
A magnetizing force of $360A{m^{ - 1}}$ produces a magnetic flux density of 0.6T in a ferromagnetic material. The susceptibility of the material is:
$
(a){\text{ 1625}} \\
(b){\text{ 1329}} \\
(c){\text{ 2105}} \\
(d){\text{ 1914}} \\
$
Answer
241.5k+ views
Hint – In this question use the direct relationship between magnetization force and magnetic flux density that is $B = {\mu _o}\left( {H + M} \right)$, where B is the magnetic flux density, H is magnetizing force, $\chi $is magnetic susceptibility and is magnetic vacuum permeability.
Complete step-by-step answer:
Given data:
Magnetization force (H) = 360 A/m.
Magnetic flux density (B) = 0.6 T
Now as we know the relation between magnetization force and magnetic flux density which is given as,
$ \Rightarrow B = {\mu _o}\left( {H + M} \right)$................ (1), where ${\mu _o} = $vacuum permeability = $4\pi \times {10^{ - 7}}$(H/m), and M = magnetization of ferromagnetic material in (A/m).
Now as we know that magnetization is susceptibility ($\chi $) time’s magnetic field strength.
$ \Rightarrow M = \chi H$, both M and H have equal units therefore $\chi $ has dimensionless.
Now substitute this value in equation (1) we have,
$ \Rightarrow B = {\mu _o}\left( {H + \chi H} \right)$
\[ \Rightarrow 0.6 = \left( {4\pi \times {{10}^{ - 7}}} \right)\left( {360 + 360\chi } \right)\]
\[ \Rightarrow 1 + \chi = \dfrac{{0.6}}{{\left( {4\pi \times {{10}^{ - 7}}} \right) \times 360}}\]
\[ \Rightarrow \chi = \dfrac{{0.6}}{{\left( {4 \times \dfrac{{22}}{7} \times {{10}^{ - 7}}} \right) \times 360}} - 1\] $\left[ {\because \pi = \dfrac{{22}}{7}} \right]$
\[ \Rightarrow \chi = \dfrac{{0.6}}{{\left( {4 \times \dfrac{{22}}{7} \times {{10}^{ - 7}}} \right) \times 360}} - 1 = 1325.75 - 1\]
\[ \Rightarrow \chi = 1324.75 \simeq 1329\]
So this is the required answer.
Hence option (B) is the correct answer.
Note – When a material is placed in a magnetic field it tends to get magnetized, so magnetic susceptibility is the measure of how much an object in that field will be able to get magnetized. Magnetic susceptibility may also be written as $\dfrac{M}{H}$ where M is the magnetic moment per unit volume and where H is the intensity of the magnetizing field.
Complete step-by-step answer:
Given data:
Magnetization force (H) = 360 A/m.
Magnetic flux density (B) = 0.6 T
Now as we know the relation between magnetization force and magnetic flux density which is given as,
$ \Rightarrow B = {\mu _o}\left( {H + M} \right)$................ (1), where ${\mu _o} = $vacuum permeability = $4\pi \times {10^{ - 7}}$(H/m), and M = magnetization of ferromagnetic material in (A/m).
Now as we know that magnetization is susceptibility ($\chi $) time’s magnetic field strength.
$ \Rightarrow M = \chi H$, both M and H have equal units therefore $\chi $ has dimensionless.
Now substitute this value in equation (1) we have,
$ \Rightarrow B = {\mu _o}\left( {H + \chi H} \right)$
\[ \Rightarrow 0.6 = \left( {4\pi \times {{10}^{ - 7}}} \right)\left( {360 + 360\chi } \right)\]
\[ \Rightarrow 1 + \chi = \dfrac{{0.6}}{{\left( {4\pi \times {{10}^{ - 7}}} \right) \times 360}}\]
\[ \Rightarrow \chi = \dfrac{{0.6}}{{\left( {4 \times \dfrac{{22}}{7} \times {{10}^{ - 7}}} \right) \times 360}} - 1\] $\left[ {\because \pi = \dfrac{{22}}{7}} \right]$
\[ \Rightarrow \chi = \dfrac{{0.6}}{{\left( {4 \times \dfrac{{22}}{7} \times {{10}^{ - 7}}} \right) \times 360}} - 1 = 1325.75 - 1\]
\[ \Rightarrow \chi = 1324.75 \simeq 1329\]
So this is the required answer.
Hence option (B) is the correct answer.
Note – When a material is placed in a magnetic field it tends to get magnetized, so magnetic susceptibility is the measure of how much an object in that field will be able to get magnetized. Magnetic susceptibility may also be written as $\dfrac{M}{H}$ where M is the magnetic moment per unit volume and where H is the intensity of the magnetizing field.
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