The absolute magnetic permeability `mu` of a specimen of magnetic material is related to magnetic intensity `H` according to the relation as `mu=(0.6)/(H)+8.0 xx 10^(-4) T A m^(-1)`. Find the value of `H` (in `Am^(-1)`) for which magnetic induction of `0.22 T` can be produced.

Answer

To solve the problem, we need to find the value of magnetic intensity \( H \) for which the magnetic induction \( B \) is \( 0.22 \, T \). The relationship given is:

\[ \mu = \frac{0.6}{H} + 8.0 \times 10^{-4} \, T \, A \, m^{-1} \]

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

\[ B = \mu H \]

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

Since we know \( B = 0.22 \, T \), we can substitute this into the equation:

\[ 0.22 = \mu H \]

### Step 2: Express \( \mu \) in terms of \( H \)

From the first equation, we can express \( \mu \):

\[ \mu = \frac{0.6}{H} + 8.0 \times 10^{-4} \]

### Step 3: Substitute \( \mu \) into the \( B \) equation

Now, substitute this expression for \( \mu \) into the equation for \( B \):

\[ 0.22 = \left(\frac{0.6}{H} + 8.0 \times 10^{-4}\right) H \]

### Step 4: Simplify the equation

Distributing \( H \) gives:

\[ 0.22 = 0.6 + 8.0 \times 10^{-4} H \]

### Step 5: Rearrange the equation

Rearranging the equation to isolate \( H \):

\[ 0.22 - 0.6 = 8.0 \times 10^{-4} H \]

\[ -0.38 = 8.0 \times 10^{-4} H \]

### Step 6: Solve for \( H \)

Now, divide both sides by \( 8.0 \times 10^{-4} \):

\[ H = \frac{-0.38}{8.0 \times 10^{-4}} \]

### Step 7: Calculate \( H \)

Calculating the value:

\[ H = \frac{-0.38}{0.0008} = -475 \]

Since magnetic intensity cannot be negative, we take the absolute value:

\[ H \approx 475 \, A/m \]

### Final Answer

The value of \( H \) for which the magnetic induction of \( 0.22 \, T \) can be produced is approximately:

\[ H \approx 475 \, A/m \]