Question

A 200-turn solenoid having a length of 25 cm and a diameter of 10 cm carries a current of 0.29 A. Calculate the magnitude of the magnetic field B inside the solenoid

Answer 1

Hi there!

We can use the following equation to calculate the magnetic field inside of a solenoid:
$B = \mu_0 ni$

B =  Magnetic Field Strength (T)
n = number of loops PER LENGTH

i = current through solenoid (A)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

First, we can solve for 'n' given 'N' (total # of loops) and L (length of solenoid).

$n = \frac{N}{L} = \frac{200}{0.25} = 800$

Now, we can calculate the magnetic field:

$B = (4\pi \times 10^{-7}) (800)(.29) = \boxed{2.915 \times 10^{-4} T}$