Moving Charges and Magnetism ISC Class 12: Notes, Derivations and Numericals

What Moving Charges and Magnetism is about

Weightage: Moving Charges and Magnetism is part of the unit Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carries 16 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 3 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Biot-Savart law and the field at the centre of a circular coil

A current element $Id\vec{l}$ produces at a point at displacement $\vec{r}$ a field $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2}$, magnitude $\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$, where ${\mu }_0=4\pi \times 10^{-7}Tm{A}^{-1}$. Its direction is perpendicular to both $d\vec{l}$ and $\vec{r}$. Centre of a coil of radius R: every element is at distance R and at $\theta =90^{\circ }$ to $\vec{r}$, and every $d\vec{B}$ points the same way along the axis. So $B=\frac{{\mu }_{0}I}{4\pi R^2}\oint dl=\frac{{\mu }_{0}I}{4\pi R^2}(2\pi R)=\frac{{\mu }_{0}I}{2R}$; with N turns, $B=\frac{{\mu }_{0}NI}{2R}$. Direction by the right-hand rule: curl the fingers along the current, the thumb gives B.

Worked example

A coil of 50 turns and radius 10 cm carries 2 A. Find the field at its centre.

Try it first, then show the solutionSolution

1. $B=\frac{{\mu }_{0}NI}{2R}=\frac{4\pi \times 10^{-7}\times 50\times 2}{2\times 0.10}$
2. $=6.28\times 10^{-4}T$

Answer: $6.28\times 10^{-4}T$, along the axis (direction by the right-hand rule)

Field on the axis of a circular coil; the loop as a magnetic dipole

At a point P on the axis at distance x from the centre, each element is at distance $\sqrt{R^2+x^2}$ and $d\vec{l}\perp \vec{r}$. The components of $d\vec{B}$ perpendicular to the axis cancel in pairs from opposite elements; the axial components add: $d{B}_x=dB\sin \alpha$ with $\sin \alpha =\frac{R}{\sqrt{R^2+x^2}}$. Integrating, $B=\frac{{\mu }_{0}NI{R}^2}{2(R^2+x^2{)}^{3/2}}$. At x = 0 this gives $\frac{{\mu }_{0}NI}{2R}$. For $x\gg R$, $B\approx \frac{{\mu }_0}{4\pi }\frac{2NIA}{x^3}=\frac{{\mu }_0}{4\pi }\frac{2m}{x^3}$ with $m=NIA$, which has the same form as the axial field of a dipole. So a current loop is a magnetic dipole of moment $\vec{m}=NI\vec{A}$ (unit A m²), directed along the thumb of the right hand.

Worked example

For the 50-turn, 10 cm coil carrying 2 A, find the field on the axis 10 cm from the centre, and its ratio to the field at the centre.

Try it first, then show the solutionSolution

1. $B=\frac{{\mu }_{0}NI{R}^2}{2(R^2+x^2{)}^{3/2}}=\frac{4\pi \times 10^{-7}\times 50\times 2\times 0.01}{2\times (0.02{)}^{3/2}}$
2. $(0.02{)}^{3/2}=2.83\times 10^{-3}$, so $B=\frac{1.257\times 10^{-6}}{5.66\times 10^{-3}}=2.22\times 10^{-4}T$
3. Ratio $=\frac{R^3}{(R^2+x^2{)}^{3/2}}=\frac{1}{2\sqrt{2}}=0.354$

Answer: $2.22\times 10^{-4}T$, about 0.35 of the field at the centre

Ampère's circuital law: long straight wire and solenoid

Ampère's law: the line integral of $\vec{B}$ round any closed loop equals ${\mu }_0$ times the net current threading the loop: $\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$. Long straight wire: take a circle of radius r centred on the wire; by symmetry B is tangential and constant on it, so $B(2\pi r)={\mu }_{0}I$ and $B=\frac{{\mu }_{0}I}{2\pi r}$. Field lines are concentric circles (right-hand thumb rule). For a wire of finite length, only the formula is needed: $B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\varphi }_1+\sin {\varphi }_2)$. Long solenoid (qualitative treatment): the field inside is strong, uniform and along the axis; outside it is nearly zero. A rectangular Ampèrian loop with one side inside gives $Bl={\mu }_0(nl)I$, so $B={\mu }_{0}nI$, where n is turns per unit length. At the ends it is about half this value. Unlike an electric field line, a magnetic field line forms a closed loop: there are no magnetic monopoles.

Worked example

(a) Find the field 5 cm from a long straight wire carrying 10 A. (b) Find the field inside a long solenoid with 1000 turns per metre carrying 2 A.

Try it first, then show the solutionSolution

1. (a) $B=\frac{{\mu }_{0}I}{2\pi r}=\frac{2\times 10^{-7}\times 10}{0.05}=4\times 10^{-5}T$
2. (b) $B={\mu }_{0}nI=4\pi \times 10^{-7}\times 1000\times 2=2.51\times 10^{-3}T$

Answer: (a) $4\times 10^{-5}T$ (b) $2.51\times 10^{-3}T$

Force on a moving charge; Lorentz force; circular motion

The magnetic field is defined by the force it exerts on a moving charge: $\vec{F}=q(\vec{v}\times \vec{B})$, magnitude $qvB\sin \theta$ (not by force on a 'unit pole'). The force is perpendicular to $\vec{v}$, so it does no work and cannot change the speed. Special cases: $\vec{v}\parallel \vec{B}$: F = 0 (straight line); $\vec{v}\perp \vec{B}$: F = qvB, giving uniform circular motion with $\frac{m{v}^2}{r}=qvB$, so $r=\frac{mv}{qB}$, period $T=\frac{2\pi m}{qB}$ and frequency $f=\frac{qB}{2\pi m}$ (independent of speed). At any other angle the path is a helix. With an electric field as well, the Lorentz force is $\vec{F}=q(\vec{E}+\vec{v}\times \vec{B})$. Crossed fields with $qE=qvB$ pass particles of speed $v=E/B$ undeviated, whatever their charge or mass (a velocity selector).

Worked example

A proton moves at $2\times 10^{6}m{s}^{-1}$ perpendicular to a field of 0.5 T. Find the force on it, the radius of its path and its frequency of revolution. ($m_p=1.67\times 10^{-27}kg$)

Try it first, then show the solutionSolution

1. $F=qvB=1.6\times 10^{-19}\times 2\times 10^6\times 0.5=1.6\times 10^{-13}N$
2. $r=\frac{mv}{qB}=\frac{1.67\times 10^{-27}\times 2\times 10^6}{1.6\times 10^{-19}\times 0.5}=0.0418m$
3. $f=\frac{qB}{2\pi m}=\frac{1.6\times 10^{-19}\times 0.5}{2\pi \times 1.67\times 10^{-27}}=7.62\times 10^{6}Hz$

Answer: $1.6\times 10^{-13}N$; 4.18 cm; $7.62\times 10^{6}Hz$

Force on a current-carrying conductor

Replacing q by $Idt$ and $\vec{v}$ by $d\vec{l}/dt$ in $q(\vec{v}\times \vec{B})$ gives $d\vec{F}=I(d\vec{l}\times \vec{B})$; for a straight conductor of length l in a uniform field, $\vec{F}=I(\vec{l}\times \vec{B})$, magnitude $BIl\sin \theta$. Maximum when the wire is perpendicular to B, zero when parallel. The direction follows the cross product (Fleming's left-hand rule gives the same result, but the ISC syllabus asks for the right-hand rule and vector product).

Worked example

A 0.5 m wire carrying 4 A lies at $30^{\circ }$ to a uniform field of 0.2 T. Find the force on it.

Try it first, then show the solutionSolution

1. $F=BIl\sin \theta =0.2\times 4\times 0.5\times \sin 30^{\circ }$
2. $=0.4\times 0.5=0.2N$, perpendicular to both the wire and the field

Answer: 0.2 N

Force between two parallel currents and the definition of the ampere

Wire 1 (current $I_1$) produces at wire 2, a distance d away, a field $B_1=\frac{{\mu }_0{I}_1}{2\pi d}$ perpendicular to wire 2. Wire 2 then feels a force per unit length $\frac{F}{l}=I_2{B}_1=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$. By the right-hand rule, currents in the same direction attract; opposite currents repel. The ampere: if two infinitely long, thin, parallel wires 1 m apart in vacuum carrying equal currents experience a force of $2\times 10^{-7}N$ per metre of length, each current is 1 A. Then 1 coulomb is the charge carried by 1 A in 1 s ($Q=It$).

Worked example

Two long parallel wires 10 cm apart carry 5 A and 8 A in the same direction. Find the force per metre between them and say whether it attracts or repels.

Try it first, then show the solutionSolution

1. $\frac{F}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{2\times 10^{-7}\times 5\times 8}{0.1}$
2. $=8\times 10^{-5}N{m}^{-1}$, attractive (same direction)

Answer: $8\times 10^{-5}N$ per metre, attractive

Torque on a current loop; orbital magnetic moment

A rectangular coil (N turns, sides l and b, area A = lb) carrying I in a uniform field B, with its normal at angle φ to B: forces on the two sides of length b are equal, opposite and collinear (they cancel). The forces on the sides of length l are $NIlB$ each, equal and opposite but separated by $b\sin \varphi$, forming a couple: $\tau =NIlB\times b\sin \varphi =NIAB\sin \varphi$, i.e. $\vec{\tau }=\vec{m}\times \vec{B}$ with $\vec{m}=NI\vec{A}$. Net force zero, torque maximum when the plane of the coil is parallel to B (φ = 90°). An electron in a Bohr orbit is a tiny current loop: $I=\frac{ev}{2\pi r}$, so its orbital magnetic moment is ${\mu }_l=IA=\frac{evr}{2}$; for the first orbit this is $\frac{eh}{4\pi m_e}\approx 9.27\times 10^{-24}A{m}^2$ (the Bohr magneton).

Worked example

A square coil of side 2 cm has 100 turns and carries 50 mA. It hangs in a field of 0.1 T with its plane parallel to the field. Find its magnetic moment and the torque on it.

Try it first, then show the solutionSolution

1. $A=(0.02{)}^2=4\times 10^{-4}{m}^2$; $m=NIA=100\times 0.05\times 4\times 10^{-4}=2\times 10^{-3}A{m}^2$
2. Plane parallel to B means the normal is at $90^{\circ }$: $\tau =mB\sin 90^{\circ }=2\times 10^{-3}\times 0.1=2\times 10^{-4}Nm$

Answer: $2\times 10^{-3}A{m}^2$; $2\times 10^{-4}Nm$ (the maximum)

Moving coil galvanometer: radial field, sensitivity and conversion

Construction: a rectangular coil on a light frame between concave pole pieces, with a soft-iron cylinder inside, suspended by a phosphor-bronze strip (or pivoted with a hairspring). The concave poles and core make the field radial: the plane of the coil is always parallel to B, so $\tau =NIAB$ at every deflection. At equilibrium this balances the restoring torque $k\varphi$ (k = torsion constant): $NIAB=k\varphi$, so $I=\frac{k}{NAB}\varphi$, and the scale is linear. Current sensitivity $\frac{\varphi }{I}=\frac{NAB}{k}$; voltage sensitivity $\frac{\varphi }{V}=\frac{NAB}{kG}$. Doubling N doubles current sensitivity but also roughly doubles G, so voltage sensitivity may not rise. Ammeter: a small shunt $S=\frac{I_{g}G}{I-I_g}$ in parallel (low total resistance). Voltmeter: a large resistance $R=\frac{V}{I_g}-G$ in series (high total resistance).

Worked example

A galvanometer of resistance 50 Ω gives full-scale deflection for 2 mA. How would you convert it into (a) an ammeter of range 0-2 A and (b) a voltmeter of range 0-20 V?

Try it first, then show the solutionSolution

1. (a) Shunt in parallel: $S=\frac{I_{g}G}{I-I_g}=\frac{0.002\times 50}{2-0.002}=0.050\Omega$
2. (b) Series resistance: $R=\frac{V}{I_g}-G=\frac{20}{0.002}-50=9950\Omega$

Answer: (a) about 0.05 Ω in parallel (b) 9950 Ω in series

Physics in real life: Moving Charges and Magnetism

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Biot-Savart Law and the Field of a Circular Coil
• Motion of Charges in a Magnetic Field
• Moving Coil Galvanometer and Its Conversion

Moving Charges and Magnetism formula sheet

Biot-Savart law

$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2},dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

${\mu }_0=4\pi \times 10^{-7}Tm{A}^{-1}$.

Centre of a circular coil

$B=\frac{{\mu }_{0}NI}{2R}$

Along the axis; right-hand rule for direction.

Axis of a circular coil

$B=\frac{{\mu }_{0}NI{R}^2}{2(R^2+x^2{)}^{3/2}}$

For $x\gg R$: $B=\frac{{\mu }_0}{4\pi }\frac{2m}{x^3}$.

Ampère's circuital law

$\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$

I is the net current threading the loop.

Long straight wire

$B=\frac{{\mu }_{0}I}{2\pi r}$

Concentric circular field lines.

Finite straight conductor (formula only)

$B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\varphi }_1+\sin {\varphi }_2)$

Angles measured at the point from the perpendicular.

Long solenoid

$B={\mu }_{0}nI$

n = turns per metre; uniform inside, about half at the ends.

Force on a moving charge

$\vec{F}=q(\vec{v}\times \vec{B}),F=qvB\sin \theta$

No work done; speed constant.

Lorentz force

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

Undeviated in crossed fields if v = E/B.

Circular motion in B

$r=\frac{mv}{qB},T=\frac{2\pi m}{qB},f=\frac{qB}{2\pi m}$

T and f do not depend on v.

Force on a conductor

$\vec{F}=I(\vec{l}\times \vec{B}),F=BIl\sin \theta$

Zero when the wire is along B.

Parallel wires

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

Same direction attract; opposite repel.

Torque on a coil

$\vec{\tau }=\vec{m}\times \vec{B},\tau =NIAB\sin \varphi ,\vec{m}=NI\vec{A}$

φ is between the normal and B.

Orbital magnetic moment

${\mu }_l=\frac{evr}{2},{\mu }_B=\frac{eh}{4\pi m_e}=9.27\times 10^{-24}A{m}^2$

Electron in the first Bohr orbit.

Galvanometer

$I=\frac{k}{NAB}\varphi ;\frac{\varphi }{I}=\frac{NAB}{k};\frac{\varphi }{V}=\frac{NAB}{kG}$

Radial field gives a linear scale.

Conversion

$S=\frac{I_{g}G}{I-I_g}(ammeter);R=\frac{V}{I_g}-G(voltmeter)$

Shunt in parallel; resistance in series.

Moving Charges and Magnetism: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Moving Charges and Magnetism

The question patterns to expect, and the method that works for each.

Tricks to remember Moving Charges and Magnetism

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Centre is half, wire is two-pi

  Coil centre: $\frac{{\mu }_{0}I}{2R}$ ('two R'). Long wire: $\frac{{\mu }_{0}I}{2\pi r}$ ('two pi r'). Solenoid: ${\mu }_{0}nI$ (no denominator).

  Use it when: Not mixing the three field formulas, a mistake examiners report.

• F = q v B order

  'Very Big' comes after q: $q(\vec{v}\times \vec{B})$. Swapping the order reverses the force.

  Use it when: Vector form of the magnetic force.

• Likes attract for currents

  Charges: likes repel. Currents: likes (same direction) attract. Opposite of what you'd expect.

  Use it when: Parallel-wire MCQs.

• Magnetic force never speeds you up

  Force ⊥ velocity, so no work: speed and KE stay constant, only direction changes.

  Use it when: Assertion-reason items on KE in a magnetic field.

• SAG and VSR

  Shunt for Ammeter, Galvanometer in parallel; Voltmeter: Series Resistance.

  Use it when: Conversion questions.

• Radial means always parallel

  In a radial field the coil's plane is always parallel to B, so sin φ = 1 at every deflection.

  Use it when: Explaining why the galvanometer scale is linear.

Moving Charges and Magnetism: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Moving Charges and Magnetism

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing the cross product in the wrong order

  CISCE examiners' comments, 2024: for the Lorentz force in vector form, candidates chose $q(\vec{B}\times \vec{v})$, showing confusion about the order of a cross product.

  Wrong working

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

  Right working

  $\vec{F}=q(\vec{v}\times \vec{B})$; reversing the order reverses the force.

  ✓Do this: Learn the vector forms exactly: $q(\vec{v}\times \vec{B})$, $I(\vec{l}\times \vec{B})$, $\vec{m}\times \vec{B}$.

  Usually costs: 1 mark.

• ✗Mistake: Incomplete Biot-Savart derivation

  CISCE examiners' comments, 2024: for the field at the centre of a coil, diagrams lacked $d\vec{l}$, θ = 90° or the direction of B; some used Ampère's law instead; limits of integration were missing; some changed R to r or a.

  Wrong working

  $B=\oint \frac{{\mu }_{0}Idl}{4\pi r^2}$ with no figure, giving $\frac{{\mu }_{0}I}{2r}$.

  Right working

  Figure with $d\vec{l}\perp \vec{r}$, the law, $\oint dl=2\pi R$, result $\frac{{\mu }_{0}I}{2R}$ with direction.

  ✓Do this: Use the question's symbols and show every vector in the diagram.

  Usually costs: 1-2 marks in a 3-mark derivation.

• ✗Mistake: Giving a field direction as just 'inward'

  CISCE examiners' comments, 2024: for the field of a long wire, candidates wrote only 'inward' without saying perpendicular to the plane of the paper, or used the wrong current symbol in the formula.

  Wrong working

  B at P is inward.

  Right working

  B at P is perpendicular to the plane of the paper, directed into it (⊗).

  ✓Do this: Name the plane and use ⊗ or ⊙.

  Usually costs: 1 mark.

• ✗Mistake: Using the straight-wire formula for a coil

  CISCE examiners' comments, 2025: asked how the field at the centre of a coil changes when I, N and R are all doubled, candidates used the long-wire formula or did not relate the new field to B.

  Wrong working

  $B=\frac{{\mu }_{0}I}{2\pi R}$, so the field is unchanged.

  Right working

  $B=\frac{{\mu }_{0}NI}{2R}$; doubling I, N and R gives $\frac{2\times 2}{2}=2$, so the field becomes 2B.

  ✓Do this: Write the formula for the right shape before substituting.

  Usually costs: 1-2 marks.

• ✗Mistake: Adding magnitudes without directions for two wires

  CISCE examiners' comments, 2025: for the field at a point due to two long wires, candidates computed magnitudes without considering directions and could not add the fields vectorially.

  Wrong working

  $B=B_1+B_2$ at a point between two wires with currents in the same direction.

  Right working

  Between wires with currents in the same direction the fields oppose: $B=|B_1-B_2|$.

  ✓Do this: Mark ⊗ or ⊙ for each wire's field at the point first.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong path or missing formula for circular motion

  CISCE examiners' comments, 2025: for an electron moving perpendicular to B, candidates wrote 'helical' or 'linear' instead of circular, omitted the sense of rotation, did not know $f=qB/2\pi m$, omitted the unit or rounded off twice.

  Wrong working

  Path is a helix; $f=\frac{qB}{m}$.

  Right working

  Velocity ⊥ B gives a circle; $f=\frac{qB}{2\pi m}$ in Hz. A helix needs a velocity component along B.

  ✓Do this: Check the angle between v and B; round only at the final step.

  Usually costs: 1-2 marks.

Practice questions on Moving Charges and Magnetism

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The SI unit of magnetic field B is the tesla, which equals:
   (a) $NA{m}^{-1}$
   (b) $N{A}^{-1}{m}^{-1}$
   (c) $N{A}^{-1}m$
   (d) $A{m}^{-1}$

   Hint

   F = BIl.

   Answer

   (b) $N{A}^{-1}{m}^{-1}$

2. Q2 · 1 mark · MCQBoard level

   The magnetic field inside a long current-carrying solenoid is:
   (a) zero
   (b) uniform and along the axis
   (c) maximum at the ends
   (d) radial

   Hint

   Think of the parallel field lines inside.

   Answer

   (b) uniform and along the axis

3. Q3 · 1 mark · MCQBoard level

   Two long parallel wires carry currents in opposite directions. They:
   (a) attract
   (b) repel
   (c) exert no force
   (d) twist

   Hint

   Unlike currents.

   Answer

   (b) repel

4. Q4 · 1 mark · Very short answerBoard level

   Why is the magnetic field in a moving coil galvanometer made radial?

   Hint

   Torque at every position.

   Answer

   So that the plane of the coil is always parallel to the field, the torque is NIAB at all deflections, and the deflection is proportional to the current (linear scale).

5. Q5 · 2 marks · Short answerBoard level

   A long wire carries 10 A. Find B at 5 cm from it. At what distance is the field $2\times 10^{-5}T$?

   Hint

   $B\propto 1/r$.

   Answer

   $B=4\times 10^{-5}T$ at 5 cm; half that field is at 10 cm.

6. Q6 · 2 marks · Short answerHarder board level

   A proton moves at $2\times 10^{6}m{s}^{-1}$ at right angles to a 0.5 T field. Find the radius of its path and its frequency of revolution.

   Hint

   $r=mv/qB$, $f=qB/2\pi m$.

   Answer

   r = 4.18 cm; $f=7.62\times 10^{6}Hz$

7. Q7 · 2 marks · Short answerHarder board level

   A 50 cm wire carrying 4 A lies at $30^{\circ }$ to a 0.2 T field. Find the force on it. At what angle would the force be maximum, and what would it be?

   Hint

   $F=BIl\sin \theta$.

   Answer

   F = 0.2 N; maximum at 90°: $0.2\times 4\times 0.5=0.4N$.

8. Q8 · 2 marks · Short answerHOTS / competency

   A beam of charged particles passes undeviated through crossed electric and magnetic fields of $3\times 10^{4}V{m}^{-1}$ and 0.15 T. Find their speed. If the particles' charge were doubled at the same speed, would they still pass undeviated? Explain.

   Hint

   Balance qE and qvB.

   Answer

   $v=E/B=2\times 10^{5}m{s}^{-1}$. Yes: both forces double, so they still balance; the condition v = E/B does not involve q or m.

9. Q9 · 3 marks · Short answerHarder board level

   A galvanometer of 50 Ω gives full-scale deflection at 2 mA. Find the shunt for a 0-2 A ammeter and the series resistance for a 0-20 V voltmeter.

   Hint

   Shunt carries $I-I_g$.

   Answer

   Shunt 0.050 Ω in parallel; 9950 Ω in series.

10. Q10 · 3 marks · Short answerHarder board level

    A coil of 50 turns and radius 10 cm carries 2 A. Find the field (a) at its centre and (b) on its axis 10 cm from the centre.

    Hint

    Use the axial formula with x = R.

    Answer

    (a) $6.28\times 10^{-4}T$ (b) $\frac{{\mu }_{0}NI{R}^2}{2(R^2+x^2{)}^{3/2}}=2.22\times 10^{-4}T$

11. Q11 · 3 marks · Short answerHarder board level

    State Ampère's circuital law. Explain qualitatively why the field inside a long solenoid is uniform and that outside is nearly zero, and find B inside a solenoid of 1000 turns per metre carrying 2 A.

    Hint

    Fields of neighbouring turns add inside and cancel outside.

    Answer

    $\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$. Inside, the fields of all turns point the same way along the axis and add; outside, the fields of the near and far sides of each turn largely cancel, and the field lines spread over a huge area, so B is nearly zero. $B={\mu }_{0}nI=2.51\times 10^{-3}T$.

12. Q12 · 5 marks · Long answerHarder board level

    (a) State the Biot-Savart law in vector form. [1]
    (b) Derive the field at the centre of a circular coil of N turns. [2]
    (c) Show that an electron revolving in a circular orbit of radius r with speed v has a magnetic moment $\frac{evr}{2}$, and evaluate it for the first Bohr orbit using $mvr=\frac{h}{2\pi }$. [2]

    Hint

    Treat the orbit as a current loop of current $e/T$.

    Answer

    (a) $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2}$. (b) $B=\frac{{\mu }_{0}NI}{2R}$ (all $d\vec{B}$ along the axis, $\oint dl=2\pi R$). (c) $I=\frac{e}{T}=\frac{ev}{2\pi r}$, $\mu =I\pi r^2=\frac{evr}{2}$. With $vr=\frac{h}{2\pi m}$: $\mu =\frac{eh}{4\pi m}=\frac{1.6\times 10^{-19}\times 6.63\times 10^{-34}}{4\pi \times 9.1\times 10^{-31}}=9.28\times 10^{-24}A{m}^2$.

13. Q13 · 2 marks · Short answerHarder board level

    A square coil of side 2 cm and 100 turns carries 50 mA in a 0.1 T field. Find its magnetic moment and the torque when (a) its plane is parallel to B, (b) its plane is perpendicular to B.

    Hint

    φ is measured from the normal.

    Answer

    $m=2\times 10^{-3}A{m}^2$. (a) $\tau =mB=2\times 10^{-4}Nm$ (b) normal along B, τ = 0.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State the Biot-Savart law.Show answer

   $d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \hat{r}}{r^2}$: the field of a current element is proportional to I dl sin θ and inversely to r².

2. 2.State Ampère's circuital law.Show answer

   $\oint \vec{B}\cdot d\vec{l}={\mu }_{0}I$, where I is the net current through the closed loop.

3. 3.Field at the centre of a coil of N turns?Show answer

   $\frac{{\mu }_{0}NI}{2R}$.

4. 4.Expression for the Lorentz force?Show answer

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

5. 5.Define one ampere.Show answer

   The current which, in each of two infinitely long thin parallel wires 1 m apart in vacuum, produces a force of $2\times 10^{-7}N$ per metre between them.

6. 6.Torque on a current loop?Show answer

   $\vec{\tau }=\vec{m}\times \vec{B}$, $\tau =NIAB\sin \varphi$.

7. 7.Define current sensitivity of a galvanometer.Show answer

   Deflection per unit current, $\varphi /I=NAB/k$.

8. 8.How is a galvanometer converted into an ammeter?Show answer

   By connecting a low resistance (shunt) $S=\frac{I_{g}G}{I-I_g}$ in parallel.

9. 9.Frequency of a charge circling in a field B?Show answer

   $f=\frac{qB}{2\pi m}$, independent of its speed.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you derive $B={\mu }_{0}NI/2R$ and the axial field with a complete diagram?
• ☐Can you write the long-wire, finite-wire and solenoid formulas?
• ☐Can you find the net field due to two long wires with directions?
• ☐Can you write $q(\vec{v}\times \vec{B})$ in the right order and describe the path for θ = 0°, 90° and any other angle?
• ☐Can you find r, T and f for a charge in B?
• ☐Can you derive the force between parallel wires and define the ampere?
• ☐Can you derive $\tau =NIAB\sin \varphi$ and the orbital magnetic moment $evr/2$?
• ☐Can you explain the radial field and convert a galvanometer to an ammeter and a voltmeter?

Moving Charges and Magnetism: frequently asked questions