Question

A solenoid 10.0 cm in diameter and 62.1 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 6.40 mT at its center?

Answer 1

Magnetic field inside solenoid is given by:

B = u0*n*i = u0*N*i/L

Current in solenoid will be

i = B*L/(uo*N)

Using given values:

N = number of turns = 62.1 cm/0.1 cm = 621 turns

i = 6.40*10^-3*0.621/(4*pi*10^-7*621) = 5.09 Amp

Resistance of wire will be

R = rho*L/A

rho for copper wire = 1.7*10^-8 ohm-m

L = 2*pi*r*N = 2*pi*0.05*621 = 195.09 m

A = pi*d^2/4 = pi*(0.1*10^-2)^2/4 = 7.85*10^-7 m^2

R = 1.7*10^-8*195.09/(7.85*10^-7) = 4.22 ohm

Now power is given by:

P = i^2*R

P = 5.09^2*4.22

P = 109.33 W

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