = 1ne.
The sign of V_H tells the sign of the carriers.

Practise PYQs on this topicMoving Charge in a Magnetic Field · 134 questions with solutions

CONCEPT

Force on Current-Carrying Conductors

In one line
A current is many moving charges, so a wire in a magnetic field feels a sideways force.

STRAIGHT CONDUCTOR

21.9.1 Length L in a uniform field

Fleming’s left-hand rule gives F

F = i(L × B)

F = BiL sin θ

θ = 0° or 180°: F = 0.
θ = 90°: F = BiL (maximum).
Only the part of B perpendicular to the wire counts.
Right-hand palm rule: thumb along i, fingers along B, push of the palm gives F.

BENT WIRES AND LOOPS

21.9.2 Uniform field

Curved wire = straight line joining its ends

Curved wire: L = straight line from start to end.
Closed loop in a uniform field: net force = 0.
Each side of the loop may still feel its own force.
Triangle in a field along one side: F on another side = iℓB sin(angle between them).

PARALLEL WIRES

21.9.3 Force per unit length

Like currents attract, unlike currents repel

Fl = μ₀4π · 2i₁i₂a = μ₀i₁i₂2πa

Each wire feels the same force (third law).
Same direction: attract. Opposite direction: repel.
One ampere: two wires 1 m apart, each current 1 A, force 2 × 10⁻⁷ N per metre.
Third wire with no force on it: i₁x = i₂d − x.

BALANCING

21.9.4 Magnetic force against weight

Upper wire floats: repulsion = weight

μ₀i₁i₂l2πh = mg

horizontal rod, slack strings: i = mgBl

The long wire is fixed; the free wire settles at height h.
Square loop with one side in a field: iLB = Mg.

LOOP NEAR A LONG WIRE

21.9.5 Sides at different distances

Near side wins

F_net = μ₀i₁i₂l2π (1a − 1a + b)

Top and bottom sides cancel.
Near side parallel to i₁: the loop is pulled towards the wire.

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Force between moving charges

Two like charges moving side by side: electric force repels, magnetic force attracts.

F_mF_e = (vc)²

So F_m is always far smaller than F_e for v ≪ c.

Current in a loop or a spring

A flexible loop in a field opens out into a circle.
A flexible loop opens even without an outside field: opposite sides repel.
A hanging spring carrying current contracts: its neighbouring turns have parallel currents and attract.

Rod on an incline, vertical B

Horizontal magnetic force balances the slope

B = mgil tan θ

Component of F up the incline = mg sin θ.

Suspended wire pushed down

T = 2π √(hg)

A wire balanced at height h does SHM when slightly displaced.

Wire perpendicular to a long wire

F = μ₀i₁i₂2π log_e (x + lx)

Length l starts at distance x from the long wire.
B varies along the length, so add dF over it.

Practise PYQs on this topicForce on Current-Carrying Conductors · 72 questions with solutions

CONCEPT

Current Loop as a Magnetic Dipole

In one line
A current loop behaves as a small magnet with moment M = NiA. A uniform field gives it a torque but no net force.

MAGNETIC MOMENT

21.10.1 Of a current loop

M = NiA

Unit: A m².
Direction: right-hand thumb, along the area vector.
Depends on N, i and A.

M is perpendicular to the plane

Shape (wire length L, one turn)	M
circle	iL²4π
square	iL²16
equilateral triangle	√3 iL²36

For a given perimeter, the circle has the largest area, hence the largest M.
For a circular coil, M ∝ L².
Same wire rewound to n turns: M ∝ 1n.
N turns of diameter d ⇒ doubling d halves N and doubles M.

M AND B OF A COIL

21.10.2 Linking the two

BM = μ₀2πR³

M = 2BR³·πμ₀ = 2BA^3/2μ₀√π

The current cancels out in B over M.
Solenoid: M = NiA, flux φ = BA.

solenoid: M = φlμ₀

ORBITING ELECTRON

21.10.3 Moment and gyromagnetic ratio

For an electron, M is opposite to L

i = ev2πr · M = iπr² = evr2

M = e2m L · ML = e2m

e2m is the gyromagnetic ratio; it depends only on e and m.
As vectors, M = −RL, where R is this ratio.
For any charge: ML = q2m.

Bohr orbit: L = nh2π ⇒ M = n eh4πm

eh4πm is the Bohr magneton, μ_B = 9.27 × 10⁻²⁴ A m².
So M ∝ n.
Electron going round f times per second: M = efπr².

TORQUE

21.10.4 Coil in a uniform field

Forces on opposite sides form a couple

τ = M × B

τ = NBiA sin θ

θ = angle between the normal to the coil (M) and B.
θ = 90° (plane of the coil parallel to B): τ is maximum.
θ = 0° (plane of the coil perpendicular to B): τ = 0.
Net force = 0, so the coil turns but does not move.
τ depends on area, not on the shape.
Wire of length L: τ_max is largest for one circular turn, iL²B4π.

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Work and potential energy

W = MB(cos θ₁ − cos θ₂)

From the stable position (θ₁ = 0) to θ: W = MB(1 − cos θ).

U = −M · B = −MB cos θ

θ	τ	U	Equilibrium
0°	0	−MB	stable
90°	MB	0	none
180°	0	+MB	unstable

Practise PYQs on this topicCurrent Loop as a Magnetic Dipole · 86 questions with solutions

CONCEPT

Moving Coil Galvanometer, Ammeter & Voltmeter

In one line
A galvanometer measures small currents by the turning of a coil. A shunt makes it an ammeter; a series resistance makes it a voltmeter.

MOVING COIL GALVANOMETER

21.11.1 Construction and principle

Concave poles and an iron core make the field radial

A coil of N turns hangs in a strong field; the suspension resists its turning.
In a radial field the plane of the coil stays parallel to B, so sin θ = 1.
The soft-iron core makes the field strong and radial.

NBAi = Cα

i = CNBA α = Gα

C = torsional constant; α = deflection.
G = CNBA is the galvanometer constant.
i ∝ α: the scale is linear.

SENSITIVITY

21.11.2 Deflection per unit input

current sensitivity S_i = αi = NBAC

voltage sensitivity S_V = αV = NBACR

More turns raise S_i, but the resistance rises too.
Doubling N doubles both S_i and R, so S_V stays the same.

AMMETER

21.11.3 Low resistance in parallel

Shunt S carries most of the current

S = i_gGi − i_g

S = Gn − 1 (n = ii_g)

Fraction of the current through G: i_gi = SS + G.
Connected in series; ideal ammeter: zero resistance.
Ammeter resistance = GSG + S = Gn.

VOLTMETER

21.11.4 High resistance in series

Series R drops most of the voltage

R = Vi_g − G

R = G(n − 1) (n = VV_g)

V_g = i_gG is the voltage range of the galvanometer.
Connected in parallel; ideal voltmeter: infinite resistance.
Voltmeter resistance = R + G = nG.

Practise PYQs on this topicMoving Coil Galvanometer, Ammeter & Voltmeter · 33 questions with solutions

EXAM-DAY CARD

Just see this before the exam

B = μ₀ni2R

centre of a coil

B = μ₀i2πr

long wire

B = μ₀ni

solenoid

r = mvqB

charge in B

Fl = μ₀i₁i₂2πa

parallel wires

τ = NBiA sin θ

coil in B

FORMULA SHEET

Whole chapter on one page

Coils

centre: μ₀ni2R; arc: μ₀4π · iθR

axis: μ₀niR²2(R² + x²)^3/2

rewound: B ∝ n²

charge: i = qf; orbit: i = ev2πr

Straight wires

μ₀4π · ir(sin φ₁ + sin φ₂)

infinite: μ₀i2πr

semi-infinite: μ₀i4πr

on its line: 0

Ampere, solenoid

∮B·dl = μ₀I_enc

inside wire: μ₀ir2πR²

solenoid, toroid: μ₀ni; end: μ₀ni2

H = ni

Moving charge

F = qvB sin θ; v = EB

r = mvqB = √(2mK)qB

T = 2πmqB

K_max = q²B²r²2m

Force, torque

F = BiL sin θ

Fl = μ₀i₁i₂2πa

M = NiA; τ = NBiA sin θ

ML = e2m; M = evr2

Meters

i = CNBA α; S_i = NBAC

S_V = NBACR

shunt: S = Gn − 1

series: R = G(n − 1)

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Rewound coil: B ∝ n

θ measured from the plane of the coil

Electron: using the rule as for + charge

Magnetic force does work on a charge

CORRECT ✓

Same wire: B ∝ n², M ∝ 1n

θ is between the normal (M) and B

Electron: current is opposite to v

F ⊥ v, so W = 0 and speed is constant