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Question: The magnetic susceptibility of annealed iron at saturation is \[4224\] . Find the permeability of an...

The magnetic susceptibility of annealed iron at saturation is 42244224 . Find the permeability of annealed iron at saturation. (μ0=4π×10−7 SI unit)\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,SI\,unit} \right)
A. 5.31×10−3TmA5.31 \times {10^{ - 3}}T\dfrac{m}{A}
B. 5.71×10−4TmA5.71 \times {10^{ - 4}}T\dfrac{m}{A}
C. 6.8×10−3TmA6.8 \times {10^{ - 3}}T\dfrac{m}{A}
D. 3.78×10−3TmA3.78 \times {10^{ - 3}}T\dfrac{m}{A}

Explanation

Solution

Susceptibility to magnetic fields When an external magnetic field is applied to a material, it is measured in terms of how much it will be magnetised. It's a dimensional-less quantity created by electrons and nuclei interacting with an external magnetic field. It is the ratio of magnetization field intensity HH in mathematics. The final answer will be obtained by substituting the value of susceptibility in the form of permeability.

Formula used:
Relation between permeability and susceptibility:
μm=μ0(1+χ){\mu _m} = {\mu _0}\left( {1 + \chi } \right)
Where μ\mu is absolute permeability, μ0{\mu _0} is the permeability in free space that is equal to μ0=4π×10−7 {\mu _0} = 4\pi \times {10^{ - 7}}\, and χ\chi is the magnetic susceptibility.

Complete step by step answer:
Given that susceptibility of annealed iron at saturation is 42244224 .i.e. χ=4224\chi = 4224
We have formula for magnetic permeability of material as-
μm=μ0(1+χ){\mu _m} = {\mu _0}\left( {1 + \chi } \right)
⇒μm=4π×10−7(1+4224)\Rightarrow {\mu _m} = 4\pi \times {10^{ - 7}}\left( {1 + 4224} \right)
Simplifying the expression we get,
{\mu _m}= 16900\pi \times {10^{ - 7}} \\\ \Rightarrow {\mu _m}= 169 \times 3.142 \times {10^{ - 5}} \\\
carrying out the calculations we get
μm=[log⁡(169)+log⁡(3.142)]×10−5 ⇒μm=[2.2279+0.4972]×10−15 ⇒μm=antilog⁡ (2.7251)×10−5 ⇒μm=531×10−5 ∴μm=5.31×10−3TmA{\mu _m}= \left[ {\log \left( {169} \right) + \log \left( {3.142} \right)} \right] \times {10^{ - 5}} \\\ \Rightarrow {\mu _m}= \left[ {2.2279 + 0.4972} \right] \times {10^{ - 15}} \\\ \Rightarrow {\mu _m}= anti\log \,\left( {2.7251} \right) \times {10^{ - 5}} \\\ \Rightarrow {\mu _m}= 531 \times {10^{ - 5}} \\\ \therefore {\mu _m}= 5.31 \times {10^{ - 3}}T\dfrac{m}{A}
Therefore, the permeability of annealed iron at saturation 5.31×10−3TmA5.31 \times {10^{ - 3}}T\dfrac{m}{A}.

So, the correct option is A.

Note: To answer the problem, one must know the formula for permeability of a material in terms of magnetic susceptibility. We need also to understand the physics underlying the formula and what magnetic susceptibility means. When a material is held in a magnetic field, its susceptibility tells us whether it will be attracted or repelled. We should be careful with our calculations when doing such numerical calculations so that we can be certain of the end result. Remember to mention units in your final response.