Answer:
19559303watts
Explanation:
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 = 7.8*10^-3 T, d = 0.001 m, µ0 = 4*π*10^-7 H/m
I = 8*10^-3*0.001/(4*π*10^-17))
I = 6.366A
For power
We need to know ρ, the resistivity of the wire. The resistance of the coil R = ρ*L/A where L is the total wire length and A the area.
L = N*π*D, were D = diameter of the solenoid and N the no of turns. As above, N = h/d so L = h*π*D/d. The wire area is π*d²/4 so the resistance is
R = ρ*h*π*D/d*(π*d²/4) = 4*ρ*h*D/d³
The power is I²*R = (B*d/µ0)²*4*ρ*h*D/d³ = (B/µ0)²*4*ρ*h*D/d
For copper ρ = 16.8*10^-9 Ω-m and
D = 0.10 m, h = 0.722 m
P = (6.366 / 4 * π * 10^-7)^2 * 4 * 16.8 * 10 ^-9 * 0.722 * 0.1/0.001
P = 19559303watts
P = 20MW
