Question

A solenoid 10.0 cm in diameter and 75.3 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 8.80 mT at its center?

Answer 1

Given that -

Diameter of the solenoid, d = 10 cm = 0.10 m

Hence, radius of the solenoid = r = d / 2 = 0.05 m

No. of turns n = 75.3 / 0.1 = 753

B = 8.80 mT = 8.80 x10^-3 T

B = μo n I / 2 r
Current = I = (B*2r) / (μo n) = (8.80 x10^-3 x 2 x 0.05) / (4*pi*10^-7 x 753)

= 0.93 A

Power = I^2 R
where R is the resistance of the total length of the copper wire wound round the solenoid, which has to be given .

If the resistivity of copper is assumed then ,
1.7x10^-8 = pi r^2 R / L
L = 2 pi r x 750 m
=> R = (1.7 x 10^-8 x 2 x pi x 0.05 x 750) / (3.141 x 0.05^2)

= 5.10 x 10^-4 Ohm

Therefore, the power delivered to the solenoid = I^2 * R

= 0.93^2 * 5.10 x 10^-4

= 4.41 x 10^-4 W = 0.441 mW (Answer)

Hope, you understand the solution!