Question

A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 4 A. If the number of turns is 500 per metre, then the magnetizing field is:

• A. \( 2\pi \times 10^3 \, \text{Am}^{-1} \)
• B. \( 1 \times 10^3 \, \text{Am}^{-1} \)
• C. \( 4 \times 10^3 \, \text{Am}^{-1} \)
• D. \( 2 \times 10^3 \, \text{Am}^{-1} \)

Hint:

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

Answer 1

The Correct Option is D

The magnetizing field \( H \) is given by the formula: \[ H = \frac{N \cdot I}{L} \] where:

• \( N = 500 \, \text{turns/m} \) is the number of turns per metre,
• \( I = 4 \, \text{A} \) is the current,
• \( L = 1 \, \text{m} \) is the length of the solenoid.

Substituting the values into the formula: \[ H = \frac{500 \times 4}{1} = 2000 \, \text{Am}^{-1} \] Thus, the magnetizing field is \( 2 \times 10^3 \, \text{Am}^{-1} \).

Answer 2

Step 1: Understand the concept of magnetizing field.
The magnetizing field (also called magnetic field intensity) \( H \) inside a solenoid is given by the formula:
\[ H = nI \] where:
- \( n \) = number of turns per unit length (turns/m),
- \( I \) = current through the solenoid (A).

Step 2: Use the given data.
- Number of turns per metre, \( n = 500 \, \text{turns/m} \)
- Current, \( I = 4 \, \text{A} \)

Step 3: Calculate the magnetizing field.
\[ H = nI = 500 \times 4 = 2000 \, \text{A/m} \]

Note: Relative permeability is not needed here because the magnetizing field \( H \) is independent of the material and depends only on coil parameters.

Step 4: Conclusion.
The magnetizing field is \( \boxed{2 \times 10^3 \, \text{A/m}} \).