Question

A solenoid 10.0 cm in diameter and 84.4 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 6.50 mT at its center?

Answer 1

The field inside a solenoid is

B =¨µ0*N*I/h

where N is the number of turns, I the current in each turn, and h the length.

If the wire's diameter is d, then the number of turns is N = h/d so the equation becomes

B = ¨µ0*I/d

I = B*d/µ0

B = 6.5*10^-3 T, d = 0.001 m, µ0 = 4*?*10^-7 H/m

I = 6.5*10^-3*0.001/4*?*10^-7 = 5.1725

This gives you the current. If you really need power, you need to know ?, the resistivity of the wire. The resistance of the coil R = ?*L/A where L is the total wire length and A the area.

L = N*?*D, were D = diameter of the solenoid and N the no of turns. As above, N = h/d so L = h*?*D/d. The wire area is ?*d²/4 so the resistance is

R = ?*h*?*D/d*(?*d²/4) = 4*?*h*D/d³

The power is I²*R = (B*d/µ0)²*4*?*h*D/d³ = (B/µ0)²*4*?*h*D/d = (6.5*10^-3/4*pi)^2*4*16.8*10^-9*0.844*0.10/0.001
For copper ? = 16.8*10^-9 ?-m and
D = 0.10 m, h = 0.844 m

The power is = 151.746N