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

2653 turns

Step-by-step explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm=
`8* 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* 10^(-2) m`

Magnetic field of solenoid

`B=(\mu_0NI)/(l)`

Using the formula

`0.1=(4\pi* 10^(-7)* 2.4* N)/(8* 10^(-2))`

Where
`\mu_0=4\pi* 10^(-7)`

`N=(0.1* 8* 10^(-2))/(4\pi* 10^(-7)* 2.4)=2653 turns`