How fast does it take to create a magnetic field in a solenoid?

If a solenoid/electromagnet has current flowing, it creates a magnetic field. Electricity is very very fast, I believe close to the speed of light?

So, when power is given to a solenoid with $n$ turns, how fast is the magnetic field created? Similarly the same question, but for the case of a wire.

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.

• Cool :-) Though it's possible the magnet's control circuit is slowly ramping up the power. It may not just be the inductance of the coil. – John Rennie Apr 2 '14 at 14:21
• Well, the thing is I'm trying to use your method to calculate the time manually. Sorry deleted my comment by mistake. – Pupil Apr 2 '14 at 14:23
• Hows is it that L = 1 H and V = 10 , is 0.1 Seconds? Im confused a bit. – Pupil Apr 2 '14 at 14:24
• $B \propto Vt/L$, so if $V = 10$ and $L = 1$, $B \propto 10t$ and the field increases at ten times the rate time does. The time constant really comes from the solution to the equation when resistance is present, in which case $B$ tends to its maximum value exponentially. – John Rennie Apr 2 '14 at 14:28
• John, for the case of current flowing in a conductor its much faster than creating the magnetic field correct? Let assume, a conductor(any size,shape,etc...) has 10(or... 100,1000,10000) Amps flowing within, it's much quicker than the magnetic field being induced due to the self inductance and the time needed for it to stabilize. – Pupil Apr 13 '14 at 19:22