# I want to find the force on a circular wire around a solenoid using $d\mathbf{F} = Id\mathbf{l}\times \mathbf{B}$? What magnetic field do I use?

I've found the current induced in the wire due to the solenoid by using $$I = \frac{\int \vec{E} \cdot \vec{\mathrm dl}}{R}$$, but I'm confused as to whether I should use the magnetic field due to the induced current in the wire to find the force, or the field inside the solenoid ($$\vec{B} = \mu_0 n I \hat{z}$$). The solenoid current varies linearly with time, so I believe there will be a force exerted on the wire.

Any help would be greatly appreciated.

• The force must be an external force. If you take an element $d\bf{l}$ with the current $I$ in it, it must be the sum of forces created by the other elements/wires. Mar 15 at 11:42
• Where is the circular wire in relation to the solenoid? Around the outside? And around the middle of the solenoid or towards one end? Mar 15 at 11:46
• It is outside the solenoid, away from the ends. Mar 15 at 11:47

General method.

use biot savart to find the magnetic field of the solenoid in question at a position r.
Then find the dforce element on each wire element Idl in your circular wire due to that magnetic field via df=Idl x B

then integrate df where the bounds span from the circular wire in question e.g your wire element dl is dr/dt dt and you have a vector function r(t) which traces out your circular wire from e.g t=0 to t=1

The ring (circle) will tend to expand due to repulsive forces between opposite sides of the ring, as it is carrying an induced current. But there is no resultant force on the ring due just to the current that is carries.

Any resultant force on the ring must arise from an external field in which it is sitting. But outside the solenoid, around its middle, the solenoid's field is negligible! Things are different towards either end of the solenoid, because here there is a field with a component radially outwards from, or inwards towards, the solenoid's axis, and such a field does give a resultant force on a current-carrying ring placed there. [This is famously shown in the 'jumping ring' demonstration.]

This answer assumes that you are using 'solenoid' correctly – to mean a tube-shaped coil, not a 'flat' coil.

• Thank you for answering. I'm still unsure as to how I can show this mathematically. Using $d\mathbf{F} = Id\mathbf{l}\times \mathbf{B}$ gives me a non-zero expression for the force, in the radial direction. Mar 15 at 12:32
• What value are you using for $\mathbf B$? Mar 15 at 12:34
• I know that the field is zero outside the solenoid, but the field still induces an azimuthal electric field around the wire, and thus a current, so I used $\vec{B} = \mu_0 n I \hat{z}$. I'm not sure if this is correct, but the question is asking for an expression for the force. The solenoid current isn't constant either Mar 15 at 12:39
• The formula you quote is for the field inside the solenoid near the middle. So it's not the field local to your ring – which is zero. This doesn't stop the changing solenoid field inducing a current in the ring (because the flux through the ring is changing), but it does mean that there isn't a force on the ring. Mar 15 at 12:50
• Note that even if the ring were inside the solenoid near its middle, the magnetic forces on it would be radial and there would be no resultant force! Mar 15 at 13:00