JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2023
- Medium
- Numerical
The current required to be passed through a solenoid of \(15\,cm\) length and 60 turns in order to demagnetise a bar magnet of magnetic intensity \(2.4 \times 10^3\,Am ^{-1}\) is \(.........A\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
\(I = H\)
Given, \(I =2.4 \times 10^3\,A / m\)
\(2.4 \times 10^3= H = ni\)
\(n =\frac{ N }{\ell}\)
\(2.4 \times 10^3=\frac{60}{15 \times 10^{-2}} i\)
\(i =\frac{2.4 \times 15 \times 10}{60}=\frac{36}{6}=6\,A\)
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