Magnetic fields and electromagnets

Electric current is ordered motion of charges, and every moving charge creates a magnetic field $\vec{B}$ around itself, so a current-carrying wire is always surrounded by a magnetic field.

Near a long straight wire the field is described by Ampère's law:

$$B = \frac{\mu_0 I}{2\pi r},$$

where $I$ is the current, $r$ is the distance from the wire, and $\mu_0 = 4\pi \times 10^{-7}\ \text{T}\cdot\text{m/A}$ is the magnetic constant. The direction of the field is found with the right-hand rule: point the thumb along the current, and the curled fingers show the direction of the field lines — concentric circles around the wire.

If the wire is wound into a coil of $N$ turns of length $l$, the fields of the individual turns add up, and inside a long solenoid the field is nearly uniform:

$$B = \mu_0 n I, \qquad n = \frac{N}{l}.$$

An iron core inside the coil turns it into an electromagnet: the iron domains align with the external field and multiply it many times over, while the field remains controllable.

Because Ohm's law gives $I = \frac{U}{R}$, the strength of an electromagnet can be adjusted smoothly by the source voltage, the coil resistance, or the number of turns — which is exactly why electromagnets are used in relays, lifting cranes, and loudspeakers.

A long straight wire carries a current $I = 2.0\ \text{A}$. Find the magnetic field magnitude at a distance $r = 5.0\ \text{cm}$ from the wire.

The winding of an electromagnet — a coil of length $l = 20\ \text{cm}$ with $N = 400$ turns — is connected to a source with $U = 12\ \text{V}$. The winding resistance is $R = 6.0\ \Omega$. Find the current in the winding and the field inside the coil without a core.

The winding of an electromagnet with resistance $R = 4.0\ \Omega$ is powered by a source with $U = 24\ \text{V}$ and has $N = 300$ turns. The average magnetic path length in the iron core is $l_c = 20\ \text{cm}$, the relative permeability of the iron is $\mu_r = 1000$, and the core has an air gap $l_g = 2.0\ \text{mm}$. In the gap a straight wire of length $L = 10\ \text{cm}$ carries a current $I_2 = 5.0\ \text{A}$ perpendicular to the field. Find the magnetic field in the gap and the force acting on this wire.

Model the electromagnet in a simulation (for example, PhET "Magnets and Electromagnets" or "Faraday's Electromagnetic Lab") so that you never work with large currents.

1. Build a circuit: source, winding (coil), switch; place a compass or field sensor nearby.
2. Vary the source voltage and record the current $I$ and the compass deflection (or the sensor reading $B$).
3. Repeat for different numbers of turns $N$.
4. Compare the coil field with and without an iron core.

Make a table and plot $B$ against $I$: for a solenoid you should see direct proportionality $B \propto I$. Conclude how Ohm's law, the number of turns, and the strength of the electromagnet are related.

Safety: do not use the $220\ \text{V}$ mains; if you work with real equipment, use only a battery of up to $9\ \text{V}$ and connect the coil through a resistor to limit the current.

Show that magnetic field lines are closed, surround a current-carrying wire, and concentrate inside a coil with a core, while their direction is set by the right-hand rule.

A two-part diagram.

(1) Cross-section of a straight wire perpendicular to the page: the current points toward the observer ($\odot$), surrounded by concentric field-line circles with counter-clockwise arrows; $I$ and $r$ are labelled.

(2) Longitudinal section of a solenoid with an iron core: $N$ turns, nearly parallel field lines along the axis inside, with poles $N$ and $S$ labelled.

Below is a mini-graph of $B$ versus $I$: a straight line through the origin, as predicted by $B = \mu_0 n I$.

A scrap-yard crane lowers a steel disc onto a flattened car body and switches on the coil. The coil has $N = 800$ turns, is fed at $U = 220\ \text{V}$ and has resistance $R = 2.2\ \Omega$, so $I = U/R = 100\ \text{A}$ flows. The iron core concentrates the flux until it saturates near $B \approx 1.8\ \text{T}$; the disc must hold a load of $m = 1200\ \text{kg}$, that is $mg \approx 11.8\ \text{kN}$.

The coil is an electromagnet: its field follows $B = \mu_0 \mu_r \dfrac{N I}{L}$, so the strength is set by the current that Ohm's law fixes, and a ferromagnetic core multiplies $B$ by a factor of hundreds over air until saturation — which is why such a crane drops its load the instant the current is switched off.

A doorbell is powered by a $6\ \text{V}$ battery. Its coil has $N = 500$ turns and resistance $R = 30\ \Omega$, so $I = U/R = 0.2\ \text{A}$. The coil sits on a soft-iron core and faces a spring-loaded armature across an air gap of about $1\ \text{mm}$; the field pulls the armature, its hammer strikes the gong, the contact opens and closes again about $20$ times each second.

Only current creates the field here, $B \propto N I$, and the soft-iron core keeps $B$ large enough across the millimetre gap for the pull to be felt. Switching the current off removes the field at once, so the spring wins and the armature returns — a magnetic field that exists only while charge flows.

The loudspeaker of a phone is a coil of $N = 50$ turns glued to a paper cone and sitting in the permanent field $B_0 = 0.4\ \text{T}$ of a small magnet. Music sends an alternating current of up to $0.5\ \text{A}$ through the coil, which has an impedance of $8\ \Omega$. The coil's own field in the gap reaches $B_{\text{coil}} \approx 0.6\ \text{mT}$ and reverses sign with every half-cycle of the audio signal.

The coil is a small electromagnet whose field lines follow the right-hand grip rule; they add to the permanent field on one side of the coil and subtract on the other, so the total field $B_0 \pm B_{\text{coil}}$ is non-uniform in the gap — the superposition of two magnetic fields is what the cone responds to.