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 2.4 AA , how many turns of wire would you need

Answer 1

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm=$8\times 10^{-2} m$

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid=$8\times 10^{-2} m$

Magnetic field of solenoid

$B=\frac{\mu_0NI}{l}$

Using the formula

$0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}$

Where $\mu_0=4\pi\times 10^{-7}$

$N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns$