Question

A 40 cm-long solenoid, 1.8 cm in diameter, is to produce a 0.40 T Part A magnetic field at its center. If the maximum current is 4.4 A, how many turns must the solenoid have? Express your answer using two significant figures.

Answer 1

The number of turns in the solenoid is approximately 28,937 turns.

How to determine the number of turns the solenoid must have?

The magnetic field within a solenoid is represented by the formula
`\( B = \mu_0 \cdot n \cdot I \)`, where
`\( B \)` is the magnetic field strength,
`\( \mu_0 \)` is the permeability of free space,
`\( n \)` stands for the number of turns in the solenoid, and
`\( I \)` denotes the current passing through the coil.

Given a magnetic field strength
`\( B = 0.40 \, \text{T} \), \( I = 4.4 \, \text{A} \)`, and the length of the solenoid
`\( L = 0.4 \, \text{m} \)`, we can determine the number of turns
`\( n \)` using the formula
`\( B = \mu_0 \cdot n \cdot I \)`.

By rearranging the formula and solving for
`\( n \)`, we find:

`\[ n = (B \cdot L)/(\mu_0 \cdot I) \]`

Substituting the given values:

`\[ n = \frac{0.40 \, \text{T} \cdot 0.4 \, \text{m}}{\mu_0 \cdot 4.4 \, \text{A}} \]`

The value of
`\( \mu_0 \) is \( 4\pi * 10^(-7) \, \text{T} \cdot \text{m/A} \)`.

`\[ n = (0.4 \cdot 0.4)/(4\pi * 10^(-7) \cdot 4.4) \]`

Calculating this yields approximately 28,937.3 turns.

Therefore, the number of turns in the solenoid is approximately 28,937 turns.