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Question: A rod of ferromagnetic material with dimensions \(10cm\times 0.5cm\times 0.2cm\) is placed in a magn...

A rod of ferromagnetic material with dimensions 10cm×0.5cm×0.2cm10cm\times 0.5cm\times 0.2cm is placed in a magnetic field of strength 0.5cm×104Am−10.5cm\times {{10}^{4}}A{{m}^{-1}} as a result of which a magnetic moment of 5Am−25A{{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

Explanation

Solution

The value of magnetic induction is given as B=μ0(H+I)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=MVI=\dfrac{M}{V}. M is a magnetic moment and V is the volume of the material.

Formula used:
B=μ0(H+I)B={{\mu }_{0}}\left( H+I \right)
I=MVI=\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=μ0(H+I)B={{\mu }_{0}}\left( H+I \right) …. (i).
Here, μ0{{\mu }_{0}} is a constant called the permeability of free space. The value of μ0{{\mu }_{0}}is equal to 4π×10−7NA−24\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×104Am−1H=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=MVI=\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=5Am−2M=5A{{m}^{-2}} and the volume of the cuboidal material is equal to V=10cm×0.5cm×0.2cm=1cm3V=10cm\times 0.5cm\times 0.2cm=1c{{m}^{3}}
1cm=10−2m⇒1cm3=10−6m31cm={{10}^{-2}}m\Rightarrow 1c{{m}^{3}}={{10}^{-6}}{{m}^{3}}
Therefore, V=10−6m3V={{10}^{-6}}{{m}^{3}}.
This gives us that I=MV=5Am−210−6m3=5×106Am−5I=\dfrac{M}{V}=\dfrac{5A{{m}^{-2}}}{{{10}^{-6}}{{m}^{3}}}=5\times {{10}^{6}}A{{m}^{-5}}
Substitute the value of μ0{{\mu }_{0}}, H and I in equation (i).
Therefore, we get
B=4π×10−7(0.5×104+5×106)=6.28TB=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.