First slide

The bar magnet

Question

A short bar magnet of magnetic moment $20 A m^{2}$ has cross sectional area of $1.5 \times 10^{- 4} m^{2}$ . If the intensity of magnetization of the magnet is $10^{5}$ A/m. The length of magnet is

Easy

A

0.33 m
B

0.13 cm
C

1.33 m
D

1.33 cm

Solution

$\begin{array}{l} I = \frac{M}{V} = \frac{M}{A \times 2 l} \\ 2 l = \frac{M}{I A} = \frac{20}{10^{5} \times 1.5 \times 10^{- 4}} = 1.33 m \end{array}$

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