Relation between permeability' `mu` and magnetizing field `H` for a sample of iron is `mu=left((0.4)/(H)+12 xx 10^(-4)right)` henery! meter, where unit of `H` is `(Am)`. Find value of `H((in) (A) / (m))` for which magnetic induction of `1.0 . (Wb) / (m)^(2)` can, be produced.

Answer

To solve the problem, we need to find the value of the magnetizing field \( H \) for which the magnetic induction \( B \) is \( 1.0 \, \text{Wb/m}^2 \). The relationship between permeability \( \mu \) and magnetizing field \( H \) is given by:

\[ \mu = \left( \frac{0.4}{H} + 12 \times 10^{-4} \right) \, \text{H/m} \]

We also know that the magnetic induction \( B \) is related to permeability \( \mu \) and the magnetizing field \( H \) by the equation:

\[ B = \mu H \]

### Step 1: Substitute \( B \) into the equation for \( \mu \)

Given \( B = 1.0 \, \text{Wb/m}^2 \), we can express \( \mu \) in terms of \( H \):

\[ \mu = \frac{B}{H} = \frac{1.0}{H} \]

### Step 2: Set up the equation

Now we can set the two expressions for \( \mu \) equal to each other:

\[ \frac{1.0}{H} = \frac{0.4}{H} + 12 \times 10^{-4} \]

### Step 3: Clear the fractions

To eliminate the fractions, multiply through by \( H \):

\[ 1.0 = 0.4 + 12 \times 10^{-4} H \]

### Step 4: Rearrange the equation

Rearranging gives:

\[ 1.0 - 0.4 = 12 \times 10^{-4} H \]

\[ 0.6 = 12 \times 10^{-4} H \]

### Step 5: Solve for \( H \)

Now, isolate \( H \):

\[ H = \frac{0.6}{12 \times 10^{-4}} = \frac{0.6}{0.0012} \]

Calculating this gives:

\[ H = 500 \, \text{A/m} \]

### Final Answer

Thus, the value of \( H \) for which the magnetic induction of \( 1.0 \, \text{Wb/m}^2 \) can be produced is:

\[ H = 500 \, \text{A/m} \]

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