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JEE Main 2023 (10 Apr, Shift 1) · Hysteresis, Soft Iron vs Steel & Shielding

Hysteresis, Soft Iron vs Steel & Shielding question from JEE Main 2023 (10 Apr, Shift 1)

Q7JEE Main 2023 · 10 APR · SHIFT 1Easy

Hysteresis, Soft Iron vs Steel & Shielding

The current required to be passed through a solenoid of 15 cm length and 60 turns in order of demagnetise a bar magnet of magnetic intensity 2.4×103 Am−12.4 \times 10^{3}\text{ Am}^{-1} is ___________ A.

Numerical answer type. Work it out, then open the answer.

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Answer: 6

  1. The solenoid's H=nIH = nI must equal the given magnetic intensity (the coercivity): I=HnI = \dfrac{H}{n}.
  2. n=600.15=400 m−1n = \dfrac{60}{0.15} = 400\text{ m}^{-1}
  3. I=2.4×103400=6 AI = \dfrac{2.4 \times 10^{3}}{400} = 6\text{ A}
  4. So the answer is 6 A.
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