electromagnetisminductancemagnetic fieldsmagnetostatics
If a current is passed through bundle of loosely coiled, insulated wire the space between the wires rapidly contracts, crushing it down into a disk. How can this crushing force be calculated?
I know that in a vacuum the magnetic field created by a solenoid is B = μIN/l (μ = permeability of free space, N = turns, I = current, l = length), so reducing the length increases the magnetic field strength. I'm not sure how to describe the force in favour of a smaller L though.
$n$ in this case isn't the number of loops, it's the number of loops per unit length. This is why you have to multiply it by $L$ to get $N$.
Electromagnetic fields propagate at the speed of light, however in a real circuit the rate at which the electromagntic field grows is controlled by the inductance of the solenoid.
You probably know that a changing magnetic field through a conductor creates a current - this is after all the way all our electricity is generated. When you apply a voltage to a solenoid the magnetic field starts growing, but the growing magnetic field through the solenoid creates a back EMF that opposes the applied voltage. The end result is that the formula for the current in the coil is:
$$ \frac{dI}{dt} = \frac{V}{L} $$
integrating and assuming $I = 0$ when $t = 0$ gives:
$$ I = \frac{V}{L} t $$
where $L$ is the inductance of the coil. The magnetic field is proportional to the current so we get:
$$ B \propto \frac{V}{L} t $$
So the rate of increase of the field is determined by the applied voltage and the inductance of the solenoid. The value of $V/L$ gives you an idea of the timescale. It's not unusual to have inductances of $1$ H, so with a $10$ V supply the timescale can be as long as a tenth of a second.
Note that in this case the current keeps growing continually, but that's because we've assumed the coil has no resistance. Real coils have a resistance $R$, and the current plateaus as it approaches the steady state of $V/R$.
You ask about a straight wire: even a straight wire has an inductance. According to this inductance calculator a 10 cm length of copper wire has an inductance of about $10^{-7}$H, so if you apply $10$ V to the wire the value of $V/L \approx 10^8$. This is around the speed at which electrical signals in copper wire propagate (i.e. a few tenths of $c$) so in this case the field will be limited partly by the inductance and partly by the rate the signals propagate through the wire.
Response to comment:
To calculate the field for a real solenoid you need to take into account the resistance of the solenoid, so the circuit would look like:
Where I've shown the internal resistance of the solenoid as a resistor in series with a pure inductor. The inductance of the solenoid is $L$ and the internal resistance is $R$.
If you turn on a voltage $V_0$ at time $t = 0$, then the current as a function of time is given by:
$$ I = \frac{V_0}{R} \left( 1 - \exp (-t \tfrac{R}{L}) \right) $$
If we consider a simple solenoid then the flux density is related to the inductance by:
$$ B = \frac{I}{nA} = \frac{V_0}{RnA} \left( 1 - \exp (-t \tfrac{R}{L}) \right) $$
where $n$ is the number of turns per unit length and $A$ is the area of the coil. This equation allows you to feed in your inductance and internal resistance and calculate how the flux density changes with time.
Best Answer
(a) First we give an approximate force-based treatment for a solenoid whose length, $l$, is several times greater than its radius, $r$.
Consider the North end of the solenoid. There's a very simple argument to show that half the flux in the main part of the solenoid (that is half of $\pi r^2\ \mu_0 N I/l$) escapes through the sides of the solenoid before reaching the geometrical end of the solenoid. Therefore if $B_{rad}$ is the radial component of flux density at the surface of the solenoid at distance $z$ from its end, $$\int_0^{z_0} B_{rad}\ 2\pi r dz =\tfrac 12 \frac{\pi r^2\ \mu_0 N I}l.$$ in which $z_0$ is the notional distance from the geometrical end at which there is no more radial escape of flux.
However in length $dz$ of solenoid there are $dz\ (N/l)$ turns, so a length $2\pi r\ dz\ (N/l)$ of wire. Therefore the total (axial) force on the North end part of the solenoid due to the radial field will be $$\int_0^{z_0} B_{rad}\ \ I\ 2\pi r \tfrac {N}l dz\ \ =\ \tfrac {NI}l \int_0^{z_0} B_{rad}\ 2\pi r dz\ =\ \frac{\pi r^2\ \mu_0 N^2 I^2}{2l^2}.$$
This force acts towards the middle (lengthways) of the solenoid; a similar force acts on the South Pole end of the solenoid, so there is a compressive stress on the solenoid.
[It might be argued that, rather than $2\pi r$, the length of a turn is $\sqrt{(2\pi r)^2+(l/N)^2}$. So indeed it is, but the axial component, $(l/N)\hat z$, of the turn gives rise in the presence of $B_{rad}$ to tangential forces (that cancel over a complete turn anyway) rather than to an axial magnetic force!]
(b) The result just obtained agrees with what we get using $|F_{\text{axial}}|= -\frac {\partial U}{\partial l}$ in which $U$ is the magnetic energy stored in the solenoid. So, (ironically) ignoring end-effects and treating $B$ as uniform throughout the interior volume of the solenoid, $$U=\frac 1{2\mu_0} B^2 \pi r^2 l=\frac 1{2\mu_0} \left( \frac {\mu_0 NI}l\right)^2 \pi r^2 l=\frac{\mu_0 N^2 I^2}{2l}\pi r^2.$$
$$\text{so}\ \ \ \ |F_{\text{axial}}|= -\frac {\partial U}{\partial l}=\frac{\mu_0 N^2 I^2}{2l^2}\pi r^2.$$ The irony is that in this method, (b), we have ignored what was the very basis of the force method, (a) – end effects! The reason it is permissible in this method is our assumption that $l>>r$. The end effects (escape of flux through the sides of the solenoid) are pretty much confined to end lengths $z_0$ of solenoid of the same order as the solenoid radius, $r$.