Question

A rod of ferromagnetic material with dimensions $10cm\times 0.5cm\times 0.2cm$ is placed in a magnetic field of strength $0.5cm\times {{10}^{4}}A{{m}^{-1}}$ as a result of which a magnetic moment of $5A{{m}^{-2}}$ is produced in the rod. The value of magnetic induction will be
A. 0.358 T
B. 0.54 T
C. 6.28 T
D. 2.519 T

Answer 1

Hint: The value of magnetic induction is given as $B={{\mu }_{0}}\left( H+I \right)$. Use this formula to find the magnetic induction. H is the magnetic field strength of the external magnetic field. I is the magnetization and given as $I=\dfrac{M}{V}$. M is a magnetic moment and V is the volume of the material.

Formula used:
$B={{\mu }_{0}}\left( H+I \right)$
$I=\dfrac{M}{V}$

Complete step-by-step answer:
When a ferromagnetic material is placed in an external magnetic field. The external magnetic induces a magnetic field inside the ferromagnetic material.
The magnetic induction is denoted by B. The value of the magnitude of the induced magnetic field is equal to $B={{\mu }_{0}}\left( H+I \right)$ …. (i).
Here, ${{\mu }_{0}}$ is a constant called the permeability of free space. The value of ${{\mu }_{0}}$is equal to $4\pi \times {{10}^{-7}}N{{A}^{-2}}$.
H is the magnetic field strength of the external magnetic field. I is the magnetization of the material.
In the given case, the value of magnetic field strength is equal to $H=0.5cm\times {{10}^{4}}A{{m}^{-1}}$ .
Magnetization of a material is defined as the magnetic moment induced in one unit volume of the material. Therefore, magnetization $I=\dfrac{M}{V}$, where M is the magnetic moment produced in the material and V is the volume of the material.
Here, the magnetic moment $M=5A{{m}^{-2}}$ and the volume of the cuboidal material is equal to $V=10cm\times 0.5cm\times 0.2cm=1c{{m}^{3}}$
$1cm={{10}^{-2}}m\Rightarrow 1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}$
Therefore, $V={{10}^{-6}}{{m}^{3}}$.
This gives us that $I=\dfrac{M}{V}=\dfrac{5A{{m}^{-2}}}{{{10}^{-6}}{{m}^{3}}}=5\times {{10}^{6}}A{{m}^{-5}}$
Substitute the value of ${{\mu }_{0}}$, H and I in equation (i).
Therefore, we get
$B=4\pi \times {{10}^{-7}}\left( 0.5\times {{10}^{4}}+5\times {{10}^{6}} \right)=6.28T$.
The SI unit of magnetic field is tesla (T).
Hence, the correct option is C.

Note: Do not confuse between magnetic field strength and magnetic field.
Magnetic field is the magnetic force exerted on a unit charge placed. When we divide the magnetic field by the permeability of free space, we get the magnetic field strength.