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

Answer:

$18.8\times 10^{-3} T$

Explanation:

We are given that

Number of turns,N=500

Radius,r=0.04 m

Length of solenoid,L=40 cm=$\frac{40}{100}=0.4 m$

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=$\frac{500}{0.4}=1250$

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

Where $\mu_=0=4\pi\times 10^{-7}Tm/A$

$B=4\pi\times 10^{-7}\times 1250\times 12=18.8\times 10^{-3} T$

$B=18.8\times 10^{-3} T$