Question

A solenoid 10.0 cm in diameter and 83.3 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 9.15 mT at its center? (Assume the resistivity of the copper wire is 1.70 ✕ 10−8 Ω · m.)

___________ W

Answer 1

The total number of turns on the solenoid is, N = nl = (1.0 times 10^3/m) (0.833m) = 8.33 times 10^2 The expression for the magnetic field inside the solenoid is, B = mu_0 nI Here, I is the current in the solenoid, n is the number of turns per unit length of the solenoid, and mu_0 is the permeability of free space. Solve the equation for current. B = mu_0 nI I = B/mu_0 n = 9.15 times 10^-3 T/(4 pi times 10^-7 T middot m/A) (1.0 times 10^3/m) = 7.28 A The length of the wire required for the solenoid is, L = N (pi d_solenoid) = (8.33 times 10^2) pi (0.10m) = 83.3 pi m The resistance of the wire is, R = pL/A = pL/(pi (d_wire^2/4) = 4 (1.70 times 10^-8Ohm middot m) (83.3 pi m)/pi (1.0 times 10^-3 m)^2 = 5.664 Ohm The power delivered to the solenoid is, P = I^2