Question

Consider a solenoid made of copper wire. The solenoid is 5.00 cm in diameter and 30.0 cm in length. The copper wire has a diameter of 0.50 mm and is wound with 20 turns per cm. The resistivity of Cu is 1.72 Times 10-8 Ohm *m.

a) what is the resistance of the wire in the solenoid?

b) how much voltage would need to be applied to the solenoid to create a 5.00 mT magnetic field?

Answer 1

a)
L = N*2*pi*r = (30*20)*2*3.14*0.025 = 94.2 m

cross sectional artae, A = pi*r^2 = 3.14*2.5^2*10^-8 = 1.9625*10^-7 m^2

R = rho*L/A
= 1.72*10^-8*94.2/1.9625*10^-7

= 8.256 ohms

B)
B = mue*n*I

I = B/(mue*n)
n = 20 turns/cm = 2000 turns/m

= 5*10^-3/(4*pi*10^-7*2000)

= 1.989 A

V = I*R = 1.989*8.256 = 16.42 volts