# Current in wire to set magnetic flux

Discussion in 'Homework Help' started by smarch, Jul 4, 2010.

1. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
A solenoid with a circular cross-section has a length of 10 cm, a radius of 3
cm and is wound with 2,000 turns per metre of wire. How much current must
flow in the wire to set up a magnetic flux density of 1 T at the centre of the
solenoid, assuming that the inside of the solenoid is filled with air?

I know I use the formulas H= nI/2 [cosθ_1- cosθ_2 ] and B = μ0μrH.

But where do I get θ from?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
Presumably at the geometric centre θ_1=θ_2=arctan(3/5)

3. ### JimmyB Member

Jun 1, 2010
38
0
Hi

I tried it like this

L=0.1m CSA=0.0028m2 N=200

Found S 1st S= L = 0.1 =28420525 A/Wb
.....................μ A (4∏x10-7)(0.0028)

Wb=BA = 1x0.0028 = 0.0028Wb

m.m.f = Wb x S = 0.0028 x 28420525 = 79577 A/t

therefore I = F = 79577 = 397A
...................N 200

or

H = B = 1 =795774.7 A/m
......U 4pi x 10-7

F = HL = 795774.7 x 0.1 = 79577 A/t

I = F = 79577 = 397 A
.....N 200

I think this is correct, but I would imagine a 2.5mm2 wire with nearly 400A flowing being quite hot!!!!!!
you might want to check the answer with someone...

Last edited: Jul 8, 2010
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
I believe the formula for B along the coil axis would be of the form

$B=\frac{\mu_o NI}{2L}$cos(\theta_1)-cos(\theta_2)$$

Where L is the coil length and the angles are those subtended from the coil axis by a line extended from the point of interest to either extremity of the coil end edges

Contrary to my earlier post I would have

θ1=Pi-atan(3/5) and θ2=atan(3/5)

I get an answer of 46.4A to achieve the required flux density of 1T.