he 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 1.9 AA, how many turns of wire would you need? Express your answer using two significant figures.

Answer: 3400

Explanation:

Given

Magnetic field, B = 0.1 T

Diameter of magnet, d = 2 cm = 0.02 m

Length of magnet, l = 8 cm = 0.08 m

Current of the magnet, I = 1.9 A

Number of turns needed, N = ?

To solve this problem, we would use the formula,

N = (LB) / (μI), where

μ = 1.257*10^-6 Tm/A, so that

N = (0.08 * 0.1) / (1.257*10^-6 * 1.9)

N = 0.008 / 2.388*10^-6

N = 3350

N ~ 3400

Therefore, the number of turns of wire needed is 3400