
The correct relation between.$B,H\text{ and }I$.is
$\begin{align}
& \text{A}\text{. }B={{\mu }_{0}}I\times H \\
& \text{B}\text{. }B=\dfrac{{{\mu }_{0}}I}{H} \\
& \text{C}\text{. }B={{\mu }_{0}}\left( I+H \right) \\
& \text{D}\text{. }B={{\mu }_{0}}I-H \\
\end{align}$
Answer
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Hint: \[B\] represents the magnetic induction, $H$ represents the magnetizing field intensity, and $I$ represents the intensity of magnetization, for a given system. $I$ is also represented as $M$. They are all interrelated to each other. Write the expression for each of them and do some substitution to get the relationship between $B,H\text{ and }I$.
Formula used:
$\begin{align}
& B={{\mu }_{0}}ni \\
& H=ni \\
& M\text{ or }I=\dfrac{\text{magnetic moment developed}}{\text{Volume}} \\
& \text{Magnetic moment developed, } \\
& m=\text{ pole strength developed}\times \text{distance between the poles} \\
\end{align}$
Where,
$\begin{align}
& {{\mu }_{0}}=\text{permeability} \\
& \text{n=number or rotation of a solenoid}\text{.} \\
& i=\text{ current through the solenoid}\text{.} \\
\end{align}$
Complete step by step answer:
Magnetic induction ($B$): When current flows inside the wire of a solenoid some magnetic moments developed. This current $i$ induces some magnetic field inside solenoid called magnetic induction and is given by $B={{\mu }_{0}}ni$, where $n=\text{ number of turns of solenoid}$.
Magnetizing field intensity ($H$): The ability of the magnetizing field to magnetize a material medium is called magnetic field intensity. Its magnitude is defined as the number of turns of solenoid per unit length required to produce a given magnetic field,
So $H=ni$, but ${{B}_{0}}={{\mu }_{0}}ni={{\mu }_{0}}H$
The intensity of magnetization ($I$).When a magnetic material is placed in the magnetizing field it gets magnetized. The magnetic moment developed per unit volume of the material is called intensity of magnetization, thus $M=\dfrac{m}{V}$
If magnetic induction ${{B}_{m}}$ produced due to magnetizing field then magnetic field intensity.
$I={{\mu }_{0}}M$
As the total magnetic field or the magnetic induction $\overrightarrow{B}$inside a magnetic material is the resultant of magnetizing field $\overrightarrow{{{B}_{o}}}$and the field $\overrightarrow{{{B}_{m}}}$produced due to magnetization of the material. So
$\begin{align}
& B={{B}_{0}}+{{B}_{m}}={{\mu }_{0}}H+{{\mu }_{0}}I \\
& \Rightarrow B={{\mu }_{0}}(H+I) \\
\end{align}$
Hence, the correct answer is option C.
Note:
Note that the S.I unit of magnetic induction is $tesla(T)\text{ or }weber\times mete{{r}^{-2}}(Wb{{m}^{-2}})$ which is equivalent to $N{{m}^{-1}}{{A}^{-1}}\text{ or }Joule{{A}^{-1}}{{m}^{-2}}$. And the S.I unit of Magnetizing field intensity is
$Ampere\times meter{{r}^{-1}}$ which is equivalent to $\dfrac{N}{{{m}^{2}}T}\text{ or }\dfrac{Joule}{Weber\times meter}$.
Formula used:
$\begin{align}
& B={{\mu }_{0}}ni \\
& H=ni \\
& M\text{ or }I=\dfrac{\text{magnetic moment developed}}{\text{Volume}} \\
& \text{Magnetic moment developed, } \\
& m=\text{ pole strength developed}\times \text{distance between the poles} \\
\end{align}$
Where,
$\begin{align}
& {{\mu }_{0}}=\text{permeability} \\
& \text{n=number or rotation of a solenoid}\text{.} \\
& i=\text{ current through the solenoid}\text{.} \\
\end{align}$
Complete step by step answer:
Magnetic induction ($B$): When current flows inside the wire of a solenoid some magnetic moments developed. This current $i$ induces some magnetic field inside solenoid called magnetic induction and is given by $B={{\mu }_{0}}ni$, where $n=\text{ number of turns of solenoid}$.
Magnetizing field intensity ($H$): The ability of the magnetizing field to magnetize a material medium is called magnetic field intensity. Its magnitude is defined as the number of turns of solenoid per unit length required to produce a given magnetic field,
So $H=ni$, but ${{B}_{0}}={{\mu }_{0}}ni={{\mu }_{0}}H$
The intensity of magnetization ($I$).When a magnetic material is placed in the magnetizing field it gets magnetized. The magnetic moment developed per unit volume of the material is called intensity of magnetization, thus $M=\dfrac{m}{V}$
If magnetic induction ${{B}_{m}}$ produced due to magnetizing field then magnetic field intensity.
$I={{\mu }_{0}}M$
As the total magnetic field or the magnetic induction $\overrightarrow{B}$inside a magnetic material is the resultant of magnetizing field $\overrightarrow{{{B}_{o}}}$and the field $\overrightarrow{{{B}_{m}}}$produced due to magnetization of the material. So
$\begin{align}
& B={{B}_{0}}+{{B}_{m}}={{\mu }_{0}}H+{{\mu }_{0}}I \\
& \Rightarrow B={{\mu }_{0}}(H+I) \\
\end{align}$
Hence, the correct answer is option C.
Note:
Note that the S.I unit of magnetic induction is $tesla(T)\text{ or }weber\times mete{{r}^{-2}}(Wb{{m}^{-2}})$ which is equivalent to $N{{m}^{-1}}{{A}^{-1}}\text{ or }Joule{{A}^{-1}}{{m}^{-2}}$. And the S.I unit of Magnetizing field intensity is
$Ampere\times meter{{r}^{-1}}$ which is equivalent to $\dfrac{N}{{{m}^{2}}T}\text{ or }\dfrac{Joule}{Weber\times meter}$.
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