Question

A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of magnetic field inside the solenoid?

Hint:

The magnetic field inside a solenoid is directly proportional to the current and the number of turns per unit length.

Answer 1

Step 1: Formula for magnetic field inside a solenoid.
The magnetic field \( B \) inside a solenoid is given by the formula: \[ B = \mu_0 n I \] where: - \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \)), - \( n \) is the number of turns per unit length of the solenoid, - \( I \) is the current passing through the solenoid.
Step 2: Finding the number of turns per unit length.
The number of turns per unit length is given by: \[ n = \frac{N}{L} \] where:
- \( N = 500 \) is the total number of turns, - \( L = 0.5 \, \text{m} \) is the length of the solenoid.
Substitute the values: \[ n = \frac{500}{0.5} = 1000 \, \text{turns/m} \] Step 3: Substituting values into the formula for \( B \).
Now substitute the values of \( n \) and \( I \) into the formula for \( B \): \[ B = (4 \pi \times 10^{-7}) \times (1000) \times (5) \] \[ B = 6.28 \times 10^{-3} \, \text{T} = 6.28 \, \text{milliTesla} \] Step 4: Conclusion.
The magnetic field inside the solenoid is \( 6.28 \times 10^{-3} \, \text{T} \), or 6.28 milliTesla.