First slide

Magnetic properties of matter

Question

A bar magnet of magnetic moment $10 {Am}^{2}$ has a cross sectional area of $2.5 \times 10^{- 4} m^{2}$ . If the intensity of magnetization of the magnet is $10^{6} A / m$ , the length of magnet is

Easy

A

0.4m
B

0.04cm
C

0.04m
D

40 cm

Solution

Given
Magnetic moment $M = 10 {Am}^{2}$
Cross sectional Area $A = 2.5 \times 10^{- 4} m^{2}$
$∴$ Intensity of magnetization $I = \frac{M}{V} (∵ V = A L)$
$I = \frac{M}{A L}$
$\Rightarrow I = \frac{10}{2.5 \times 10^{- 4} \times L}$
$\Rightarrow L = \frac{10}{2.5 \times 10^{- 4} \times 10^{6}}$
$∴ L = 4 \times 10^{- 2} m$

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