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The relation between magnetic permeability , $\mu $ and field strength, $H$ for a specimen of iron is as follows $\mu $ = $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$ $Henry/meter$ . The value of $H$ which produces flux density of $1T$ will be?
A) $250$ $A/m$
B) $500$ $A/m$
C) $750$ $A/m$
D) ${10^3}$ $A/m$

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Last updated date: 17th Jul 2024
Total views: 62.1k
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Answer
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Hint: Given question has three terms magnetic permeability, field strength and flux density thus we will have to use a mathematical relation between these three which is, Flux density is a product of permeability and field strength.

Formula Used:
$B$ = $\mu$ $H$

Complete step by step answer:
In the question we have been a relation between Magnetic permeability and field strength but we need to find the value of field strength which will produce 1T flux density , Thus we will be using a direct relation between above three terms which is given by :
$B$ = $\mu$ $H$
Where B is flux density
Now we have given a relation is $\mu $= $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$ with B = 1 ,
Therefore substituting the value of B and expression of $\mu $ in the main formula 1
$ \Rightarrow $1 = $\left( {\dfrac{{0.4}}{H} + 12 \times {{10}^{^{ - 4}}}} \right)$H
$ \Rightarrow $1 = $\left( {\dfrac{{0.4 + 12 \times {{10}^{ - 4}}H}}{H}} \right)H$
Cancelling H we get
$ \Rightarrow $$1 - 0.4 = 12 \times {10^{ - 4}}H$
$ \Rightarrow $$H = \dfrac{{.6}}{{12 \times {{10}^{ - 4}}}}$
$ \Rightarrow $$H = \dfrac{{1000 \times 6}}
  {12} \\
    \\
$
$ \Rightarrow H = 500$ $A/m$

Hence option B is the correct answer.

Note: Most important here is to remember the relation between the terms properly so that we don’t waste a lot of in straight formula based questions . It is also important that we have an idea of all the terms used in the problem so as to understand the process. It is important not to get confused within the terms themselves that is we should be very clear with denotion.