Question

A solenoid 86.0 cm long has a radius of 2.40 cm and a winding of 1000 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.

Answer 1

Hello!

We can use the following equation:

`B = \mu_0 ni`

B = Magnetic field strength (T)
μ₀ = Permeability of free space (Tm/A)

n = # of loops per unit length

We are given the total number of loops (N), so divide by length:

`n = (N)/(l) = (1000)/(.86) = 1162.79`

Now, plug this value into the equation for the magnetic field strength:

`B = (4\pi * 10^(-7))(1162.79)(3.6) = \boxed{0.00526 T}`