Question

A solenoid has length \(l = 0.5\) m, radius \(r = 1\) cm \(= 0.01\) m, total turns \(N = 500\), and carries current \(I = 5\) A. What is the turns per unit length?

Options

• \(n = 500\) turns/m

• \(n = 50{,}000\) turns/m

• \(n = 1000\) turns/m

• \(n = 2500\) turns/m

Answer 1

The turns per unit length is \(n = \frac{N}{l} = \frac{500}{0.5} = 1000\) turns/m. This value is then used directly in \(B = \mu_0 nI\) to compute the interior magnetic field of the solenoid.