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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 \(.........A\).

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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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