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

Answer:

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

Explanation:

Pls see attached file for step by step calculation

Answer 2

Answer:

B = 0.025T

Explanation:

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

$B=\frac{\mu N i}{L}$         (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):

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

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