The Biot-Savart law gives the magnetic field produced at a point by a small current-carrying element — analogous to how Coulomb's law gives the electric field due to a point charge.
where $\mu_0$ = permeability of free space ($4\pi\times10^{-7}\ \text{T m/A}$), $I$ = current, $d\vec{l}$ = current element vector, $r$ = distance from the element to the point, $\hat{r}$ = unit vector along that direction.
Since the law involves a cross product, the magnetic field is zero along the direction of current flow itself (when $d\vec{l}$ and $\hat{r}$ are parallel, $\sin\theta = 0$), and maximum perpendicular to the current element.
Explain why the magnetic field due to a straight current-carrying wire is zero at points lying exactly on the wire's own axis (extension of the wire).
At any point along the extended axis of the wire, the position vector $\hat{r}$ (from any current element to that point) is parallel (or antiparallel) to the current element $d\vec{l}$ itself. Since the Biot-Savart law involves $d\vec{l}\times\hat{r}$, and the cross product of two parallel vectors is zero, the contribution from every current element to the field at such a point is zero — hence the total magnetic field along the wire's own axis is zero.
The direction of the magnetic field around a current-carrying conductor can be quickly found using the right-hand thumb rule: if the thumb points in the direction of conventional current flow, the curled fingers indicate the direction of the circular magnetic field lines around the conductor.
A straight wire carries current vertically upward. Using the right-hand thumb rule, determine the direction of the magnetic field at a point directly to the east of the wire.
Pointing the right thumb upward (direction of current) and curling the fingers, the field circles counterclockwise when viewed from above. At a point to the east of the wire, the curling fingers point towards the north — so the magnetic field at that point is directed northward (horizontally, perpendicular to the line joining the wire and the point).
where $R$ = radius of the loop. For $N$ turns: $B = \dfrac{\mu_0 NI}{2R}$.
Calculate the magnetic field at the centre of a circular coil of 50 turns, radius 0.1 m, carrying a current of 2 A. ($\mu_0 = 4\pi\times10^{-7}$ T m/A)
$B = \dfrac{\mu_0 NI}{2R} = \dfrac{(4\pi\times10^{-7})(50)(2)}{2(0.1)} = \dfrac{4\pi\times10^{-7}\times100}{0.2}$
$= \dfrac{1.257\times10^{-4}}{0.2} \approx 6.28\times10^{-4}\ \text{T}$
The line integral of the magnetic field around any closed loop equals $\mu_0$ times the total current enclosed by that loop — the magnetic analogue of Gauss's law in electrostatics.
Using Ampere's law, derive the magnetic field at a distance $r$ from a long, straight current-carrying wire.
Choose a circular Amperian loop of radius $r$, centred on the wire. By symmetry, $B$ has the same magnitude everywhere on this loop and is always tangent to it (parallel to $d\vec{l}$ at every point).
$\oint \vec{B}\cdot d\vec{l} = B(2\pi r) = \mu_0 I \Rightarrow B = \dfrac{\mu_0 I}{2\pi r}$
A solenoid is a long, tightly-wound helical coil of wire. Inside a long solenoid, the magnetic field is remarkably uniform and parallel to the solenoid's axis (much like the field between the plates of a parallel-plate capacitor), while the field outside is nearly zero.
where $n$ = number of turns per unit length, $I$ = current.
A solenoid has 1000 turns over a length of 0.5 m and carries a current of 3 A. Calculate the magnetic field inside it.
$n = \dfrac{1000}{0.5} = 2000\ \text{turns/m}$
$B = \mu_0 nI = (4\pi\times10^{-7})(2000)(3) \approx 7.54\times10^{-3}\ \text{T}$
A toroid is essentially a solenoid bent into a closed, ring (doughnut) shape. Its key advantage is that the magnetic field is confined entirely within the ring-shaped core, with essentially zero field outside (both inside the central hole and outside the outer surface) — since there are no "ends" like an ordinary solenoid, there's no field leakage at open ends to worry about.
Using Ampere's law, explain why the magnetic field outside a toroid (both in the central hole and beyond the outer edge) is zero.
Consider an Amperian loop drawn either through the toroid's central hole or entirely outside its outer radius. In both cases, this loop encloses zero net current — for a loop through the central hole, no current-carrying turns pass through the loop's interior at all; for a loop outside the outer radius, each turn's current enters and exits the enclosed region an equal number of times, contributing zero net enclosed current. By Ampere's law, zero enclosed current directly implies the field along such a loop must be zero.
The Lorentz force is the total electromagnetic force on a moving charged particle, combining both electric and magnetic contributions:
$$\vec{F} = q\vec{E} + q\vec{v}\times\vec{B}$$$\vec{F} = q\vec{v}\times\vec{B}$, with magnitude $F = qvB\sin\theta$ — this force is always perpendicular to both the particle's velocity and the magnetic field, meaning it can change the particle's direction but never its speed (since it does no work — the force is always perpendicular to displacement).
Since the magnetic force provides exactly the centripetal force needed (always perpendicular to velocity), a charged particle moving perpendicular to a uniform magnetic field travels in a circular path, with radius $r = \dfrac{mv}{qB}$.
An electron moves with a speed of $2\times10^6$ m/s perpendicular to a magnetic field of 0.5 T. Calculate the radius of its circular path. ($m_e = 9.1\times10^{-31}$ kg, $e = 1.6\times10^{-19}$ C)
$r = \dfrac{mv}{qB} = \dfrac{(9.1\times10^{-31})(2\times10^6)}{(1.6\times10^{-19})(0.5)} = \dfrac{1.82\times10^{-24}}{8\times10^{-20}} \approx 2.28\times10^{-5}\ \text{m}$
A cyclotron is a particle accelerator that uses a combination of a magnetic field (to keep charged particles moving in a circular/spiral path) and an alternating electric field (to repeatedly accelerate the particles each time they cross a gap), progressively increasing their speed and the radius of their circular path — used to accelerate charged particles to high energies for research and medical applications.
Explain why the time period of a charged particle's circular motion in a cyclotron's magnetic field remains constant, even as the particle's speed and orbital radius continuously increase.
The time period of circular motion is $T = \dfrac{2\pi m}{qB}$ (derived from $T = 2\pi r/v$ combined with $r = mv/qB$) — notice that $v$ cancels out entirely from this expression. Since $T$ depends only on the particle's mass, charge, and the magnetic field strength (all constant, assuming non-relativistic speeds), the time period remains exactly the same regardless of how much the particle's speed or radius increases — this constancy is precisely what allows the cyclotron's alternating electric field to be applied at a fixed, unchanging frequency throughout the acceleration process.
where $L$ = length of the conductor in the field, $\theta$ = angle between the conductor's length direction and the field.
A straight wire of length 0.5 m carries a current of 4 A, placed perpendicular to a magnetic field of 0.3 T. Calculate the force on the wire.
$F = BIL\sin90° = (0.3)(4)(0.5)(1) = 0.6\ \text{N}$
Two parallel current-carrying wires exert a force on each other — like currents (same direction) attract, and unlike currents (opposite directions) repel. This force per unit length was historically used to define the SI unit of current, the ampere.
where $d$ = separation between the wires, $I_1, I_2$ = currents in each wire.
Two parallel wires, each carrying a current of 5 A in the same direction, are separated by 0.2 m. Calculate the force per unit length between them.
$\dfrac{F}{L} = \dfrac{\mu_0 I_1I_2}{2\pi d} = \dfrac{(4\pi\times10^{-7})(5)(5)}{2\pi(0.2)} = \dfrac{(4\times10^{-7})(25)}{2(0.2)} = \dfrac{10^{-5}}{0.4} = 2.5\times10^{-5}\ \text{N/m}$
A moving coil galvanometer detects small electric currents by measuring the torque a current-carrying coil experiences in a magnetic field, which causes the coil (attached to a pointer) to deflect by an amount proportional to the current: $I \propto \theta$ (deflection angle), for a suitably designed instrument with a radial magnetic field and restoring spring.
A galvanometer (which can only handle very small currents) is converted into an ammeter (for measuring larger currents) by connecting a small resistance, called a shunt, in parallel with it — this diverts most of the current around the galvanometer, protecting its delicate coil while still allowing it to indicate the total current.
A galvanometer is converted into a voltmeter (for measuring potential difference) by connecting a large resistance in series with it — this limits the current through the galvanometer to a safe, small value even when a large voltage is applied across the combination.
Explain why an ammeter must have very low resistance, while a voltmeter must have very high resistance, in terms of how they're connected in a circuit.
An ammeter is always connected in series with the circuit element whose current is being measured — for it to measure the true current without altering the circuit's behaviour, it must have negligible resistance, so it doesn't itself significantly reduce the current flowing through that part of the circuit. A voltmeter is always connected in parallel across the element whose voltage is being measured — for it to measure the true voltage without drawing significant current away from the main circuit, it must have very high resistance, ensuring only a negligible fraction of current is diverted through the voltmeter itself.
A bar magnet behaves like a magnetic dipole, characterised by a magnetic dipole moment ($m$), analogous to the electric dipole moment — pointing from the south pole to the north pole of the magnet.
Notice the close structural similarity to the electric dipole field formulas, with the axial field again being twice the equatorial field at the same distance.
$\tau = mB\sin\theta$ — the magnet experiences a torque tending to align it with the external field, exactly analogous to an electric dipole in a uniform electric field.
A bar magnet of dipole moment $2\ \text{A m}^2$ is placed in a uniform magnetic field of 0.5 T at an angle of 30°. Calculate the torque on the magnet.
$\tau = mB\sin\theta = (2)(0.5)\sin30° = (1)(0.5) = 0.5\ \text{N m}$
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Response to field | Weakly repelled | Weakly attracted | Strongly attracted |
| Cause | Induced opposing magnetic moment | Alignment of existing, randomly-oriented atomic dipoles | Alignment of atomic dipoles within "domains" |
| Effect of temperature | Largely unaffected | Weakens with increasing temperature | Weakens with temperature, vanishing above the Curie point |
| Examples | Bismuth, copper, water | Aluminium, sodium, platinum | Iron, cobalt, nickel |
Ferromagnetic materials contain small regions called domains, within which large numbers of atomic magnetic dipoles are naturally aligned in the same direction, even without any external field. In an unmagnetised sample, these domains point in random directions, cancelling out overall — applying an external field causes domains to align (or grow at the expense of misaligned ones), producing the very strong net magnetisation characteristic of ferromagnetism.
Explain why a diamagnetic material is only very weakly repelled by a magnetic field, and why this effect doesn't depend significantly on temperature.
Diamagnetism arises from a magnetic moment induced in each atom by the changing external field itself (via a form of electromagnetic induction at the atomic level, opposing the applied field per Lenz's law), rather than from any pre-existing atomic magnetic moment that needs to be aligned. Since this induced effect exists in every material to some small degree and doesn't rely on the random thermal orientation of pre-existing dipoles (unlike paramagnetism), it remains weak and essentially temperature-independent, since there's no alignment process being disrupted by thermal agitation.