Question

The current required to be passed through a solenoid of 15 cm length and 60 turns in order to demagnetize a bar magnet of magnetic intensity \(2.4 × 10^3Am^{−1}\) is A.

Answer 1

Correct Answer: 6

Step 1: Formula for Magnetic Intensity

The magnetic intensity \( H \) inside a solenoid is given by:

\[ H = \frac{N I}{l} \]

Where:

• \( N \): Number of turns (60)
• \( I \): Current in Amperes
• \( l \): Length of the solenoid (15 cm = 0.15 m)

Step 2: Rearrange to Solve for \( I \)

Rearrange the formula to solve for \( I \):

\[ I = \frac{H l}{N} \]

Step 3: Substitute the Values

Substitute \( H = 2.4 \times 10^3 \, \text{A/m} \), \( l = 0.15 \, \text{m} \), and \( N = 60 \):

\[ I = \frac{(2.4 \times 10^3)(0.15)}{60} \]

Simplify the expression:

\[ I = \frac{360}{60} = 6 \, \text{A} \]

Final Answer:

The current required to demagnetize the bar magnet is \( 6 \, \text{A}. \)