Question

A current of \(2.00 \mathrm{~A}\) is flowing through a 1000 -turn solenoid of length \(L=40.0 \mathrm{~cm} .\) What is the magnitude of the magnetic field inside the solenoid?

Short answer

Answer: The magnitude of the magnetic field inside the solenoid is approximately \(6.28 × 10^{-4} \, T\).

Step-by-step solution

Identify given variables and constants

We are given the following information: - Current (\(I\)): \(2.00 \, A\) - Number of turns (\(N\)): \(1000\) - Length of the solenoid (\(L\)): \(40.0 \, cm = 0.4 \, m\) - Permeability of free space (\(\mu_0\)): \(4 \pi × 10^{-7} \, Tm/A\)

Calculate the number of turns per unit length (n)

To find the number of turns per unit length (n), we'll use the formula \(n = \frac{N}{L}\). Using the given values: $$ n = \frac{1000}{0.4} = 2500 \, turns/m $$

Calculate the magnitude of the magnetic field (B)

Now, we can use the formula for the magnetic field inside a solenoid: \(B = \mu_0 n I\). Plug in the values and calculate B: $$ B = (4 \pi × 10^{-7} \, Tm/A) × (2500 \, turns/m) × (2 \, A) \\ B = (4 \pi × 10^{-7} \, Tm/A) × 5000 \, A/m \\ B \approx 6.28 × 10^{-4} \, T $$ So, the magnitude of the magnetic field inside the solenoid is approximately \(6.28 × 10^{-4} \, T\).