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Magnetic fields of currents
Ørsted's discovery, the magnetic field around a straight current-carrying wire and the right-hand grip rule, B = μ₀I/2πr, adding fields from two wires, the field of a single loop, the field inside a solenoid (B = μ₀NI/L), electromagnets, and the compass-deflection investigation.
In April 1820 Hans Christian Ørsted was demonstrating a battery to a lecture audience in Copenhagen when he noticed a compass needle twitch each time he connected the circuit. Within months, scientists across Europe had shown that every current is surrounded by a magnetic field. That one twitch joined electricity and magnetism into a single subject, and led within a decade to the electromagnet, the motor and the generator.
The field around a straight wire
A current in a long straight wire produces a magnetic field whose lines are concentric circles centred on the wire, in planes at right angles to it. The field is strongest close to the wire. Its direction is given by the right-hand grip rule: point your right thumb in the direction of the conventional current, and your fingers curl in the direction of the field.
In diagrams, a dot (⊙) means a current or field coming out of the page, like the tip of an arrow coming towards you, and a cross (⊗) means going into the page, like the tail feathers of an arrow moving away.
The strength of the field a distance r from a long straight wire carrying current I is
$$B = \frac{\mu_0 I}{2\pi r}$$
where $\mu_0 = 4\pi \times 10^{-7}$ T m A−1 is the permeability of free space. B is proportional to I and inversely proportional to r (not r2): twice as far away, half the field. (Since the 2019 SI redefinition μ0 is a measured constant rather than exact, but it still equals 4π × 10−7 to better than one part in a billion.)
A long straight wire carries 10 A. Find the field 5.0 cm from it and compare it with the horizontal part of Earth's field in Sydney, 2.5 × 10−5 T.
$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi \times 10^{-7})(10)}{2\pi(0.050)} = 4.0 \times 10^{-5}$ T.
That is 1.6 times Earth's horizontal field, so it would swing a nearby compass well away from north. This is why compasses are kept away from car wiring, and why Ørsted saw his needle move.
Fields from two wires
Magnetic fields add as vectors, just like electric fields. For two parallel wires, find the field of each at the point (size from μ0I/2πr, direction from the grip rule), then add.
Two long parallel wires are 10 cm apart. Each carries 5.0 A, in opposite directions. Find the field halfway between them.
Each wire is 0.050 m away: $B = \dfrac{(4\pi \times 10^{-7})(5.0)}{2\pi(0.050)} = 2.0 \times 10^{-5}$ T.
Suppose the left wire's current comes out of the page and the right wire's goes in. By the grip rule, at the midpoint (to the right of the left wire) the left wire's field points up the page; at the midpoint (to the left of the right wire) the right wire's field also points up the page.
$B = 2.0 \times 10^{-5} + 2.0 \times 10^{-5} = 4.0 \times 10^{-5}$ T up the page. With currents in the same direction, the two fields at the midpoint would cancel. (The force each wire exerts on the other is a Year 12 Module 6 topic.)
A single loop
Bend the wire into a circular loop. Every small piece of the loop makes circles of field around itself, and inside the loop all of these point the same way, so they add to give a strong field through the middle. Outside, the field spreads out and loops back. The pattern looks like the field of a short bar magnet: one face of the loop behaves as an N pole and the other as an S pole.
The solenoid
A solenoid is a long coil of many loops. The fields of the turns add, giving a strong, nearly uniform field inside, running along the axis, and a field outside like that of a bar magnet. For a long solenoid with N turns over a length L carrying current I,
$$B = \frac{\mu_0 N I}{L} = \mu_0 n I$$
where n = N/L is the number of turns per metre. The field inside doesn't depend on the solenoid's diameter or on where you are inside it (away from the ends). To find which end is N, curl the fingers of your right hand round the solenoid in the direction of the current: your thumb points to the N pole, which is the direction of the field inside.
A solenoid 25 cm long has 500 turns and carries 2.0 A. Find the field inside it. How would the field change if the same wire were wound into a solenoid 50 cm long with 1000 turns?
$B = \dfrac{\mu_0 NI}{L} = \dfrac{(4\pi \times 10^{-7})(500)(2.0)}{0.25} = 5.0 \times 10^{-3}$ T.
N/L is still 2000 turns per metre, so B is unchanged at 5.0 × 10−3 T: only the turns per metre and the current matter.
Put a soft iron core inside the solenoid and the domains in the iron line up with the coil's field, adding their own field. The total can be hundreds or thousands of times larger. This is an electromagnet (Year 7, topic 8G): strong, switchable and reversible. Electromagnets are used in relays, MRI scanners (with superconducting coils and no iron), speakers, maglev trains and the scrap-metal cranes at recycling yards.
- Straight wire, 10 A, 5 cm: B = 4.00 × 10−5 T, anticlockwise for current out of the page. The "B at 2r" readout is 2.00 × 10−5 T: inverse proportion, not inverse square. Reverse the current and watch the direction change.
- Solenoid, 2.0 A, 500 turns, 25 cm: n = 2,000 turns m−1 and B = 5.03 × 10−3 T. Double N to 1000: B = 0.0101 T. Then set L to 50 cm as well: back to 5.03 × 10−3 T.
- Reverse the solenoid's current. Which end is now the N pole? Check with the right-hand grip rule.
- Compass deflection: 5.0 A with the wire 4 cm above the needle gives Bwire = 2.50 × 10−5 T, equal to Earth's horizontal field, so θ = 45.0° west. Halve the current: θ = 26.6°, not 22.5°. Explain why θ is not proportional to I.
Test yourself
The magnetic field 3.0 cm from a long straight wire is 6.0 × 10−5 T. The field 9.0 cm from the wire is
B ∝ 1/r: three times as far gives one third of the field.
The current in a long straight wire is 15 A. The field 2.0 cm from the wire is
B = (4π × 10−7)(15)/(2π × 0.020) = 1.5 × 10−4 T.
A vertical wire carries current upwards. Looking down from above, the magnetic field around it is
Thumb up (towards you when looking down), fingers curl anticlockwise.
Which change would double the field inside a long solenoid?
B = μ0NI/L: doubling I doubles B. Doubling N and L together leaves N/L unchanged; diameter has no effect.
A solenoid 0.40 m long has 800 turns and carries 1.5 A. The field inside it is
B = (4π × 10−7)(800)(1.5)/0.40 = 3.8 × 10−3 T.
Label the solenoid's field.
Inside a long solenoid, sort each change.
The field lines around a straight current-carrying wire are concentric . Their direction is given by the -hand grip rule. The field strength is directly proportional to the and inversely proportional to the from the wire. Inside a long solenoid the field is nearly and depends on the number of turns per .
In an investigation, a long straight wire running north–south is placed 3.0 cm above a compass. With 4.0 A in the wire the needle is deflected 47° from north. (a) Calculate the magnetic field of the wire at the compass. (b) Use the result to calculate the horizontal component of Earth's magnetic field at that location. (c) Explain why the student should reverse the current and average the deflections.
(a) Bwire = μ0I/(2πr) = (4π × 10−7)(4.0)/(2π × 0.030) = 2.7 × 10−5 T, horizontal and east–west at the compass.
(b) The wire's field is perpendicular to Earth's horizontal field, and the needle lines up with the resultant, so tan θ = Bwire/BE. BE = 2.67 × 10−5/tan 47° = 2.5 × 10−5 T.
(c) Reversing the current deflects the needle the same amount in the opposite direction. If the wire is not exactly north–south, or the compass scale is slightly misaligned, one deflection reads too large and the other too small by about the same amount. Averaging the two cancels this systematic error and improves the validity of the result; repeating also improves reliability.