Question

A 10.5 cm long solenoid contains 891 turns and carries a current of 5.95 A . What is the strength of the magnetic field at the center of this solenoid?

Answer 1

Answer:

B=0.06345T

Explanation:

The magnetic field at the center of the solenoid can be calculated with the formula:

$B=\frac{\mu_0NI}{L}$

where $\mu_0=4\pi\times10^{-7}N/A^2$ is the vacuum permeability, and for our values is:

$B=\frac{(4\pi\times10^{-7}N/A^2)(891)(5.95A)}{(0.105m)}=0.06345T$