121-123 - Magnetic Flux Density, Flux and Magnetic Forces

Sideways force from a field

Picture two arrangements between the poles of a strong magnet:

The shared idea is not "the field pulls the object along". A magnetic field gives a force sideways to the motion of charge. The force is greatest when the charge motion cuts across the field at right angles, and zero when the charge motion is parallel to the field.

Magnetic flux density, $B$, is the quantity that tells us how strong the magnetic field is. Its unit is the tesla, T.

For a wire at right angles to the field:

so:

For a single moving charge at right angles to the field:

so the same tesla can also be written as:

The angle matters because only the part of the motion that cuts across the field produces the magnetic force. In the force equations later, $\theta$ is the angle between the velocity or current direction and the magnetic field direction:

Misconception contrast:

When Edexcel asks you to state what is meant by magnetic flux density, a useful answer links $B$ to force, not just to "how strong the magnet is". For example:

Flux and linkage

Magnetic flux density $B$ is about field strength at a place. Magnetic flux, $\phi$, adds another idea: how much magnetic field passes through an area.

For a uniform magnetic field passing through a flat area:

where:

The normal is an imaginary line at right angles to the surface. This angle convention prevents a common error:

Here $N$ is the number of turns. Flux linkage is not a different kind of flux; it is the single-turn flux multiplied by the number of turns linked by that flux.

Worked example:

A rectangular coil has $80$ turns. Each turn is $5.0\,\mathrm{cm}$ by $8.0\,\mathrm{cm}$. The coil is in a uniform magnetic field of flux density $0.35\,\mathrm{T}$. The normal to the coil makes an angle of $30^{\circ}$ with the magnetic field. Calculate the magnetic flux through one turn and the flux linkage.

Decision point 1: flux uses area in square metres, not square centimetres.

Decision point 2: the angle is measured between the field and the normal, so use $\cos 30^{\circ}$.

Decision point 3: flux linkage comes after the flux through one turn.

The answer makes physical sense: the angle is not zero, so the flux is a little less than $BA$, but it is not zero because the field still passes through the coil face.

[DIAGRAM: asset_name: Magnetic force perpendicular to positive-particle motion and into-page field; flux angle is measured to coil normal; asset_slug: fields_magnetic_force_flux; recommended_method: matplotlib; description: Magnetic force perpendicular to positive-particle motion and into-page field; flux angle is measured to coil normal]

Charged particles in fields

For one charged particle moving through a magnetic field, Edexcel gives:

where:

For force magnitude, use the magnitude of the charge. The sign of the charge is then used for the direction.

Fleming's left-hand rule gives the direction:

For a positive particle, conventional current is in the same direction as its velocity. For a negative particle, conventional current is opposite to its velocity. This is the trap in many electron questions.

Worked example:

An electron travels to the right at $2.5 \times 10^{6}\,\mathrm{m\,s^{-1}}$ through a uniform magnetic field of flux density $0.12\,\mathrm{T}$. The field is into the page. Calculate the force on the electron and state its direction. Use $e = 1.60 \times 10^{-19}\,\mathrm{C}$.

Decision point 1: this is a single charged particle, not a wire, so use:

Decision point 2: the velocity is perpendicular to the field, so $\theta = 90^{\circ}$ and $\sin \theta = 1$.