JEE Main 20204 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 4 September
Shift: Evening Shift
Subject: Physics
Chapter: Magnetic Properties of Matter

JEE Main 2020 · 4 September · Evening Shift · Physics · Magnetic Properties of Matter

A paramagnetic sample shows a net magnetisation of 6A / m when it...

A paramagnetic sample shows a net magnetisation of

6A / m

when it is placed in an external magnetic field of

0 . 4T

at a temperature of

4K

. When the sample is placed in an external magnetic field of

0 . 3T

at a temperature of

24K

, then the magnetisation will be :

Options

A $1A / m$
B $4A / m$
C $2 . 25A / m$
D $0 . 75A / m$

Correct answer

D. $0 . 75A / m$

Step-by-step solution

According to Curie's Law,
$χ ∝ \dfrac{1}{T}$
Therefore, ${χ}_{1}{T}_{1} = {χ}_{2}{T}_{2}$
Also, $χ = \dfrac{M}{B}$
So,
$\dfrac{{M}_{1}{T}_{1}}{{B}_{1}} = \dfrac{{M}_{2}{T}_{2}}{{B}_{2}}$
On substituting the values,
$\dfrac{6 \times 4}{0 . 4} = \dfrac{{M}_{2} \times 24}{0 . 3}$

Therefore, ${M}_{2} = 0 . 75A{m}^{- 1}$

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