Question

A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10-2 m will have what strength magnetic field at its center

Answer 1

Magnetic field strength at the center is 2.51x10^-2T

Step-by-step explanation:

Pls see attached file for step by step calculation

Answer 2

B = 0.025T

Step-by-step explanation:

In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:

`B=(\mu N i)/(L)` (1)

μ: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 500

i: current = 4.0A

L: length of the solenoid = 0.10m

You replace the values of the parameters in the equation (1):

`B=((4\pi*10^(-7)T/A)(500)(4.0A))/(0.10m)=0.025T`

The strength of the magnetic field at the center of the solenoid = 0.025T