Magnetic fields and moving charges

Core ideas

Flux density and the force on a wire

$B$ is the force per unit current per unit length on a conductor at right angles to the field. 1 T = 1 N A⁻¹ m⁻¹.

Here $\theta$ is the angle between the wire and the field. The force is greatest at 90° and zero when the wire is parallel to the field.

Fleming’s left-hand rule: First finger points along the Field (N to S), seCond finger along the conventional Current, and the thuMb gives the force (Motion). The force is perpendicular to both the field and the current.

Field patterns

Two parallel wires: currents in the same direction attract, and opposite currents repel. By Newton’s third law the forces on the two wires are equal and opposite, even when the currents are different.

Force on a moving charge

A current is moving charge. Put $I = nAvQ$ into $F = BIl$ and divide by the $nAl$ carriers to get the force on each one:

For direction, use the left-hand rule with the current in the direction a positive charge moves. For electrons, point the second finger opposite to their velocity.

Because the force is always perpendicular to the velocity, it does no work. The speed and kinetic energy never change; only the direction does.

Circular paths

The magnetic force provides the centripetal force:

• Faster or heavier particles curve less. A stronger field or a larger charge curves them more.
• The period does not depend on speed. A faster particle travels a bigger circle in the same time. This is the key to the cyclotron.
• For a particle accelerated from rest through pd $V$: $r = \dfrac{1}{B}\sqrt{\dfrac{2mV}{Q}}$.

Velocity selector

Crossed $E$ and $B$ fields are set at right angles to each other and to the beam, so that the electric force $QE$ and the magnetic force $BQv$ point in opposite directions. A particle passes undeflected when

This condition does not depend on the charge or the mass. Faster particles feel a larger magnetic force and are deflected towards the magnetic-force side. Slower particles are deflected towards the electric-force side.

Mass spectrometer

Ions leave a velocity selector with the same $v$ and enter a region of field $B$ only. With the same charge, $r = mv/BQ$ is proportional to $m$. After a semicircle, ions hit the detector a distance $2r$ from the entrance slit, so heavier isotopes land further away.

Cyclotron

Two hollow D-shaped electrodes (“dees”) sit in a uniform $B$-field, with an alternating pd across the gap between them.

• Inside a dee there is no electric field, so the particle moves in a semicircle at constant speed. In the gap the pd accelerates it, so each semicircle is larger than the last.
• The period is independent of speed, so a fixed supply frequency keeps it in step: $f = \dfrac{BQ}{2\pi m}$.
• It leaves at the dee radius $R$ with , so the maximum kinetic energy is .

The Hall effect (Cambridge, Eduqas, WJEC)

Current flows along a thin slab of width $w$ and thickness $t$, with $B$ perpendicular to its face. The carriers are pushed sideways by $BQv$ and build up on one edge. This sets up a transverse field $E_H = V_H/w$. Equilibrium is reached when . Using with :

A semiconductor with a small $n$, made into a thin slab, gives a measurable $V_H$ of a few mV. In a metal, $V_H$ would be around nanovolts. At constant , , so a calibrated Hall probe reads . Rotate the probe until the reading is a maximum, so that the face is perpendicular to the field.

Measuring $B$ with a balance (AQA RP10, OCR A)

A magnet sits on a top-pan balance, with a stiff wire clamped across the gap between its poles at right angles to the field. When a current flows, the field pushes on the wire. By Newton’s third law the wire pushes equally on the magnet, so the reading changes by $\Delta m$, and $F = \Delta m\,g = BIl$. Plot $F$ against $$: the gradient is . Zero the balance with no current, keep the current low to limit heating, and reverse the current and average the readings to remove drift.

Worked examples

Exam technique

• Direction questions: draw three perpendicular axes before you use the left-hand rule, and reverse the current finger for negative particles.
• “Explain why the path is circular” (2–3 marks): the force is perpendicular to the velocity, so the speed is constant. The force then has constant magnitude and always acts perpendicular to the motion, which gives a circle.
• Ratio reasoning: $r \propto mv/Q$ at fixed $B$. For the same accelerating pd, $r \propto \sqrt{m/Q}$. For the same kinetic energy, .

Common mistakes

Quick recap

• $F = BIl\sin\theta$ and $F = BQv\sin\theta$. 1 T = 1 N A⁻¹ m⁻¹. Use Fleming’s left-hand rule, with the current opposite to electron motion.
• The magnetic force is perpendicular to $v$: no work is done, the speed is constant and the path is circular with .