Physics · Magnetism and Matter

JEE Main 2024 — 8 April, Shift 2 — Question 48

The coercivity of a magnet is $5 \times 10^{3} \mathrm{~A} / \mathrm{m}$ . 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 $\qquad$ A.

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{H}_{\mathrm{c}}=\frac{\mu_{\mathrm{o}} \mathrm{ni}}{\mu_{\mathrm{o}}}$ $5 \times 10^{3}=\frac{150}{30} \times 100 \times i$

\frac{50}{5}=\mathrm{i}

$\mathrm{I}=10$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Physics
Chapter: Magnetism and Matter
Topic: Magnetization and Magnetic Intensity, Hysteresis