# B Solenoid self force

1. Nov 29, 2016

### RingNebula57

Hello everyone! I found in my textbook that in order to calculate the pressure experienced by the surface of a long straight solenoid with $n$ turns per unit length and current $I$ we don't use the typical magnetic field $B=\mu_0nI$ on a loop of the coil , but a corrected magnetic field $B=\frac{\mu_onI}{2}$ which excludes the contribution or "self force" of the loop. How can I prove this formula? (to calculate the pressure is not a problem if I know the formula for $B$). Furthermore, I saw that this factor $"\frac{1}{2}"$ is used in other cases also in which we have self-forces.

2. Nov 29, 2016

### Hesch

( my insertion )

1) is used for calculation of B in the center of an infinit long solenoid. So the solenoid is symmetrical around this center. The resulting B-value is a sum of the contributions from the left/right parts of the solenoid.

2) is used at the ends of the solenoid, because here only the solenoid to the left ( or to the right ) is contributing the B-value at this point. Thus the factor "½".

3. Nov 29, 2016

### RingNebula57

I am aware of the problem that you solved, but this is not my point. What I am trying to say is that we have an infinitely long solenoid and we try to calculate the tensile force acting on one of its loops ( we can assume that the loop is in the middle) due to the magnetic field of the solenoid. But when we try to do this we have to substract the contribution of the loop itself. And after substracting that we obtain $B=\frac{\mu_onI}{2}$.