Magnetic Fields

Magnetic flux density $B$ is the force acting per unit current per unit length on a straight wire placed perpendicular to the field: $B = \frac{F}{Il}$

Its SI unit is the tesla ($\text{T}$), where $1\ \text{T} = 1\ \text{N A}^{-1}\text{m}^{-1}$.

Fleming's left-hand rule gives the direction of the force on a current-carrying conductor (or a moving positive charge) in a magnetic field. With the thumb and first two fingers of the left hand held mutually at right angles:

• the First finger points in the direction of the Field (N to S);
• the seCond finger points in the direction of the (conventional) Current;
• the thuMb points in the direction of the Motion / force (thrust).

It is used to find the direction of the magnetic force on the conductor or charge.

Magnetic flux $\phi$ through a surface is the product of the magnetic flux density and the area perpendicular to the field: $\phi = BA$ Its unit is the weber ($\text{Wb}$), where $1\ \text{Wb} = 1\ \text{T m}^2$.

Magnetic flux linkage is the flux multiplied by the number of turns of the coil, $N\phi$. Its unit is the weber-turn ($\text{Wb turns}$), numerically the same as the weber.

(a) The current is horizontal (west→east) and the field is vertical (downwards), so they are at right angles ($\theta = 90°$). The force per metre is: $\frac{F}{l} = BI\sin 90° = 0.15 \times 12 \times 1 = 1.8\ \text{N m}^{-1}$

(b) Using Fleming's left-hand rule: First finger points down (field), seCond finger points east (current). The thuMb (force) then points to the north (horizontally). So the force on the wire is directed northwards, horizontally.

(a) For the proton (positive), the conventional current is in its direction of motion — to the right. With the field into the page, Fleming's left-hand rule (First finger into the page, seCond finger right) gives a force (thuMb) pointing upwards on the proton.

For the electron (negative), the conventional current is opposite to its motion — i.e. to the left. Applying the rule with the current to the left gives a force pointing downwards on the electron.

So the two particles are deflected in opposite directions, even though both move to the right, because their charges have opposite sign.

(b) The radius is $r = mv/(Bq)$. The speed $v$, field $B$ and charge magnitude $q$ are the same, so $r \propto m$. The proton is about $1800$ times more massive than the electron, so the proton's circular path has a much larger radius than the electron's.

An e.m.f. is induced whenever the flux linkage through the coil changes. Two ways to achieve this:

1. Move the magnet into (or out of) the coil. As the magnet approaches, the magnetic flux density through the coil — and hence the flux linkage $N\phi$ — increases with time, inducing an e.m.f. Moving it faster gives a larger rate of change of flux and a bigger e.m.f.

2. Rotate the coil (or the magnet) so the coil's orientation to the field changes. This changes the area presented perpendicular to the field, so the flux $\phi = BA\cos\theta$ — and the flux linkage — changes with time, again inducing an e.m.f.

In both cases the quantity that changes is the magnetic flux linkage; a stationary magnet held inside a stationary coil produces no e.m.f. because the flux is constant.

(a) The magnetic force provides the centripetal force, $Bqv = mv^2/r$, which rearranges to: $r = \frac{mv}{Bq} = \frac{p}{Bq}$ so the radius is proportional to the momentum $p = mv$ (for a fixed field and charge).

(b) Since $r = mv/(Bq)$ with $B$, $q$ and $m$ unchanged, the radius is directly proportional to the speed, $r \propto v$. Doubling the speed therefore doubles the radius — the faster particle is deflected less sharply and follows a wider circle.

(a) Area of the coil: $A = (0.080)^2 = 6.4 \times 10^{-3}\ \text{m}^2$ Initial flux through one turn: $\phi = BA = 0.40 \times (6.4 \times 10^{-3}) = 2.56 \times 10^{-3}\ \text{Wb}$.

The coil is fully removed, so the final flux is zero. The change in flux linkage is: $\Delta(N\phi) = N \times \Delta\phi = 250 \times (0 - 2.56 \times 10^{-3}) = -0.64\ \text{Wb turns}$

(b) Faraday's law (magnitude): $|\varepsilon| = \left| \frac{\Delta(N\phi)}{\Delta t} \right| = \frac{0.64}{0.020} = 32\ \text{V}$

(a) Lenz's law: the direction of an induced current (or e.m.f.) is always such that it opposes the change producing it. This is the origin of the minus sign in $\varepsilon = -N\,\mathrm{d}\phi/\mathrm{d}t$.

(b) As the north pole approaches, the flux through the coil increases. By Lenz's law the induced current must oppose this increase, so it flows to make the near face of the coil a north pole — repelling the incoming magnet. Looking at that face from the magnet's side, the induced current flows anticlockwise.

(c) Because the induced current opposes the motion, the person pushing the magnet must do work against this repulsive force. That mechanical work is the source of the electrical energy dissipated in the circuit. If the induced current instead attracted the magnet, the magnet would accelerate in by itself, generating electrical energy from nothing — creating energy and violating conservation of energy. So the opposing direction demanded by Lenz's law is exactly what conservation of energy requires.

(a) Each moving ion experiences a magnetic force $F = Bqv$ that is always perpendicular to its velocity. A constant-magnitude force at right angles to the motion provides a centripetal force, so the ion moves in a circular arc at constant speed.

(b) The magnetic force provides the centripetal force: $Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}$ $r = \frac{(3.8 \times 10^{-26}) \times (1.9 \times 10^{5})}{0.35 \times (1.6 \times 10^{-19})} = \frac{7.22 \times 10^{-21}}{5.6 \times 10^{-20}} = 1.29 \times 10^{-2}\ \text{m} \approx 1.3\ \text{cm}$

(c) Since $B$, $q$ and $v$ are the same for all the ions, the radius $r = mv/(Bq)$ is proportional to the mass $m$. Heavier ions follow larger-radius arcs and lighter ions follow smaller-radius arcs, so ions of different mass land at different positions on the detector and are thereby separated.