Question

A solenoid with 500 turns has a radius of 0.040 m and is 40 cm long. If this solenoid carries a current of 12 A, what is the magnitude of the magnetic field near the center of the solenoid? (μ 0 = 4π × 10-7 T · m/A)

Answer 1

`18.8* 10^(-3) T`

Step-by-step explanation:

We are given that

Number of turns,N=500

Radius,r=0.04 m

Length of solenoid,L=40 cm=
`(40)/(100)=0.4 m`

1 m=100 cm

Current,I=12 A

We have to find the magnitude of magnetic field near the center of the solenoid.

Number of turns per unit length,n=
`(500)/(0.4)=1250`

Magnetic field near the center of the solenoid,B=
`\mu_0 nI`

Where
`\mu_=0=4\pi* 10^(-7)Tm/A`

`B=4\pi* 10^(-7)* 1250* 12=18.8* 10^(-3) T`

`B=18.8* 10^(-3) T`