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.
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.
Correct Answer: 6
Solution and Explanation
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 \( 6 \, \text{A}. \)
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