Question

A solenoid 10.0 cm in diameter and 82.7 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 8.35 mT at its center? (Assume the resistivity of the copper wire is 1.70 ✕ 10−8 Ω · m.)

Answer 1

for making solenoid, number of turns required,
n = length of solenoid / diameter of copper
= 82.7 cm / 0.100 cm
= 827 turns

B = n*miu*I/L
8.35*10^-3 = 827*(1.2566*10^-6)*I/(0.817)
I = 6.56 A

This is the current flowing in solenoid

Length of copper wire required = n*2*pi*R
= 827*2*pi*0.1
= 519.62 m

Area of copper wire = pi*(d/2)^2
= pi*(0.1*10^-2/2)^2
= 7.85*10^-7 m^2

R= rho*L/A
R = (1.7*10^-8) * (519.62) / (7.85*10^-7)
= 11.25 ohm

P = i^2 * R
= 6.56^2 * 11.25
= 484 W
Answer: 484 W