The coercivity of a magnet is 5 × 10³ A/m. The amount of current required to be passed in a solenoid of length 30 cm and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A. Numerical answer type. Work it out, then open the answer.

Answer: 10. The solenoid's H = nI must equal the coercivity: I = H/n. n = (150)/(0.30) = 500 m⁻¹ I = 5 × 10³500 = 10 A So the answer is 10 A.

The coercivity of a magnet is 5 × 10³ A/m. The amount of current… | JEE Main 2024 (8 Apr, Shift 2)

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Hysteresis, Soft Iron vs Steel & Shielding

The coercivity of a magnet is $5 \times 10^{3}\text{ A/m}$ . The amount of current required to be passed in a solenoid of length $30\text{ cm}$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

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

Show answer and solution

Answer: 10

The solenoid's $H = nI$ must equal the coercivity: $I = \dfrac{H}{n}$ .
$n = \dfrac{150}{0.30} = 500\text{ m}^{-1}$
$I = \dfrac{5 \times 10^{3}}{500} = 10\text{ A}$
So the answer is 10 A.

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