Question

A bar magnet has pole strength 5 ampere-meter,
magnetic length 20 cm and cross-sectional area
2.5 cm? then its magnetic dipole moment and
magnetisation are
(1) 1 A m2, 4 x 104 A/m
(2) 1 A m2, 2 x 104 A/m
(3) 100 A m2, 2 A/m
(4) 7.5 A m2, 5 x 104 A/m​

Answer 1

The magnetic dipole moment is 2 Am²

The magnetization is $4\times10^{4}\ A/m$

Explanation:

Given that,

Pole strength = 5 Am

Length = 20 cm

Area = 2.5 cm²

We need to calculate the magnetic dipole moment

Using formula of magnetic moment

$m=\dfrac{M}{2l}$

$M=m\times 2l$

Where, m = pole strength

M = magnetic moment

l = length

Put the value into the formula

$M=5\times2\times20\times10^{-2}$

$M=2\ Am^2$

We need to calculate the magnetization

Using formula of magnetization

$M_{z}=\dfrac{M}{l\times A}$

$M_{z}=\dfrac{2}{20\times10^{-2}\times2.5\times10^{-4}}$

$M_{z}=40000\ A/m$

$M_{z}=4\times10^{4}\ A/m$

Hence, The magnetic dipole moment is 2 Am²

The magnetization is $4\times10^{4}\ A/m$

Learn more :

Topic : magnetization

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