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 \(6.28 \times 10^{-4} \mathrm{T}\).

Step-by-step solution

Identify the given values

We are given the following values: - Current (I) = 2.00 A - Number of turns (n) = 1000 turns - Length (L) = 40.0 cm = 0.4 m (converted to meters) We now need to find the magnetic field (B) inside the solenoid.

Calculate the number of turns per unit length (n')

To find the magnetic field inside the solenoid, we need to find the number of turns per unit length (n'), which can be calculated as follows: n' = n / L n' = 1000 turns / 0.4 m = 2500 turns/m

Calculate the magnetic field

Now that we have the number of turns per unit length (n'), we can find the magnetic field (B) inside the solenoid using the formula: B = 𝜇ₒ * n' * I Here, 𝜇ₒ (magnetic permeability of free space) is a constant with the value \(4\pi \times 10^{-7} \mathrm{Tm/A}\). Now, substitute the values: B = \((4\pi \times 10^{-7} \mathrm{Tm/A})\) * (2500 turns/m) * (2.00 A) B = \(2\pi \times 10^{-4} T\) = \(6.28 \times 10^{-4} T\) Hence, the magnitude of the magnetic field inside the solenoid is \(6.28 \times 10^{-4} \mathrm{T}\).