Question

The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same field with a solenoid of the same size, carrying a current of 1.8 AA , how many turns of wire would you need

Answer 1

Answer:

The number of turns of wire needed is 3536 turns.

Explanation:

Given;

length of the wire, L = 8 cm = 0.08 m

magnetic field on the wire, B = 0.1 T

current in the wire, I = 1.8 A

The magnetic field produced by a solenoid is calculated as;

B = μ₀ n I

where;

n is the number of turns per length = N / L

μ₀ is permeability of free space = 4π x 10⁻⁷ N/A²

$B = \frac{\mu_o N I}{L} \\\\N = \frac{BL}{\mu_o I} \\\\N = \frac{0.1 *0.08}{4\pi*10^{-7} *1.8} \\\\N = 3536.32 \ turns$

Therefore, the number of turns of wire needed is 3536 turns.