Question

A solenoid of length \(0.256 \mathrm{m}\) and radius \(2.0 \mathrm{cm}\) has 244 turns of wire. What is the magnitude of the magnetic field well inside the solenoid when there is a current of \(4.5 \mathrm{A}\) in the wire?

Short answer

Answer: The magnitude of the magnetic field inside the solenoid is approximately 5.39 x 10^{-3} T.

Step-by-step solution

Identify the given values

We are given the following values: - Length of the solenoid \((L) = 0.256 \,\text{m}\) - Radius of the solenoid \((R) = 2.0 \,\text{cm} = 0.020 \,\text{m}\) - Number of turns of wire \((n) = 244\) - Current in the wire \((I) = 4.5 \,\text{A}\)

Calculate the number of turns per unit length

To find the magnetic field inside the solenoid, we first need to find the number of turns per unit length \((n')\). This value is found by dividing the number of turns by the length of the solenoid: \(n' = \frac{n}{L}\) Plugging in the given values, we find: \(n' = \frac{244}{0.256 \,\text{m}} \approx 953.13 \, \text{turns/m}\)

Apply the formula for the magnetic field inside a solenoid

The formula for the magnetic field inside a solenoid is given by: \(B = \mu_0 n' I\) Where \(B\) is the magnetic field, \(\mu_0\) is the magnetic permeability of free space, and \(I\) is the current in the wire. The magnetic permeability of free space is a constant equal to \(\mu_0 = 4\pi \times 10^{-7} \, \text{Tm/A}\). Now, we plug in the values: \(B = (4\pi \times 10^{-7} \, \text{Tm/A})(953.13 \, \text{turns/m})(4.5 \,\text{A})\)

Calculate the magnetic field

Performing the multiplication, we obtain the magnitude of the magnetic field inside the solenoid: \(B \approx (4\pi \times 10^{-7} \, \text{Tm/A})(953.13 \, \text{turns/m})(4.5 \,\text{A})\) \(B \approx 5.39 \times 10^{-3} \, \text{T}\) Thus, the magnitude of the magnetic field well inside the solenoid when there is a current of \(4.5 \,\text{A}\) in the wire is approximately \(5.39 \times 10^{-3} \, \text{T}\).