Posted by Mary on .
A piece of copper wire has a resistance per unit length of 6.50 10-3 /m. The wire is wound into a thin, flat coil of many turns that has a radius of 0.170 m. The ends of the wire are connected to a 12.0 V battery. Find the magnetic field strength at the center of the coil.
B = (4pi x 10^-7)(12.0) / (6.5 x 10^-3)(4pi (0.170^2))
B = 0.000015082 / 0.002360899
B = 6.39 x 10^-3T
Physic please check -
B = mu H = (4 pi*10^-7)*i N/(2 r)
N is the number of turns.
r is the coil radius
i is the current
i = V/R
R = N* 2 pi r * (6.5*10^-3 ohm/m)
B = [(4 pi*10^-7)*V*N]/[(N*4*pi*r^2*(6.5*10^-3)]
= 10^-7 V/[r^2*(6.5*10^-3 ohm/m)]
The 4 pi factors and the number of turns N both cancel out. I agree with your formula and answer.