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JEE Mains · Physics · STD 12 - 5. Magnetism and matter

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

  1. A \(3\)
  2. B \(2\)
  3. C \(8\)
  4. D \(6\)
Verified Solution

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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