Question

A solenoid 10.0 cm in diameter and 63.3 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 8.00 mT at its center

Answer 1

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

Answer 2

Answer:

The power is $174W$

Explanation:

diameter:   $d=0.1cm=0.1*10^{-2} m$

Radius:  $r=\frac{d}{2} ,r=5*10^{-4} m$

Length:  $l=63.3cm=0.633m$

The n will be:

$n=\frac{1m}{0.1*10^{-2} m} \\n=1000turns/m$

N=n*l

N=1000*0.633

$N=6.33*10^{2} turns$

To find current:

$I=\frac{B}{n_{0}*n } \\I=\frac{8*10^{3} }{4\pi*1000 } \\I=6.37A$

To find L:

$L=N*\pi d\\L=(6.33*10^{2} )*(\pi .10*10^{-2}m) \\L=63.3\pi$

Now we can apply resistivity formula to find resistance:

$R=p\frac{L}{A} \\R=\frac{(1.7*10^{-8}) *(63.3\pi )}{\pi(5*10^{-4} \\)^{2} }$

$R=4.3$Ω

Now to find the power:

$Power=I^{2} R\\Power=6.37^{2}*4.3\\ Power=174W$