A bar magnet is demagnetized by inserting it inside a solenoid of length $0.2 \mathrm{~m}, 100$ turns, and carrying a current of 5.2 A . The coercivity of the bar magnet is

(B) $2600 \mathrm{~A} / \mathrm{m}$

(A) $285 \mathrm{~A} / \mathrm{m}$

(C) $520 \mathrm{~A} / \mathrm{m}$

(D) $1200 \mathrm{~A} / \mathrm{m}$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Magnetic Properties of Matter

A bar magnet is demagnetized by inserting it inside a solenoid of length

0.2 m, 100

turns, and carrying a current of 5.2 A . The coercivity of the bar magnet is

(A) $285 A / m$
(B) $2600 A / m$ ✓ Correct
(C) $520 A / m$
(D) $1200 A / m$

Answer: (B)

Solution

Inside a solenoid the magnetising field is

H = ni

, where

n = \frac{N}{L} = \frac{100}{0.2} = 500

turns per metre. The field needed to wipe out the magnetisation sets the coercivity, so

H = 500 \times 5.2 = 2600

A/m.

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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