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

What Moving Charges and Magnetism is about

Weightage: This chapter is part of Units III and IV, Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carry 17 of the 70 theory marks together with Magnetism and Matter, Electromagnetic Induction and Alternating Current.Source: CBSE Physics syllabus 2026-27, Units III and IV Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.4 marks per paper. See the questions

Key concepts

Magnetic field and the Lorentz force

NCERT section 4.2

A charge q moving with velocity $\vec{v}$ in electric field $\vec{E}$ and magnetic field $\vec{B}$ feels the Lorentz force:

$\vec{F}=q[\vec{E}+\vec{v}\times \vec{B}]$

The magnetic part, $q\vec{v}\times \vec{B}$, has magnitude $qvB\sin \theta$ and these features:

• It is zero for a charge at rest, and for motion along the field ($\theta =0$ or $180^{\circ }$).
• It is perpendicular to both $\vec{v}$ and $\vec{B}$, so it does no work and cannot change the speed or kinetic energy; it only changes direction.
• For a negative charge it acts opposite to the direction for a positive charge.

The SI unit of B is the tesla: 1 T is the field in which a charge of 1 C moving at 1 m s⁻¹ perpendicular to it feels 1 N. A smaller unit is the gauss, $1G=10^{-4}T$.

Worked example

A proton moves at $2\times 10^{6}m{s}^{-1}$ along the $+x$ axis in a uniform field $\vec{B}=0.5\hat{j}T$. Find the magnitude and direction of the force on it. ($e=1.6\times 10^{-19}C$) [2 marks]

Try it first, then show the solutionSolution

1. $\vec{F}=q\vec{v}\times \vec{B}=e(2\times 10^6\hat{i})\times (0.5\hat{j})$
2. $\hat{i}\times \hat{j}=\hat{k}$, so $\vec{F}=1.6\times 10^{-19}\times 10^6\hat{k}$
3. $F=1.6\times 10^{-13}N$ along $+z$

Answer: $1.6\times 10^{-13}N$ along the $+z$ direction

Force on a current-carrying conductor

NCERT section 4.2.3

A wire of length l and cross-section A has $nAl$ free electrons, each drifting with ${\vec{v}}_d$. Adding the Lorentz forces: $\vec{F}=(nAl)(-e){\vec{v}}_d\times \vec{B}$. Since the current density is $\vec{j}=-ne{\vec{v}}_d$ and $jA=I$:

$\vec{F}=I\vec{l}\times \vec{B},F=IlB\sin \theta$

where $\vec{l}$ points along the current. The direction is given by the vector product (or Fleming's left-hand rule). For a curved wire, add $Id\vec{l}\times \vec{B}$ along its length.

Worked example

A horizontal wire of length 0.5 m and mass 60 g lies at right angles to a horizontal uniform field of 0.4 T. What current must flow in it so that it stays suspended in mid-air? Take $g=9.8m{s}^{-2}$. [2 marks]

Try it first, then show the solutionSolution

1. For balance, the magnetic force IlB must equal the weight mg and point upward.
2. $I=\frac{mg}{lB}=\frac{0.060\times 9.8}{0.5\times 0.4}=\frac{0.588}{0.2}$
3. $I=2.94A$, flowing in the direction for which $\vec{l}\times \vec{B}$ points up

Answer: About 2.9 A

Motion of a charge in a uniform magnetic field

NCERT section 4.3

If $\vec{v}\perp \vec{B}$, the force $qvB$ is always perpendicular to the velocity, so it provides a centripetal force and the charge moves in a circle:

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

The angular frequency $\omega =\frac{qB}{m}$ and the period $T=\frac{2\pi m}{qB}$ do not depend on the speed: faster particles go round bigger circles in the same time.

If $\vec{v}$ makes angle θ with $\vec{B}$, split it: $v_{\perp }=v\sin \theta$ gives a circle of radius $\frac{mv\sin \theta }{qB}$, and $v_{\parallel }=v\cos \theta$ carries the charge along the field. The path is a helix with pitch $p=v_{\parallel }T=\frac{2\pi mv\cos \theta }{qB}$.

Worked example

An electron moves at $4\times 10^{6}m{s}^{-1}$ at right angles to a field of $2\times 10^{-3}T$. Find the radius of its path and its frequency of revolution. ($m=9.1\times 10^{-31}kg$, $e=1.6\times 10^{-19}C$) [2 marks]

Try it first, then show the solutionSolution

1. $r=\frac{mv}{eB}=\frac{9.1\times 10^{-31}\times 4\times 10^6}{1.6\times 10^{-19}\times 2\times 10^{-3}}=\frac{3.64\times 10^{-24}}{3.2\times 10^{-22}}$
2. $r\approx 1.14\times 10^{-2}m=1.14cm$
3. $\nu =\frac{eB}{2\pi m}=\frac{3.2\times 10^{-22}}{2\pi \times 9.1\times 10^{-31}}\approx 5.6\times 10^{7}Hz$

Answer: r ≈ 1.14 cm; ν ≈ $5.6\times 10^{7}Hz$

Biot-Savart law

NCERT section 4.4

The magnetic field $d\vec{B}$ at a point P due to a small element $d\vec{l}$ of a wire carrying current I is

$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \vec{r}}{r^3},dB=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

where $\vec{r}$ runs from the element to P and θ is the angle between $d\vec{l}$ and $\vec{r}$. ${\mu }_0=4\pi \times 10^{-7}Tm{A}^{-1}$ is the permeability of free space, and $\frac{{\mu }_0}{4\pi }=10^{-7}Tm{A}^{-1}$.

Similarities with Coulomb's law: both fall as $\frac{1}{r^2}$ and both are long range. Differences: the source of B is a vector (Idl) not a scalar; dB is perpendicular to both $d\vec{l}$ and $\vec{r}$; and dB is zero along the line of the element ($\theta =0$).

Worked example

A 1 cm element of wire carrying 10 A lies along the x-axis at the origin. Find the field it produces at a point 0.5 m away on the y-axis. [2 marks]

Try it first, then show the solutionSolution

1. Here $\theta =90^{\circ }$, so $dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{r^2}$
2. $dB=10^{-7}\times \frac{10\times 0.01}{(0.5{)}^2}=10^{-7}\times \frac{0.1}{0.25}$
3. $dB=4\times 10^{-8}T$, along $+z$ (direction of $\hat{i}\times \hat{j}$)

Answer: $4\times 10^{-8}T$ along the $+z$ axis

Field on the axis of a circular current loop

NCERT section 4.5

Derivation. Take a loop of radius R carrying current I, and a point P on its axis at distance x from the centre. Every element dl is perpendicular to the line joining it to P, at distance $r=\sqrt{x^2+R^2}$, so $dB=\frac{{\mu }_0}{4\pi }\frac{Idl}{x^2+R^2}$.

dB has a component along the axis and one perpendicular to it. Elements on opposite ends of a diameter give perpendicular components that cancel, so only the axial components survive: $d{B}_x=dB\cos \theta$ with $\cos \theta =\frac{R}{\sqrt{x^2+R^2}}$.

Summing over the loop ($\oint dl=2\pi R$):

$B=\frac{{\mu }_{0}I{R}^2}{2(x^2+R^2{)}^{3/2}}$, along the axis

At the centre (x = 0): $B=\frac{{\mu }_{0}I}{2R}$; for N turns, multiply by N. The direction is given by the right-hand thumb rule: curl the fingers along the current and the thumb points along B. Far away ($x\gg R$), $B\approx \frac{{\mu }_{0}I{R}^2}{2x^3}=\frac{{\mu }_0}{4\pi }\frac{2m}{x^3}$ with $m=I\pi R^2$, the same form as the axial field of an electric dipole.

Worked example

A coil of 50 turns and radius 10 cm carries 2 A. Find the field (a) at its centre, (b) on its axis 10 cm from the centre. (${\mu }_0=4\pi \times 10^{-7}Tm{A}^{-1}$) [3 marks]

Try it first, then show the solutionSolution

1. (a) $B_0=\frac{{\mu }_{0}NI}{2R}=\frac{4\pi \times 10^{-7}\times 50\times 2}{2\times 0.1}=4\pi \times 10^{-7}\times 500$
2. $B_0=6.28\times 10^{-4}T$
3. (b) With x = R, $B=\frac{{\mu }_{0}NI{R}^2}{2(2R^2{)}^{3/2}}=\frac{B_0}{2^{3/2}}=\frac{6.28\times 10^{-4}}{2.83}$
4. $B\approx 2.22\times 10^{-4}T$

Answer: (a) $6.28\times 10^{-4}T$ (b) about $2.22\times 10^{-4}T$, both along the axis

Ampere's circuital law and the long straight wire

NCERT section 4.6

Ampere's circuital law: the line integral of $\vec{B}$ round any closed loop equals ${\mu }_0$ times the total current passing through the surface bounded by the loop:

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

It plays the role for magnetism that Gauss's law plays for electrostatics, and is useful when symmetry makes B constant along a well-chosen loop (an 'Amperian loop').

Long straight wire. By symmetry the field lines are circles round the wire, and B has the same size at all points at distance r. Take a circle of radius r: $B\times 2\pi r={\mu }_{0}I$, so

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

The direction follows the right-hand thumb rule (thumb along the current, fingers curl along B). For a thick wire of radius a carrying current uniformly, inside it ($r<a$) the enclosed current is $I\frac{r^2}{a^2}$, so $B=\frac{{\mu }_{0}Ir}{2\pi a^2}$, rising linearly to a maximum at the surface.

Worked example

A long straight wire carries 10 A. Find the magnetic field 5 cm from it. [1 mark]

Try it first, then show the solutionSolution

1. $B=\frac{{\mu }_{0}I}{2\pi r}=\frac{2\times 10^{-7}\times 10}{0.05}$
2. $B=4\times 10^{-5}T$, in circles round the wire

Answer: $4\times 10^{-5}T$

The straight solenoid (qualitative)

NCERT section 4.7

A solenoid is a long wire wound in a closely spaced helix. The syllabus treats it qualitatively: you should know the field pattern and the result, not a derivation.

• Inside a long solenoid, the fields of neighbouring turns add along the axis and the field is strong and nearly uniform, parallel to the axis.
• Just outside, the fields of turns on opposite sides tend to cancel, so the field is very weak, nearly zero for a long solenoid.
• The field pattern outside resembles that of a bar magnet: one end acts as a north pole and the other as a south pole.
• The field inside is $B={\mu }_{0}nI$, where n is the number of turns per unit length. It depends on n, not on the length or radius separately. At the ends it falls to about half this value.

A soft iron core inside a solenoid greatly strengthens the field; this is an electromagnet.

Worked example

A long solenoid has 500 turns wound uniformly over 0.5 m and carries 2 A. (a) Write the field inside it. (b) If the same 500 turns are spread over 1 m, what happens to the field inside? [2 marks]

Try it first, then show the solutionSolution

1. (a) $n=\frac{500}{0.5}=1000m^{-1}$; $B={\mu }_{0}nI=4\pi \times 10^{-7}\times 1000\times 2\approx 2.5\times 10^{-3}T$
2. (b) Now $n=500m^{-1}$, half as many turns per metre, so the field halves to about $1.26\times 10^{-3}T$.

Answer: (a) about $2.5\times 10^{-3}T$, along the axis (b) it halves

Force between two parallel currents and the definition of the ampere

NCERT section 4.8

Wire a (current $I_a$) produces at wire b, a distance d away, a field $B_a=\frac{{\mu }_0{I}_a}{2\pi d}$ perpendicular to b. A length L of b then feels $F_{ba}=I_{b}L{B}_a$, so the force per unit length is

$f=\frac{F}{L}=\frac{{\mu }_0{I}_a{I}_b}{2\pi d}$

By the right-hand rule, parallel currents (same direction) attract and antiparallel currents repel. The forces on the two wires are equal and opposite.

Definition of the ampere: one ampere is the steady current which, flowing in two very long, straight, parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces between them a force of $2\times 10^{-7}N$ per metre of length.

Worked example

Two long parallel wires 4 cm apart carry currents of 5 A and 8 A in the same direction. Find the force per unit length between them and say whether it is attractive or repulsive. [2 marks]

Try it first, then show the solutionSolution

1. $f=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{2\times 10^{-7}\times 5\times 8}{0.04}$
2. $f=\frac{8\times 10^{-6}}{0.04}=2\times 10^{-4}N{m}^{-1}$
3. Currents in the same direction: attractive

Answer: $2\times 10^{-4}N{m}^{-1}$, attractive

Torque on a current loop; the loop as a magnetic dipole

NCERT section 4.9

Rectangular loop (sides a and b, current I) in a uniform field $\vec{B}$, with the area vector at angle θ to $\vec{B}$. Let the two sides of length b be perpendicular to $\vec{B}$. The forces on them are $IbB$ each, equal and opposite, but not along the same line; their perpendicular separation is $a\sin \theta$. The forces on the other two sides are equal, opposite and collinear, and cancel. So the net force is zero and the torque is

$\tau =IbB\times a\sin \theta =IAB\sin \theta$, and for N turns $\vec{\tau }=\vec{m}\times \vec{B}$, with $\vec{m}=NI\vec{A}$

$\vec{m}$ is the magnetic dipole moment (unit A m²), directed along the area vector by the right-hand rule. The torque turns the loop so that $\vec{m}$ lines up with $\vec{B}$: zero torque at θ = 0 (stable) and at θ = 180° (unstable), maximum NIAB at θ = 90° (plane of the loop parallel to B).

A current loop therefore behaves exactly like a small bar magnet (a magnetic dipole): its far field has the same form as that of an electric dipole, with $\vec{p}\to \vec{m}$ and $\frac{1}{{\epsilon }_0}\to {\mu }_0$.

Worked example

A coil of 20 turns and area $0.01m^2$ carries 3 A in a uniform field of 0.5 T. Its normal makes $30^{\circ }$ with the field. Find (a) its magnetic moment, (b) the torque on it, (c) the maximum torque. [2 marks]

Try it first, then show the solutionSolution

1. (a) $m=NIA=20\times 3\times 0.01=0.6A{m}^2$
2. (b) $\tau =mB\sin \theta =0.6\times 0.5\times 0.5=0.15Nm$
3. (c) ${\tau }_{max}=mB=0.3Nm$ (when the plane of the coil is parallel to B)

Answer: (a) $0.6A{m}^2$ (b) 0.15 N m (c) 0.3 N m

Moving coil galvanometer; conversion to ammeter and voltmeter

NCERT section 4.10

Construction and principle. A rectangular coil of N turns, area A, is pivoted between the concave poles of a permanent magnet, with a soft iron cylinder inside it. The curved poles and the iron core make the field radial, so the plane of the coil is always parallel to B and the deflecting torque is always $NIAB$, whatever the deflection. A spring provides a restoring torque $k\varphi$. At equilibrium:

$NIAB=k\varphi \Rightarrow \varphi =\left(\frac{NAB}{k}\right)I$

so the deflection is proportional to the current and the scale is linear. The soft iron core also strengthens the field.

Current sensitivity $\frac{\varphi }{I}=\frac{NAB}{k}$ (deflection per unit current). Voltage sensitivity $\frac{\varphi }{V}=\frac{NAB}{k{R}_G}$. Doubling N doubles the current sensitivity but also doubles the coil's resistance $R_G$, so the voltage sensitivity need not rise.

Ammeter: connect a small shunt S in parallel: $I_{g}G=(I-I_g)S$, so $S=\frac{I_{g}G}{I-I_g}$; the ammeter's resistance is very low. Voltmeter: connect a large resistance R in series: $V=I_g(G+R)$, so $R=\frac{V}{I_g}-G$; the voltmeter's resistance is very high.

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 0-1 A, (b) a voltmeter of range 0-10 V? [3 marks]

Try it first, then show the solutionSolution

1. (a) Shunt in parallel: $S=\frac{I_{g}G}{I-I_g}=\frac{0.002\times 50}{1-0.002}=\frac{0.1}{0.998}$
2. $S\approx 0.1\Omega$ (0.1002 Ω)
3. (b) Series resistance: $R=\frac{V}{I_g}-G=\frac{10}{0.002}-50$
4. $R=5000-50=4950\Omega$

Answer: (a) a shunt of about 0.1 Ω in parallel (b) 4950 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

• Motion of a Charge in a Magnetic Field
• Torque on a Current Loop and Magnetic Dipole
• Moving Coil Galvanometer

Moving Charges and Magnetism formula sheet

Lorentz force

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

Force on a current-carrying wire

$\vec{F}=I\vec{l}\times \vec{B},F=IlB\sin \theta$

Circular motion in B

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

T and ν do not depend on speed.

Helical path

$r=\frac{mv\sin \theta }{qB},\text{pitch}=\frac{2\pi mv\cos \theta }{qB}$

Biot-Savart law

$d\vec{B}=\frac{{\mu }_0}{4\pi }\frac{Id\vec{l}\times \vec{r}}{r^3}$

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

Circular loop: on the axis

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

Circular loop: at the centre

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

Ampere's circuital law

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

Long straight wire

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

Inside a uniform thick wire: $B=\frac{{\mu }_{0}Ir}{2\pi a^2}$.

Long solenoid (result only)

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

n = turns per unit length; qualitative treatment only.

Force between parallel currents

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

Same direction: attract; opposite: repel.

Magnetic moment of a loop

$\vec{m}=NI\vec{A}$

Torque on a loop

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

Galvanometer

$\varphi =\frac{NAB}{k}I,\frac{\varphi }{I}=\frac{NAB}{k},\frac{\varphi }{V}=\frac{NAB}{k{R}_G}$

Shunt for an ammeter

$S=\frac{I_{g}G}{I-I_g}$

Series resistance for a voltmeter

$R=\frac{V}{I_g}-G$

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.

• Right-hand thumb rule, two ways

  Straight wire: thumb along current, fingers curl along B. Loop or solenoid: fingers curl along current, thumb gives B (the north face).

  Use it when: Direction of B for wires, loops and solenoids.

• Fleming's left hand: FBI

  Thumb = Force, First finger = B (field), seCond finger = Current (I). Left hand for motors and forces.

  Use it when: Direction of force on a wire.

• Parallel currents are friends

  Same direction: they hug (attract). Opposite: they push away. The opposite of like charges.

  Use it when: Force between wires, assertion-reason items.

• Shunt Small, Series Big

  Ammeter needs a Small Shunt in parallel; a voltmeter needs a Big resistance in series.

  Use it when: Conversion questions.

• Mag force never works

  Magnetic force is always sideways to the motion, so it does no work: speed and kinetic energy stay the same.

  Use it when: MCQs on energy of a charged particle in B.

• 2 × 10⁻⁷ is your friend

  $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}$ and $\frac{{\mu }_0}{4\pi }=10^{-7}$: use these to avoid π altogether in wire and Biot-Savart problems.

  Use it when: Fast, accurate arithmetic without a calculator.

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 4-mark question split into 1 + 1 + 1 + 1 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: Using $r=mv/qB$ with the full speed when the velocity is at an angle

  Missing the helix.

  Wrong working

  v at 60° to B: $r=\frac{mv}{qB}$

  Right working

  $r=\frac{mv\sin 60^{\circ }}{qB}$; pitch uses $v\cos 60^{\circ }$

  ✓Do this: Split v into parts across and along B before using the circle formula.

  Usually costs: 1-2 marks.

• ✗Mistake: Not showing the cancelling components in the circular-loop derivation

  Jumping to the answer.

  ✓Do this: State that $dB\sin \theta$ components from diametrically opposite elements cancel, and only $dB\cos \theta$ adds up.

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

• ✗Mistake: Forgetting why the field is radial in a galvanometer

  Describing parts without purpose.

  ✓Do this: Write: concave poles and a soft iron core make B radial, so the plane of the coil stays parallel to B, the torque is NIAB at every deflection, and φ ∝ I (linear scale).

  Usually costs: 1 mark.

• ✗Mistake: Adding the shunt in series or the voltmeter resistance in parallel

  Mixing up the two conversions.

  ✓Do this: Ammeter: shunt in PARALLEL (low resistance). Voltmeter: resistance in SERIES (high resistance). Draw the circuit.

  Usually costs: Full marks of the conversion part.

• ✗Mistake: Wrong angle in the torque on a loop

  Using the angle between B and the plane of the coil.

  ✓Do this: In $\tau =mB\sin \theta$, θ is between $\vec{m}$ (the normal) and $\vec{B}$. If the angle with the plane is α, then θ = 90° − α.

  Usually costs: 1 mark.

• ✗Mistake: Deriving the solenoid field with Ampere's law at length

  Older books include it.

  ✓Do this: The 2026-27 syllabus treats the solenoid qualitatively: describe the field pattern and quote $B={\mu }_{0}nI$. Spend derivation time on the loop and the wire.

  Usually costs: Time.

NCERT exercise guide: Moving Charges and Magnetism

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Moving Charges and Magnetism: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.4 marks per paper; 36 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Moving Charges and Magnetism

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Moving Charges and Magnetism question, by year

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, the tesla, is equal to:
   (a) N A m
   (b) N A⁻¹ m⁻¹
   (c) N A⁻¹ m
   (d) N m A⁻¹ s

   Hint

   Use F = IlB.

   Answer

   (b) N A⁻¹ m⁻¹

2. Q2 · 1 mark · MCQBoard level

   A charged particle moves in a circle in a uniform magnetic field. If its speed is doubled and the field is halved, its radius becomes:
   (a) the same
   (b) twice
   (c) four times
   (d) half

   Hint

   $r=mv/qB$.

   Answer

   (c) four times

3. Q3 · 1 mark · Very short answerBoard level

   Define one ampere in terms of the force between two parallel conductors.

   Hint

   1 m apart, $2\times 10^{-7}N$ per metre.

   Answer

   One ampere is the steady current which, maintained in two long, straight, parallel conductors of negligible cross-section 1 m apart in vacuum, produces a force of $2\times 10^{-7}N$ per metre of length between them.

4. Q4 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): A current-carrying loop placed in a uniform magnetic field experiences no net force.
   Reason (R): The forces on opposite sides of the loop are equal and opposite.
   (a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

   Hint

   Add the four side forces as vectors.

   Answer

   (a). The side forces cancel in pairs, so the net force is zero (though there can be a torque).

5. Q5 · 2 marks · Very short answerHarder board level

   Two concentric coplanar circular loops of radii 10 cm and 20 cm carry 4 A each in opposite directions. Find the field at the common centre.

   Hint

   Fields at the centre are opposite.

   Answer

   $B=\frac{{\mu }_{0}I}{2}\left(\frac{1}{0.1}-\frac{1}{0.2}\right)=2\pi \times 10^{-7}\times 4\times 5\approx 1.26\times 10^{-5}T$, in the direction set by the smaller loop.

6. Q6 · 2 marks · Very short answerHarder board level

   Show that the time period of a charged particle moving perpendicular to a uniform magnetic field does not depend on its speed. Find the period for a proton in a 0.5 T field. ($m_p=1.67\times 10^{-27}kg$)

   Hint

   Divide the circumference by the speed.

   Answer

   $T=\frac{2\pi r}{v}=\frac{2\pi }{v}\cdot \frac{mv}{qB}=\frac{2\pi m}{qB}$, with no v. For the proton: $T=\frac{2\pi \times 1.67\times 10^{-27}}{1.6\times 10^{-19}\times 0.5}\approx 1.31\times 10^{-7}s$.

7. Q7 · 2 marks · Very short answerHarder board level

   A wire 0.2 m long lies along the x-axis and carries 3 A in the $+x$ direction in a field $\vec{B}=(0.4\hat{j}+0.3\hat{k})T$. Find the force on it in vector form and its magnitude.

   Hint

   $\hat{i}\times \hat{j}=\hat{k}$, $\hat{i}\times \hat{k}=-\hat{j}$.

   Answer

   $\vec{F}=3\times 0.2\hat{i}\times (0.4\hat{j}+0.3\hat{k})=0.6(0.4\hat{k}-0.3\hat{j})=(-0.18\hat{j}+0.24\hat{k})N$; magnitude $\sqrt{0.18^2+0.24^2}=0.3N$.

8. Q8 · 2 marks · Very short answerHOTS / competency

   A proton and an alpha particle enter the same uniform magnetic field at right angles with the same kinetic energy. Compare the radii of their paths. (Take $m_{\alpha }=4m_p$, $q_{\alpha }=2e$.)

   Hint

   $r=\sqrt{2mK}/qB$.

   Answer

   $r_p=\frac{\sqrt{2m_{p}K}}{eB}$; $r_{\alpha }=\frac{\sqrt{2(4m_p)K}}{2eB}=\frac{2\sqrt{2m_{p}K}}{2eB}=r_p$. The radii are equal.

9. Q9 · 3 marks · Short answerHarder board level

   A galvanometer of resistance 60 Ω gives full-scale deflection for 2 mA. (a) A shunt of 0.12 Ω is connected across it. What is the range of the ammeter formed? (b) What series resistance turns the original galvanometer into a 0-3 V voltmeter, and what is the voltmeter's total resistance? (c) How much more series resistance is needed to change the range to 0-30 V?

   Hint

   Ammeter: $I_{g}G=(I-I_g)S$. Voltmeter: total resistance $=V/I_g$.

   Answer

   (a) $I=\frac{I_g(G+S)}{S}=\frac{0.002\times 60.12}{0.12}\approx 1.0A$, so the range is 0-1 A. (b) $R=\frac{3}{0.002}-60=1440\Omega$ in series; total $1500\Omega$. (c) For 30 V the total must be $\frac{30}{0.002}=15000\Omega$, so add $15000-1500=13500\Omega$ more in series.

10. Q10 · 5 marks · Long answerHarder board level

    (a) Using Ampere's circuital law, derive the magnetic field of an infinitely long straight wire. (b) Two long parallel wires carry currents $I_1$ and $I_2$. Derive the force per unit length between them and state when it is attractive. (c) Two long parallel wires 5 cm apart carry 10 A and 15 A in opposite directions. Find the force on a 2 m length of either wire and say whether it is attractive or repulsive.

    Hint

    Circular Amperian loop; then F = I₂LB₁.

    Answer

    (a) Circular loop of radius r round the wire: $B\cdot 2\pi r={\mu }_{0}I$, so $B=\frac{{\mu }_{0}I}{2\pi r}$. (b) Field of wire 1 at wire 2: $\frac{{\mu }_0{I}_1}{2\pi d}$; force on length L of wire 2: $I_{2}L\frac{{\mu }_0{I}_1}{2\pi d}$, so $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$; attractive when the currents are in the same direction. (c) $F=\frac{2\times 10^{-7}\times 10\times 15\times 2}{0.05}=1.2\times 10^{-3}N$, repulsive because the currents are opposite.

11. Q11 · 3 marks · Short answerHOTS / competency

    A square loop of side 10 cm carrying 2 A lies in the same plane as a long straight wire carrying 10 A. The nearer side of the loop is parallel to the wire and 5 cm from it, with its current in the same direction as the wire's. Find the net force on the loop.

    Hint

    Forces on the two sides perpendicular to the wire cancel.

    Answer

    Near side (5 cm, same direction, attracted): $F_1=\frac{2\times 10^{-7}\times 10\times 2\times 0.1}{0.05}=8\times 10^{-6}N$. Far side (15 cm, opposite direction, repelled): $F_2=\frac{2\times 10^{-7}\times 10\times 2\times 0.1}{0.15}\approx 2.67\times 10^{-6}N$. Net $\approx 5.3\times 10^{-6}N$ towards the wire.

12. Q12 · 2 marks · Very short answerHarder board level

    A long solenoid has n turns per metre and carries current I. (a) Describe the field inside and just outside it. (b) The same wire is rewound so that the total number of turns is unchanged but the length is doubled, and the current is made three times as large. By what factor does the field inside change?

    Hint

    B depends on turns per unit length and on the current.

    Answer

    (a) Inside: strong, uniform, along the axis, $B={\mu }_{0}nI$. Just outside: very weak, nearly zero. (b) Turns per metre halve and the current triples, so $B^{\prime }={\mu }_0\frac{n}{2}(3I)=1.5B$: the field rises by a factor of 1.5.

13. Q13 · 2 marks · Very short answerHarder board level

    A circular coil of 50 turns and radius 4 cm carries 2 A in a uniform field of 0.2 T. The plane of the coil makes $30^{\circ }$ with the field. Find (a) its magnetic moment, (b) the torque on it, (c) the angle the plane must make with the field for the torque to be half its maximum value. Take $\pi =3.14$.

    Hint

    θ in $\tau =mB\sin \theta$ is the angle between $\vec{m}$ (the normal) and $\vec{B}$, not the plane.

    Answer

    (a) $m=NI\pi r^2=50\times 2\times 3.14\times (0.04{)}^2\approx 0.50A{m}^2$. (b) The normal makes $90^{\circ }-30^{\circ }=60^{\circ }$ with B: $\tau =0.50\times 0.2\times \sin 60^{\circ }\approx 0.087Nm$. (c) Half the maximum needs $\sin \theta =0.5$, so the normal is at $30^{\circ }$ to B and the plane makes $60^{\circ }$ with the field.