Welcome to HSLC GURU! This page presents a complete, exam-ready guide to Class 12 Physics Chapter 4 — Moving Charges and Magnetism for the ASSEB (Assam State School Education Board) English medium syllabus. You will find a clear summary, all the key formulas, illustrative SVG diagrams (right-hand rule, circular loop, solenoid, galvanometer), full exercise solutions, additional important questions and a glossary of terms. Use this resource to revise the chapter, practise numerical problems and master the conceptual ideas of magnetic force, Biot-Savart law, Ampere’s circuital law, force between parallel currents, torque on a current loop and the moving coil galvanometer.
Chapter Summary
A current is nothing more than charges in ordered motion, so the magnetic effects of currents and the magnetic forces on moving charges are two faces of the same physics. Hans Christian Oersted (1820) first observed that a current-carrying wire deflects a nearby magnetic compass; this single experiment unified electricity and magnetism and led to the modern theory developed by Ampere, Biot, Savart, Faraday and Maxwell.
A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences the Lorentz magnetic force $\vec{F} = q\vec{v}\times\vec{B}$. The force is perpendicular to both $\vec{v}$ and $\vec{B}$, so it does no work on the charge and only changes its direction of motion. When $\vec{v}\perp\vec{B}$ the charge moves in a circle of radius $r=mv/(qB)$ with cyclotron frequency $f=qB/(2\pi m)$, a result exploited in the cyclotron accelerator and in mass spectrometers. When $\vec{E}$ and $\vec{B}$ act together (Lorentz force) the total force is $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$; choosing $E=vB$ gives a velocity selector that lets only one speed pass undeflected.
For a current-carrying conductor of length $\vec{L}$ placed in a field $\vec{B}$, the force is $\vec{F}=I\vec{L}\times\vec{B}$. The Biot-Savart law gives the magnetic field due to a small current element: $d\vec{B}=(\mu_0/4\pi)(I\,d\vec{l}\times\hat{r}/r^2)$. Integrating it leads to several standard results: a long straight wire produces $B=\mu_0 I/(2\pi r)$; a circular loop of radius $R$ produces $B=\mu_0 I R^2/[2(R^2+x^2)^{3/2}]$ on its axis, reducing to $B=\mu_0 I/(2R)$ at the centre.
Ampere’s circuital law, $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}$, plays a role analogous to Gauss’s law in electrostatics and gives quick results for symmetrical situations: a long solenoid has $B=\mu_0 nI$ inside; a toroid has $B=\mu_0 NI/(2\pi r)$. Two long parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ exert a force per unit length $F/L=\mu_0 I_1I_2/(2\pi d)$ on each other — attractive for parallel currents, repulsive for antiparallel ones. This force defines the SI ampere.
A planar current loop of $N$ turns and area $A$ in a uniform field $\vec{B}$ behaves like a magnetic dipole of moment $\vec{m}=NI\vec{A}$ and experiences a torque $\vec{\tau}=\vec{m}\times\vec{B}$. This is the working principle of the moving coil galvanometer in which the deflection $\theta$ is proportional to the current: $I=(k/NBA)\theta$. Adding a low shunt converts it to an ammeter; adding a high series resistance converts it to a voltmeter.
Key Formulas at a Glance
| Quantity | Formula | Remarks |
|---|---|---|
| Lorentz force | $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ | Total electromagnetic force |
| Magnetic force on charge | $\vec{F}=q\vec{v}\times\vec{B}$ | Perpendicular to $\vec{v}$, no work |
| Force on conductor | $\vec{F}=I\vec{L}\times\vec{B}$ | Current carrying wire in field |
| Radius of circular motion | $r=mv/(qB)$ | $\vec{v}\perp\vec{B}$ |
| Cyclotron frequency | $f=qB/(2\pi m)$ | Independent of $v$ and $r$ |
| Velocity selector | $v=E/B$ | $\vec{E},\vec{B},\vec{v}$ mutually perpendicular |
| Biot-Savart law | $d\vec{B}=(\mu_0/4\pi)(I\,d\vec{l}\times\hat{r}/r^2)$ | Field of a current element |
| Long straight wire | $B=\mu_0 I/(2\pi r)$ | Concentric circular field lines |
| Centre of circular loop | $B=\mu_0 I/(2R)$ | $N$ turns gives $\mu_0 NI/(2R)$ |
| Axis of circular loop | $B=\mu_0 IR^2/[2(R^2+x^2)^{3/2}]$ | $x$ is distance from centre |
| Ampere’s law | $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}$ | Use closed Amperian loops |
| Solenoid (inside) | $B=\mu_0 nI$ | $n=$ turns per unit length |
| Toroid | $B=\mu_0 NI/(2\pi r)$ | Field uniform inside core |
| Force between parallel wires | $F/L=\mu_0 I_1 I_2/(2\pi d)$ | Defines the ampere |
| Magnetic moment of loop | $m=NIA$ | Direction by right-hand rule |
| Torque on loop | $\vec{\tau}=\vec{m}\times\vec{B}$ | $\tau=NIAB\sin\theta$ |
| Galvanometer | $I=(k/NBA)\theta$ | Linear current-deflection relation |
| Permeability of free space | $\mu_0=4\pi\times10^{-7}\;\text{T m A}^{-1}$ | Fundamental constant |
Visual Aids (SVG Diagrams)
1. Right-Hand Rule (force on a positive charge). The thumb shows $\vec{v}$, fingers $\vec{B}$, palm pushes in the direction of $\vec{F}=q\vec{v}\times\vec{B}$.
2. Magnetic Field on the Axis of a Circular Loop. The field at point $P$ on the axis is $B=\mu_0 IR^2/[2(R^2+x^2)^{3/2}]$ directed along the axis.
3. Solenoid. Inside a long solenoid the field is uniform, axial and equal to $B=\mu_0 nI$; outside it is nearly zero.
4. Moving Coil Galvanometer. A rectangular coil suspended in a radial magnetic field experiences a torque $\tau=NIAB$ when current flows; the deflection of the attached pointer measures the current.
NCERT / ASSEB Exercise Solutions
Q1. 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?
Answer: Using $B=\mu_0 NI/(2R)$ with $N=100$, $I=0.40$ A and $R=0.08$ m,
$$B=\frac{4\pi\times10^{-7}\times100\times0.40}{2\times0.08}=3.14\times10^{-4}\;\text{T}$$
Q2. 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?
Answer: $B=\mu_0 I/(2\pi r)$ with $I=35$ A and $r=0.20$ m,
$$B=\frac{4\pi\times10^{-7}\times35}{2\pi\times0.20}=3.5\times10^{-5}\;\text{T}$$
Q3. A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. Give the magnitude and direction of $\vec{B}$ at a point 2.5 m east of the wire.
Answer: $B=\mu_0 I/(2\pi r)=\dfrac{4\pi\times10^{-7}\times50}{2\pi\times2.5}=4.0\times10^{-6}$ T. By the right-hand rule, with current flowing N to S, the field at the eastern point is directed vertically upwards.
Q4. A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Answer: $B=\mu_0 I/(2\pi r)=\dfrac{4\pi\times10^{-7}\times90}{2\pi\times1.5}=1.2\times10^{-5}$ T. The field at the point 1.5 m below the wire is directed towards the south (right-hand rule with current flowing E to W).
Q5. What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of $30^\circ$ with the direction of a uniform magnetic field of 0.15 T?
Answer: $F/L=BI\sin\theta=0.15\times8\times\sin30^\circ=0.6$ N m$^{-1}$.
Q6. A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Answer: $F=BIL\sin\theta=0.27\times10\times0.03\times\sin90^\circ=8.1\times10^{-2}$ N. The force is perpendicular to both the wire and the solenoid axis.
Q7. Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Answer: $F/L=\mu_0 I_1 I_2/(2\pi d)=\dfrac{4\pi\times10^{-7}\times8\times5}{2\pi\times0.04}=2.0\times10^{-4}$ N m$^{-1}$. For a 10 cm length, $F=2.0\times10^{-5}$ N. Since the currents are parallel, the force is attractive.
Q8. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of $B$ inside the solenoid near its centre.
Answer: Total turns $N=5\times400=2000$; length $L=0.8$ m; $n=N/L=2500$ m$^{-1}$. $B=\mu_0 nI=4\pi\times10^{-7}\times2500\times8=2.51\times10^{-2}$ T.
Q9. A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^\circ$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Answer: $A=(0.10)^2=0.01$ m$^2$. $\tau=NIAB\sin\theta=20\times12\times0.01\times0.80\times\sin30^\circ=0.96$ N m.
Q10. Two moving coil meters $M_1$ and $M_2$ have the following particulars: $R_1=10\;\Omega$, $N_1=30$, $A_1=3.6\times10^{-3}$ m$^2$, $B_1=0.25$ T; $R_2=14\;\Omega$, $N_2=42$, $A_2=1.8\times10^{-3}$ m$^2$, $B_2=0.50$ T. Spring constants are the same for both. Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_2$ to $M_1$.
Answer: Current sensitivity $S_I=NBA/k$.
$$\frac{S_{I,2}}{S_{I,1}}=\frac{N_2 B_2 A_2}{N_1 B_1 A_1}=\frac{42\times0.50\times1.8\times10^{-3}}{30\times0.25\times3.6\times10^{-3}}=1.4$$
Voltage sensitivity $S_V=S_I/R$.
$$\frac{S_{V,2}}{S_{V,1}}=\frac{S_{I,2}/R_2}{S_{I,1}/R_1}=1.4\times\frac{10}{14}=1.0$$
Q11. In a chamber, a uniform magnetic field of 6.5 G ($1\;\text{G}=10^{-4}$ T) is maintained. An electron is shot into the field with a speed of $4.8\times10^6$ m s$^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. ($e=1.6\times10^{-19}$ C, $m_e=9.1\times10^{-31}$ kg.)
Answer: The magnetic force $\vec{F}=q\vec{v}\times\vec{B}$ is always perpendicular to the velocity, so it provides centripetal acceleration without changing the speed; therefore the path is a circle. From $qvB=mv^2/r$,
$$r=\frac{mv}{qB}=\frac{9.1\times10^{-31}\times4.8\times10^6}{1.6\times10^{-19}\times6.5\times10^{-4}}=4.2\times10^{-2}\;\text{m}=4.2\;\text{cm}$$
Q12. In Exercise 11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron?
Answer: $f=qB/(2\pi m)=\dfrac{1.6\times10^{-19}\times6.5\times10^{-4}}{2\pi\times9.1\times10^{-31}}\approx 1.82\times10^7$ Hz $=18.2$ MHz. The frequency is independent of the electron’s speed; this is the principle behind the cyclotron.
Q13. (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of 1.0 T. The field lines make an angle of $60^\circ$ with the normal to the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change if the circular coil were replaced by a planar coil of some irregular shape that encloses the same area?
Answer: (a) $A=\pi R^2=\pi(0.08)^2=0.0201$ m$^2$. $\tau=NIAB\sin\theta=30\times6\times0.0201\times1\times\sin60^\circ=3.13$ N m. (b) The torque depends only on $NIA$ and the field, not on the shape of the loop. So the answer would be the same for a coil of any shape enclosing the same area.
Additional Important Questions
Q14. State and explain Biot-Savart’s law. Use it to derive the magnetic field at the centre of a circular current loop.
Answer: Biot-Savart’s law states that the magnetic field $d\vec{B}$ produced at a point by a small current element $I\,d\vec{l}$ at distance $r$ is
$$d\vec{B}=\frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\hat{r}}{r^2}$$
For a circular loop of radius $R$ carrying current $I$, every element $d\vec{l}$ on the loop is perpendicular to the radius vector to the centre, so $|d\vec{l}\times\hat{r}|=dl$. The fields from all elements add up along the axis (here the perpendicular to the plane through the centre):
$$B=\oint\frac{\mu_0}{4\pi}\frac{I\,dl}{R^2}=\frac{\mu_0 I}{4\pi R^2}(2\pi R)=\frac{\mu_0 I}{2R}$$
Q15. Derive an expression for the magnetic field on the axis of a circular current loop.
Answer: Consider a loop of radius $R$ in the $yz$-plane carrying current $I$. A point $P$ on the axis is at distance $x$ from the centre. The distance from any element to $P$ is $r=\sqrt{R^2+x^2}$. Each $d\vec{B}$ has magnitude $(\mu_0/4\pi)(I\,dl/r^2)$ and is perpendicular to $\vec{r}$; the components perpendicular to the axis cancel by symmetry, while the axial components add. The axial component of $d\vec{B}$ is $dB\cos\alpha=dB\cdot R/r$. Integrating around the loop ($\oint dl=2\pi R$):
$$B=\frac{\mu_0 I}{4\pi}\cdot\frac{R}{r^3}\cdot 2\pi R=\frac{\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$$
Q16. State Ampere’s circuital law. Using it, derive the magnetic field inside a long current-carrying solenoid.
Answer: Ampere’s circuital law: $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}$. For a long solenoid with $n$ turns per unit length carrying current $I$, choose a rectangular Amperian loop of length $L$ with one side inside the solenoid (parallel to its axis) and the opposite side far outside. Outside, $B\approx 0$; the two short sides are perpendicular to $\vec{B}$ and contribute nothing. Hence $\oint\vec{B}\cdot d\vec{l}=BL$. The current enclosed is $nLI$ so $BL=\mu_0 nLI$, giving
$$B=\mu_0 nI$$
Q17. Two long parallel wires carry currents $I_1$ and $I_2$ separated by distance $d$. Derive the force per unit length between them. Use this to define the SI unit of current.
Answer: Wire 1 produces a field $B_1=\mu_0 I_1/(2\pi d)$ at the location of wire 2. The force on a length $L$ of wire 2 is $F_2=B_1 I_2 L$, so
$$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}$$
Parallel currents attract; antiparallel currents repel. Definition of the ampere: when two infinitely long parallel straight wires of negligible cross-section, separated by 1 m in vacuum, carry equal currents and the force per unit length between them is $2\times10^{-7}$ N m$^{-1}$, the current in each wire is 1 ampere.
Q18. Derive an expression for the torque on a current-carrying rectangular loop placed in a uniform magnetic field.
Answer: Consider a rectangular loop of sides $a$ and $b$ carrying current $I$ in a uniform field $\vec{B}$. The two sides of length $a$ parallel to $\vec{B}$ experience equal and opposite forces along the same line (no torque). The sides of length $b$ experience forces $F=BIb$ in opposite directions; these forces form a couple. If the normal to the plane of the loop makes angle $\theta$ with $\vec{B}$, the perpendicular distance between the two forces is $a\sin\theta$, so the torque is $\tau=BIb\cdot a\sin\theta=BIA\sin\theta$. For $N$ turns:
$$\vec{\tau}=N I \vec{A}\times\vec{B}=\vec{m}\times\vec{B},\quad \tau=NIAB\sin\theta$$
Q19. Describe the working of a cyclotron. Derive expressions for the cyclotron frequency and the maximum kinetic energy of the accelerated particle.
Answer: A cyclotron consists of two semicircular hollow metal “dees” placed in a uniform magnetic field $\vec{B}$ perpendicular to the plane of the dees, with an alternating high-frequency potential between them. A charged particle injected near the centre is accelerated across the gap, traces a semicircle inside one dee, returns to the gap when the polarity has reversed, is again accelerated and so on, spiralling outward. Inside each dee the magnetic force provides centripetal force: $qvB=mv^2/r$ giving $r=mv/(qB)$. The time to traverse a semicircle is $T/2=\pi m/(qB)$ and is independent of $v$ and $r$, so the frequency $f=qB/(2\pi m)$ matches the applied AC. Maximum KE at radius $R$:
$$K_{\max}=\frac{1}{2}mv^2=\frac{q^2B^2R^2}{2m}$$
Q20. Explain the principle, construction and working of a moving coil galvanometer. Define current sensitivity and voltage sensitivity.
Answer: Principle: When a current flows through a rectangular coil suspended in a magnetic field, the coil experiences a torque proportional to the current. Construction: A rectangular coil of fine insulated wire wound on a non-magnetic frame is suspended between the cylindrical concave pole pieces of a permanent magnet by a phosphor-bronze strip, with a soft iron cylinder inside the coil to make the field radial. Working: When current $I$ passes through the coil, the deflecting torque $\tau=NIAB$ is balanced by the restoring torque $k\theta$, so $I=(k/NBA)\theta$, i.e. $\theta\propto I$. Current sensitivity: $S_I=\theta/I=NBA/k$. Voltage sensitivity: $S_V=\theta/V=NBA/(kR)$.
Q21. How can a galvanometer be converted into (a) an ammeter and (b) a voltmeter?
Answer: (a) Ammeter: connect a small resistance (shunt) $S$ in parallel with the galvanometer so that most of the current bypasses it. If $I_g$ is the full-scale current of the galvanometer of resistance $G$, the shunt for full-scale current $I$ is $S=I_g G/(I-I_g)$. (b) Voltmeter: connect a high resistance $R$ in series so that for full-scale voltage $V$, $R=V/I_g – G$.
Q22. A proton and an alpha particle, both moving with the same speed, enter a uniform magnetic field perpendicular to the field. Compare the radii of their circular paths.
Answer: $r=mv/(qB)$. For proton: $r_p=m_p v/(eB)$. For alpha: $r_\alpha=4m_p v/(2eB)=2m_p v/(eB)$. Therefore $r_\alpha/r_p=2$.
Q23. A current of 1 A flows through a long solenoid having 1000 turns per metre. Find the magnetic field at the centre.
Answer: $B=\mu_0 nI=4\pi\times10^{-7}\times1000\times1=1.26\times10^{-3}$ T.
Q24. Explain the working of a velocity selector.
Answer: Crossed electric and magnetic fields are arranged perpendicular to the velocity of the entering charged particles. The electric force $qE$ and magnetic force $qvB$ act in opposite directions. For particles whose speed satisfies $qE=qvB$, i.e. $v=E/B$, the net force is zero and they pass undeflected; particles with other speeds are deflected and removed. The device thus selects a single velocity from a beam of mixed speeds.
Q25. A toroid has 600 turns wound on a non-magnetic core, mean radius 15 cm, carrying a current of 4 A. Find the magnetic field inside the core.
Answer: $B=\mu_0 NI/(2\pi r)=\dfrac{4\pi\times10^{-7}\times600\times4}{2\pi\times0.15}=3.2\times10^{-3}$ T.
Q26. Show that the magnetic dipole moment of a revolving electron is $m=evr/2$, where $v$ is its speed and $r$ the radius of its circular orbit.
Answer: An electron of charge $e$ moving in a circle of radius $r$ at speed $v$ constitutes a current $I=e/T=ev/(2\pi r)$, where $T=2\pi r/v$ is the orbital period. The magnetic moment of the orbital current loop is
$$m=IA=\frac{ev}{2\pi r}\cdot \pi r^2=\frac{evr}{2}$$
Q27. Distinguish between current sensitivity and voltage sensitivity of a galvanometer. Why is increasing the number of turns not always a good way to increase voltage sensitivity?
Answer: Current sensitivity $S_I=NBA/k$; voltage sensitivity $S_V=NBA/(kR)$. Increasing $N$ increases $S_I$ but it also increases the resistance $R$ of the coil, so $S_V$ may not change appreciably. Hence increasing $N$ is not always beneficial for voltage sensitivity.
Q28. Why does a moving charge experience no force when it moves parallel to the magnetic field?
Answer: The magnetic force is $\vec{F}=q\vec{v}\times\vec{B}$, with magnitude $qvB\sin\theta$. When $\vec{v}\parallel\vec{B}$, $\theta=0$, so $\sin\theta=0$ and $F=0$.
Q29. State two factors on which the strength of the magnetic field at the centre of a current-carrying coil depends.
Answer: $B=\mu_0 NI/(2R)$, so $B$ depends on (i) the current $I$ in the coil, (ii) the number of turns $N$ and the radius $R$ of the coil; it is directly proportional to $NI$ and inversely proportional to $R$.
Q30. A galvanometer of resistance 50 ohm gives full-scale deflection at 4 mA. How can it be converted to an ammeter of range 0-3 A?
Answer: Required shunt $S=\dfrac{I_g G}{I-I_g}=\dfrac{4\times10^{-3}\times50}{3-4\times10^{-3}}\approx0.067\;\Omega$, connected in parallel with the galvanometer.
Q31. The same galvanometer is to be converted into a voltmeter of range 0-100 V. Find the series resistance required.
Answer: $R=V/I_g – G=100/(4\times10^{-3})-50=25000-50=24950\;\Omega$, connected in series.
Q32. Why is the magnetic field inside a toroid uniform but zero outside?
Answer: Applying Ampere’s law to a circular Amperian loop of radius $r$ inside the toroid gives $B(2\pi r)=\mu_0 NI$, so $B=\mu_0 NI/(2\pi r)$, which is uniform along any one circular path. Loops drawn outside the toroid enclose no net current ($I_{\text{enc}}=0$), so $B=0$ outside.
Q33. A particle of charge $1.6\times10^{-19}$ C and mass $1.67\times10^{-27}$ kg moves in a circle of radius 0.25 m perpendicular to a magnetic field of 0.50 T. Find its kinetic energy.
Answer: $K=q^2B^2r^2/(2m)$.
$$K=\frac{(1.6\times10^{-19})^2(0.5)^2(0.25)^2}{2\times1.67\times10^{-27}}\approx1.20\times10^{-13}\;\text{J}$$
Q34. State two limitations of the cyclotron.
Answer: (i) It cannot accelerate uncharged particles such as neutrons. (ii) At very high speeds, the relativistic increase in mass changes the cyclotron frequency, so synchronisation with the AC voltage fails; electrons (very light) cannot be accelerated efficiently.
Q35. Define magnetic dipole moment of a current loop. Write its SI unit.
Answer: The magnetic dipole moment of a current loop is $\vec{m}=NI\vec{A}$, with magnitude $NIA$ and direction given by the right-hand rule (along the area vector). SI unit: ampere metre$^2$ (A m$^2$).
Q36. A wire of length 0.5 m carries a current of 4 A and is placed in a uniform magnetic field of 0.20 T. The wire makes an angle of $45^\circ$ with the field. Calculate the force on the wire.
Answer: $F=BIL\sin\theta=0.20\times4\times0.5\times\sin45^\circ=0.20\times4\times0.5\times0.707=0.283$ N.
Q37. A circular coil of radius 10 cm has 200 turns and carries a current of 5 A. Calculate the magnetic moment of the coil.
Answer: $A=\pi R^2=\pi(0.1)^2=3.14\times10^{-2}$ m$^2$. $m=NIA=200\times5\times3.14\times10^{-2}=31.4$ A m$^2$.
Q38. Show that the kinetic energy of a charged particle is unchanged by a static magnetic field.
Answer: The magnetic force is $\vec{F}=q\vec{v}\times\vec{B}$, which is always perpendicular to $\vec{v}$. The work done is
$$W=\int\vec{F}\cdot d\vec{r}=\int\vec{F}\cdot\vec{v}\,dt=0$$
Since no work is done on the particle, by the work-energy theorem its kinetic energy and hence its speed remain constant.
Q39. State the principle of superposition for magnetic fields. Two long parallel wires carry currents 4 A and 6 A in opposite directions, separated by 10 cm. Find the field at the midpoint between them.
Answer: Principle of superposition: the resultant magnetic field at any point is the vector sum of the fields produced by the individual currents. At the midpoint (distance 5 cm from each wire), since the currents are antiparallel, the fields due to both wires point in the same direction and add. $B_1=\mu_0\cdot 4/(2\pi\cdot 0.05)=1.6\times10^{-5}$ T; $B_2=\mu_0\cdot 6/(2\pi\cdot 0.05)=2.4\times10^{-5}$ T. Net field $B=B_1+B_2=4.0\times10^{-5}$ T.
Q40. Two concentric coplanar circular loops of radii 10 cm and 20 cm carry currents of 5 A and 10 A respectively in opposite directions. Find the net magnetic field at the common centre.
Answer: $B_1=\mu_0 I_1/(2R_1)=\mu_0\cdot 5/(2\cdot 0.10)=25\mu_0$ (units), $B_2=\mu_0 I_2/(2R_2)=\mu_0\cdot 10/(2\cdot 0.20)=25\mu_0$. Since the currents are opposite, the two fields point in opposite directions and the net field is zero.
Q41. A 100-turn coil of area $A=0.10$ m$^2$ in a 0.50 T field is rotated from its position of maximum flux to the position where its plane is parallel to the field. Calculate the work done if the current in the coil is 2.0 A.
Answer: Initial PE: $U_i=-mB\cos 0=-NIAB=-100\times2\times0.10\times0.50=-10$ J. Final PE: $U_f=-mB\cos 90^\circ=0$. Work done by external agent $=U_f-U_i=0-(-10)=10$ J.
Q42. Define the SI unit of magnetic field. State the dimensional formula of $\vec{B}$.
Answer: The SI unit of magnetic field is the tesla (T). One tesla is the magnetic field that produces a force of 1 N on a 1 m long conductor carrying 1 A current placed perpendicular to the field. Dimensional formula: $[B]=[M T^{-2} A^{-1}]$.
Q43. A proton enters a magnetic field of 1.0 T with a velocity of $5\times10^7$ m s$^{-1}$ at $30^\circ$ to the field. Describe its motion and find the radius of the helical path. ($m_p=1.67\times10^{-27}$ kg.)
Answer: The component $v_\parallel=v\cos 30^\circ=4.33\times10^7$ m s$^{-1}$ is unaffected by the field; the perpendicular component $v_\perp=v\sin 30^\circ=2.5\times10^7$ m s$^{-1}$ produces circular motion. Hence the proton spirals along the field in a helical path. Radius:
$$r=\frac{mv_\perp}{qB}=\frac{1.67\times10^{-27}\times2.5\times10^7}{1.6\times10^{-19}\times1.0}=0.261\;\text{m}$$
Q44. State three differences between an electric field and a magnetic field.
Answer: (i) An electric field is produced by static or moving charges; a magnetic field is produced only by moving charges or by changing electric fields. (ii) Electric force acts on a charge whether it moves or not; magnetic force acts only on a moving charge. (iii) Electric force can change the kinetic energy of a charge (does work); magnetic force is always perpendicular to velocity and does no work.
Q45. Why does a current-carrying conductor produce a magnetic field?
Answer: A current consists of moving charges. Every moving charge produces a magnetic field around itself (Biot-Savart law). The combined contributions of all moving charges in the wire give a net magnetic field surrounding the conductor.
Q46. A solenoid has 1200 turns wound on a 60 cm long tube of diameter 2 cm. If a current of 2 A flows through it, find $B$ near the centre. Compare with the field at the end.
Answer: $n=1200/0.6=2000$ m$^{-1}$. $B_{\text{centre}}=\mu_0 nI=4\pi\times10^{-7}\times2000\times2=5.03\times10^{-3}$ T. At the end, $B_{\text{end}}=\mu_0 nI/2=2.51\times10^{-3}$ T (half of the centre value for an infinitely long solenoid approximation).
Q47. A galvanometer of resistance 100 ohm gives full-scale deflection at 1 mA. (a) Convert it to an ammeter of range 0-10 A. (b) Convert it to a voltmeter of range 0-50 V.
Answer: (a) $S=I_g G/(I-I_g)=10^{-3}\times100/(10-10^{-3})\approx 0.01\;\Omega$ in parallel. (b) $R=V/I_g – G=50/10^{-3}-100=49900\;\Omega$ in series.
Q48. State why the field inside a current-carrying solenoid is independent of position along the axis (in the central region).
Answer: By Ampere’s law applied to a rectangular loop spanning the inside, the line integral $\oint\vec{B}\cdot d\vec{l}$ depends only on the enclosed current per unit length and not on where the loop is placed inside the solenoid. So the field has the same magnitude $\mu_0 nI$ and same axial direction everywhere in the central region.
Q49. The current sensitivity of a galvanometer is increased by 25% when its resistance is increased by a factor of 2. By what percentage does its voltage sensitivity change?
Answer: $S_V=S_I/R$. New $S_V’=(1.25 S_I)/(2R)=0.625 S_I/R=0.625 S_V$. Voltage sensitivity decreases by 37.5%.
Q50. Calculate the cyclotron radius for a proton with kinetic energy 5 MeV in a magnetic field of 1.5 T.
Answer: $K=5\times1.6\times10^{-13}=8\times10^{-13}$ J. $v=\sqrt{2K/m}=\sqrt{2\times8\times10^{-13}/1.67\times10^{-27}}=3.1\times10^7$ m s$^{-1}$. $r=mv/(qB)=1.67\times10^{-27}\times3.1\times10^7/(1.6\times10^{-19}\times1.5)=0.216$ m.
Q51. Define magnetic flux density. How is it different from magnetic flux?
Answer: Magnetic flux density $\vec{B}$ is the magnetic field per unit area, measured in tesla (T). Magnetic flux $\Phi=\int\vec{B}\cdot d\vec{A}$ is the surface integral of $\vec{B}$ over an area, measured in weber (Wb). Flux is a scalar; flux density is a vector field.
Q52. State the right-hand thumb rule for finding the direction of the magnetic field due to a straight current-carrying conductor.
Answer: Hold the wire in the right hand with the thumb pointing in the direction of conventional current. The curl of the four fingers gives the direction of the magnetic field lines, which form concentric circles around the wire.
Q53. A current-carrying loop is placed in a non-uniform magnetic field. Will it experience a net force? Will it experience a torque?
Answer: In a non-uniform field the loop experiences both a net force and a torque (in general). In a uniform field, however, the net force is zero, but the torque $\vec{\tau}=\vec{m}\times\vec{B}$ may be non-zero.
Q54. State Fleming’s left-hand rule and explain its use.
Answer: Stretch the thumb, forefinger and middle finger of the left hand mutually perpendicular. The forefinger points in the direction of magnetic field $\vec{B}$, the middle finger in the direction of current $\vec{I}$ and the thumb gives the direction of force on the conductor. It is used to find the direction of force on a current-carrying conductor in a magnetic field (e.g. in motors).
Q55. The plane of a circular coil is parallel to a uniform magnetic field. Find the torque on the coil if it has 50 turns of radius 4 cm and carries a current of 3 A in a 0.10 T field.
Answer: When the plane is parallel to $\vec{B}$, the area vector is perpendicular to $\vec{B}$, so $\theta=90^\circ$ and torque is maximum. $A=\pi(0.04)^2=5.03\times10^{-3}$ m$^2$. $\tau=NIAB=50\times3\times5.03\times10^{-3}\times0.10=7.54\times10^{-2}$ N m.
Q56. An electron with kinetic energy 100 eV moves in a circle of radius 10 cm in a uniform magnetic field. Find the magnitude of the field. ($m_e=9.1\times10^{-31}$ kg, $e=1.6\times10^{-19}$ C.)
Answer: $K=eV=100\times1.6\times10^{-19}=1.6\times10^{-17}$ J. $v=\sqrt{2K/m}=\sqrt{2\times1.6\times10^{-17}/9.1\times10^{-31}}=5.93\times10^6$ m s$^{-1}$. $B=mv/(er)=9.1\times10^{-31}\times5.93\times10^6/(1.6\times10^{-19}\times0.10)=3.37\times10^{-4}$ T.
Q57. A current of 10 A is flowing in a wire of length 1.5 m. A force of 15 N acts on it when placed in a uniform magnetic field. What is the magnetic flux density if the wire is perpendicular to the field?
Answer: $F=BIL$ giving $B=F/(IL)=15/(10\times1.5)=1.0$ T.
Q58. Why are soft iron pieces used in galvanometers?
Answer: A cylindrical soft iron core is placed inside the suspended coil to make the magnetic field radial. Soft iron has very high permeability and low retentivity, so the field lines pass through it and emerge perpendicular to the coil sides at all positions. As a result, the angle between $\vec{m}$ and $\vec{B}$ is always $90^\circ$ and the torque becomes $\tau=NIAB$, independent of orientation, giving a linear $\theta$-$I$ relation.
Q59. A circular coil of 0.20 m radius has 100 turns and is placed perpendicular to a uniform magnetic field of 0.50 T. The coil carries a current of 4 A. Find (a) the magnetic moment, (b) torque on the coil, (c) potential energy in this orientation.
Answer: (a) $A=\pi(0.2)^2=0.1257$ m$^2$. $m=NIA=100\times4\times0.1257=50.27$ A m$^2$. (b) Coil perpendicular to $\vec{B}$ means $\vec{m}\parallel\vec{B}$, so $\theta=0$ and $\tau=mB\sin 0=0$. (c) $U=-mB\cos 0=-50.27\times0.50=-25.13$ J (minimum, stable orientation).
Q60. State the basic principle of an electric motor.
Answer: A current-carrying coil placed in a magnetic field experiences a torque $\vec{\tau}=\vec{m}\times\vec{B}$ that rotates it. By using a commutator to reverse the current after every half rotation, continuous unidirectional rotation is obtained — this is the principle of a DC motor.
Q61. Show that a current loop of magnetic moment $\vec{m}$ in a uniform field $\vec{B}$ has potential energy $U=-\vec{m}\cdot\vec{B}$.
Answer: Work done in rotating the dipole from angle $\theta_1$ to $\theta_2$ against the torque $\tau=mB\sin\theta$ is
$$W=\int_{\theta_1}^{\theta_2}mB\sin\theta\,d\theta = -mB(\cos\theta_2-\cos\theta_1)$$
Taking $U=0$ at $\theta_1=90^\circ$, the potential energy in the orientation $\theta=\theta_2$ is $U=-mB\cos\theta=-\vec{m}\cdot\vec{B}$.
Q62. A long wire bent into a semicircle of radius $R$ carries current $I$. Find the magnetic field at the centre of the semicircle.
Answer: The full circular loop would give $B=\mu_0 I/(2R)$ at the centre. A semicircle gives half of this, plus the contribution from the two long straight portions of the wire, which is zero at the centre since $d\vec{l}\times\hat{r}=0$ along the line. Therefore $B=\mu_0 I/(4R)$, perpendicular to the plane of the semicircle.
Q63. Why is current sensitivity not the same as voltage sensitivity?
Answer: Current sensitivity ($\theta/I$) measures how much the galvanometer deflects per unit current; voltage sensitivity ($\theta/V$) measures deflection per unit applied voltage. Since $V=IR$, $S_V=S_I/R$. The two depend differently on the coil resistance: increasing $N$ raises both $S_I$ (through $NBA$) and $R$ (more wire), so $S_V$ may not increase proportionally.
Q64. State two applications of the cyclotron.
Answer: (i) Production of high-energy charged particles for nuclear reactions and the study of nuclear structure. (ii) Production of medical isotopes (e.g. for PET imaging) and radiotherapy in medicine.
Q65. State why a beam of charged particles cannot be focussed by a magnetic lens alone the way light is by a glass lens.
Answer: The magnetic force depends on the velocity of the charges, so charged particles of different speeds are deflected by different amounts even in the same field. A magnetic lens has chromatic aberration analogous to (but more severe than) ordinary optics. In practice charged-particle beams are focussed by combinations of magnetic and electric fields (e.g. solenoidal lenses, quadrupoles).
Multiple Choice Questions (MCQs)
MCQ 1. The magnetic force on a charged particle moving in a magnetic field is always: (a) parallel to velocity (b) parallel to field (c) perpendicular to velocity (d) along the displacement.
Answer: (c) perpendicular to velocity. From $\vec{F}=q\vec{v}\times\vec{B}$, the force is always perpendicular to $\vec{v}$.
MCQ 2. The SI unit of magnetic dipole moment is: (a) Wb (b) A m$^2$ (c) T m$^2$ (d) N m T$^{-1}$.
Answer: (b) A m$^2$. Also (d) is dimensionally equivalent.
MCQ 3. The cyclotron frequency depends on: (a) the speed of the particle (b) the radius of the path (c) only $q/m$ ratio and $B$ (d) the kinetic energy.
Answer: (c). $f=qB/(2\pi m)$, independent of $v$ and $r$.
MCQ 4. Two long parallel wires carry equal currents in opposite directions. The force per unit length between them is: (a) attractive (b) repulsive (c) zero (d) sometimes attractive sometimes repulsive.
Answer: (b) repulsive. Antiparallel currents repel.
MCQ 5. The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is: (a) $\mu_0 I/(2\pi R)$ (b) $\mu_0 I/(2R)$ (c) $\mu_0 I/(4\pi R)$ (d) $\mu_0 I R$.
Answer: (b) $\mu_0 I/(2R)$.
MCQ 6. The torque on a current loop placed in a uniform magnetic field is maximum when the: (a) plane of the loop is parallel to $\vec{B}$ (b) plane of the loop is perpendicular to $\vec{B}$ (c) angle between $\vec{m}$ and $\vec{B}$ is $0$ (d) angle between $\vec{m}$ and $\vec{B}$ is $180^\circ$.
Answer: (a). When the plane is parallel to $\vec{B}$, the angle between $\vec{m}$ (perpendicular to plane) and $\vec{B}$ is $90^\circ$, giving maximum $\sin\theta$.
MCQ 7. A long solenoid carries a current $I$ and has $n$ turns per unit length. If both $I$ and $n$ are doubled, the new field inside is: (a) unchanged (b) doubled (c) tripled (d) quadrupled.
Answer: (d) quadrupled. $B=\mu_0 nI \to 4B$.
MCQ 8. A galvanometer can be converted into an ammeter by connecting: (a) high resistance in parallel (b) high resistance in series (c) low resistance in parallel (d) low resistance in series.
Answer: (c) low resistance (shunt) in parallel.
MCQ 9. The magnetic field due to a long straight wire varies as: (a) $1/r$ (b) $1/r^2$ (c) $r$ (d) constant.
Answer: (a) $1/r$. $B=\mu_0 I/(2\pi r)$.
MCQ 10. The radius of the circular path of a charged particle in a magnetic field is doubled when: (a) speed is doubled (b) field is doubled (c) charge is doubled (d) mass is halved.
Answer: (a) speed is doubled. $r=mv/(qB)\propto v$.
MCQ 11. The magnetic field at the centre of a long solenoid is: (a) zero (b) infinite (c) uniform (d) decreases with radius.
Answer: (c) uniform.
MCQ 12. Two coils of radii $R$ and $2R$ carry equal currents. The ratio of fields at their centres ($B_1$:$B_2$) is: (a) 1:1 (b) 2:1 (c) 1:2 (d) 4:1.
Answer: (b) 2:1. $B\propto 1/R$.
Conceptual Short Answer Questions
C1. Can a magnetic field set a charged particle at rest into motion? Explain.
Answer: No. The magnetic force on a stationary charge is $\vec{F}=q\vec{v}\times\vec{B}=0$ since $\vec{v}=0$. A static magnetic field cannot accelerate a charge from rest. (However, a changing magnetic field induces an electric field, which can accelerate a charge.)
C2. Two identical charged particles enter a region of uniform magnetic field with the same speed but in different directions. Will they take the same time to complete one revolution?
Answer: Yes (provided they enter perpendicular to the field, or with equal pitch). The period $T=2\pi m/(qB)$ depends only on $m$, $q$ and $B$ — not on speed or direction.
C3. Why is the path of a charged particle in a magnetic field generally a helix?
Answer: The component of velocity along $\vec{B}$ is unaffected by the magnetic force; the perpendicular component produces uniform circular motion. The combination of uniform translation along the field and uniform rotation around it gives a helical path.
C4. What is the work done by a magnetic force on a charged particle?
Answer: Zero. The force is perpendicular to the displacement at every instant, so $W=\int\vec{F}\cdot d\vec{r}=0$.
C5. Why is the field inside a current-carrying toroid finite while it is zero outside?
Answer: Inside the toroid, an Amperian loop of radius $r$ encloses all the $N$ turns, so $B(2\pi r)=\mu_0 NI$. Outside, the same loop either encloses zero net current (because each turn passes through it once entering and once exiting) or no current at all, so $B=0$.
C6. Why must the magnetic field be radial in a moving coil galvanometer?
Answer: A radial field ensures that the plane of the coil is always parallel to $\vec{B}$ regardless of the coil’s angular position. Then $\sin\theta=1$ at all times and the deflecting torque is $\tau=NIAB$, independent of orientation. This makes the galvanometer scale linear in $I$.
C7. Will an ammeter and a voltmeter with the same galvanometer have the same resistance?
Answer: No. An ammeter has a very low resistance (galvanometer in parallel with a small shunt) while a voltmeter has a very high resistance (galvanometer in series with a high resistor).
C8. Why should an ideal ammeter have zero resistance and an ideal voltmeter have infinite resistance?
Answer: An ammeter is connected in series with the circuit; if it had any resistance, it would change the current it is meant to measure. A voltmeter is connected in parallel; if its resistance were finite, it would draw a current and disturb the voltage being measured.
C9. Two long parallel wires carry equal currents in the same direction and are 10 cm apart. A third parallel wire is placed midway. Will it experience a force?
Answer: If the third wire is placed exactly midway, the fields due to the two outer wires at its location are equal in magnitude but opposite in direction (since both wires carry current in the same direction, but the third wire is on opposite sides of each), so they cancel. The third wire experiences no force at the midpoint. If displaced even slightly, the symmetry is broken and a net force appears.
C10. Why does a solenoid behave like a bar magnet?
Answer: The pattern of magnetic field lines outside a long current-carrying solenoid is identical to that of a bar magnet — emerging from one end (north pole) and entering the other (south pole). Inside, however, the solenoid field lines run from S to N (in the opposite sense to outside), making it a closed-loop field — exactly like a bar magnet. Hence a solenoid acts as a magnetic dipole.
Worked Examples
Example 1. A particle of charge $q=2\times10^{-8}$ C and mass $m=4\times10^{-15}$ kg enters a magnetic field $B=0.5$ T with speed $v=10^5$ m s$^{-1}$ perpendicular to the field. Find (a) the radius of its path, (b) the period of revolution.
Solution: (a) $r=mv/(qB)=4\times10^{-15}\times10^5/(2\times10^{-8}\times0.5)=4\times10^{-2}$ m $=4$ cm.
$$T=\frac{2\pi m}{qB}=\frac{2\pi\times4\times10^{-15}}{2\times10^{-8}\times0.5}=2.51\times10^{-6}\;\text{s}$$
Example 2. A circular loop of radius 5 cm carrying a current of 0.10 A is placed in a uniform magnetic field of 0.20 T such that the angle between the area vector and the field is $60^\circ$. Find the torque.
Solution: $A=\pi(0.05)^2=7.85\times10^{-3}$ m$^2$. $\tau=IAB\sin\theta=0.10\times7.85\times10^{-3}\times0.20\times\sin 60^\circ=1.36\times10^{-4}$ N m.
Example 3. An infinitely long straight wire carries a current of 25 A. Find the magnetic field at a point 5 cm from the wire.
Solution: $B=\mu_0 I/(2\pi r)=4\pi\times10^{-7}\times25/(2\pi\times0.05)=1.0\times10^{-4}$ T.
Example 4. A solenoid 1 m long with 5000 turns is wound with a 10 A current. Find the magnetic field at its centre and at one of its ends.
Solution: $n=5000/1=5000$ m$^{-1}$. Centre: $B=\mu_0 nI=4\pi\times10^{-7}\times5000\times10=6.28\times10^{-2}$ T. End: $B/2=3.14\times10^{-2}$ T.
Example 5. Two parallel wires 10 cm apart each carry 5 A in the same direction. What is the force per metre on each wire?
$$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}=\frac{4\pi\times10^{-7}\times5\times5}{2\pi\times0.10}=5.0\times10^{-5}\;\text{N m}^{-1}$$
The force is attractive since the currents are parallel.
Example 6. A 100 turn rectangular coil of dimensions $0.05\times0.04$ m$^2$ carries a current of 2 A. Calculate the magnetic moment.
Solution: $A=0.05\times0.04=2\times10^{-3}$ m$^2$. $m=NIA=100\times2\times2\times10^{-3}=0.40$ A m$^2$.
Example 7. A galvanometer of resistance 50 ohm reads 5 mA at full scale. Calculate the shunt resistance to convert it into an ammeter of range 10 A.
Solution: $S=I_g G/(I-I_g)=5\times10^{-3}\times50/(10-5\times10^{-3})\approx 0.025\;\Omega$.
Example 8. A circular wire loop of radius 0.10 m and 50 turns carries a current of 4 A. Find the magnetic field at a point 0.05 m from the centre on the axis of the loop.
$$B=\frac{\mu_0 NIR^2}{2(R^2+x^2)^{3/2}}=\frac{4\pi\times10^{-7}\times50\times4\times(0.10)^2}{2\times(0.01+0.0025)^{3/2}}$$
$=\dfrac{2.51\times10^{-6}}{2\times(0.0125)^{3/2}}=\dfrac{2.51\times10^{-6}}{2\times1.398\times10^{-3}}=8.99\times10^{-4}$ T.
Important Derivations Summary
D1. Magnetic field due to an infinitely long straight wire (Ampere’s law). Choose a circular Amperian loop of radius $r$ around the wire. By symmetry $\vec{B}$ has the same magnitude all around and is tangential, so $\oint\vec{B}\cdot d\vec{l}=B(2\pi r)$. Setting equal to $\mu_0 I$ gives $B=\mu_0 I/(2\pi r)$.
D2. Field at the centre of a circular coil (Biot-Savart law). For each element $d\vec{l}$ on the loop, $d\vec{l}\perp\hat{r}$ so $|d\vec{l}\times\hat{r}|=dl$ and $dB=(\mu_0/4\pi)(I\,dl/R^2)$. Integrating $\oint dl=2\pi R$ gives $B=\mu_0 I/(2R)$.
D3. Field on the axis of a circular loop. By symmetry, only axial components of $d\vec{B}$ survive. From a single element at distance $r=\sqrt{R^2+x^2}$, the axial component is $dB(R/r)$. Integrating gives $B=\mu_0 IR^2/[2(R^2+x^2)^{3/2}]$.
D4. Field inside a long solenoid (Ampere’s law). Use a rectangular Amperian loop with one side of length $L$ inside, parallel to the axis, and the opposite side outside. Outside contribution is negligible; sides perpendicular to axis contribute zero. Enclosed current is $nLI$. So $BL=\mu_0 nLI$ giving $B=\mu_0 nI$.
D5. Field inside a toroid (Ampere’s law). Take a circular Amperian loop of radius $r$ along the centre line of the toroid. By symmetry $\vec{B}$ is tangential and uniform: $B(2\pi r)=\mu_0 NI$ giving $B=\mu_0 NI/(2\pi r)$.
D6. Force between two parallel currents. Wire 1 produces $B_1=\mu_0 I_1/(2\pi d)$ at wire 2. Force on length $L$ of wire 2: $F=B_1 I_2 L$, so $F/L=\mu_0 I_1 I_2/(2\pi d)$. Direction by right-hand rule: parallel currents attract; antiparallel currents repel.
D7. Cyclotron frequency. Magnetic force provides centripetal force: $qvB=mv^2/r$, so $r=mv/(qB)$. Period $T=2\pi r/v=2\pi m/(qB)$, frequency $f=qB/(2\pi m)$.
D8. Torque on a rectangular loop in uniform field. Forces on the two arms parallel to $\vec{B}$ have no moment. The arms perpendicular to $\vec{B}$ feel forces $F=BIb$ in opposite directions, separated by $a\sin\theta$, giving $\tau=BIA\sin\theta$ (with $A=ab$); for $N$ turns, $\tau=NIAB\sin\theta=mB\sin\theta$.
D9. Galvanometer. In equilibrium the deflecting torque $NIAB$ equals the restoring torque $k\theta$ of the spring, giving $\theta=(NAB/k)I$, i.e. $I=(k/NAB)\theta$, a linear relation since the field is radial.
Glossary
| Term | Meaning |
|---|---|
| Magnetic field $\vec{B}$ | Vector field describing magnetic influence; SI unit tesla (T) |
| Lorentz force | Total electromagnetic force $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ on a charge |
| Cyclotron | Device that uses magnetic field and oscillating electric field to accelerate charged particles in spiral paths |
| Cyclotron frequency | Frequency of revolution $f=qB/(2\pi m)$, independent of speed |
| Velocity selector | Region with crossed $\vec{E}$ and $\vec{B}$ that lets only particles with speed $v=E/B$ pass |
| Biot-Savart law | Gives field $d\vec{B}$ from current element $I\,d\vec{l}$ at distance $r$ |
| Ampere’s circuital law | $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{\text{enc}}$ for any closed loop |
| Solenoid | Long cylindrical helical coil; produces uniform field $\mu_0 nI$ inside |
| Toroid | Solenoid bent into a closed ring; field $\mu_0 NI/(2\pi r)$ inside the core |
| Magnetic moment $\vec{m}$ | $NI\vec{A}$ for a current loop; describes its strength as a magnetic dipole |
| Torque on loop | $\vec{\tau}=\vec{m}\times\vec{B}$, $\tau=NIAB\sin\theta$ |
| Galvanometer | Instrument that detects/measures small currents; deflection $\theta\propto I$ |
| Current sensitivity | Deflection per unit current, $\theta/I=NBA/k$ |
| Voltage sensitivity | Deflection per unit voltage, $\theta/V=NBA/(kR)$ |
| Shunt | Low resistance connected in parallel with a galvanometer to make an ammeter |
| Multiplier | High resistance connected in series with a galvanometer to make a voltmeter |
| Ampere | SI unit of current; defined via the force between parallel current-carrying wires |
| Permeability of free space | $\mu_0=4\pi\times10^{-7}$ T m A$^{-1}$, fundamental magnetic constant |
| Right-hand rule | Mnemonic for the direction of $\vec{v}\times\vec{B}$, $I\vec{L}\times\vec{B}$ and $\vec{m}$ |
Quick Revision Notes
R1. A magnetic field exerts a force only on a moving charge — and only on the component of velocity perpendicular to the field. The force does no work; it changes only the direction.
R2. Charges move in circles when they enter perpendicular to $\vec{B}$, in helices when they enter at an angle, and in straight lines when they move parallel to $\vec{B}$.
R3. The radius $r=mv/(qB)$ depends on momentum; the period $T=2\pi m/(qB)$ does not depend on speed. This is why the cyclotron works.
R4. A current produces a magnetic field. The Biot-Savart law and Ampere’s circuital law give equivalent ways of computing the field — Biot-Savart is universal but harder; Ampere is easy but only works when there is sufficient symmetry.
R5. Some standard fields to memorise:
- Long straight wire at distance $r$: $B=\mu_0 I/(2\pi r)$
- Centre of circular coil of radius $R$, $N$ turns: $B=\mu_0 NI/(2R)$
- On axis of circular loop at distance $x$ from centre: $B=\mu_0 IR^2/[2(R^2+x^2)^{3/2}]$
- Inside a long solenoid: $B=\mu_0 nI$ (uniform)
- End of a long solenoid: $B=\mu_0 nI/2$
- Inside a toroid at radius $r$: $B=\mu_0 NI/(2\pi r)$
R6. Two parallel wires carrying currents $I_1$ and $I_2$ separated by $d$ exert a force per unit length $\mu_0 I_1 I_2/(2\pi d)$ on each other. Same direction: attract. Opposite direction: repel. This force defines the SI ampere.
R7. A loop of $N$ turns and area $A$ carrying current $I$ has magnetic moment $m=NIA$ (a vector along the area vector by the right-hand rule). In a uniform field $\vec{B}$ it experiences torque $\vec{\tau}=\vec{m}\times\vec{B}$ but no net force.
R8. The galvanometer uses this torque on a coil in a radial magnetic field. Linear relation $I=(k/NBA)\theta$ allows current measurement. A shunt converts it to an ammeter; a series resistance to a voltmeter.
Continue to Chapter 5 of the ASSEB Class 12 Physics syllabus on Magnetism and Matter for an extension of these ideas to bulk magnetic materials and the Earth’s magnetism. Practise the numerical problems above thoroughly — they appear repeatedly in HS final examinations.