When arod of magnetic material of size `10cm xx 0.5 cm xx` `0,2cm ` is located in magnetizing field of `0.5 xx 10^(4)A/ m` then a magnetic moment of `5 `Am^(2)`` is induced in it. Find out the magnetic induction (in. `(Wb) / (m)^(2)` ) in the rod.

Answer

To find the magnetic induction (B) in the rod, we can use the relationship between magnetic moment (M), magnetizing field (H), and the volume of the rod. The magnetic induction can be calculated using the formula:

\[ B = \mu_0 (H + M/V) \]

Where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).
- \( H \) is the magnetizing field.
- \( M \) is the magnetic moment.
- \( V \) is the volume of the rod.

### Step 1: Calculate the Volume of the Rod
The dimensions of the rod are given as:
- Length (l) = 10 cm = 0.1 m
- Breadth (b) = 0.5 cm = 0.005 m
- Height (h) = 0.2 cm = 0.002 m

The volume \( V \) can be calculated as:
\[ V = l \times b \times h \]
\[ V = 0.1 \, \text{m} \times 0.005 \, \text{m} \times 0.002 \, \text{m} \]
\[ V = 0.000001 \, \text{m}^3 = 10^{-6} \, \text{m}^3 \]

### Step 2: Calculate the Magnetizing Field (H)
The magnetizing field is given as:
\[ H = 0.5 \times 10^4 \, \text{A/m} = 5000 \, \text{A/m} \]

### Step 3: Calculate the Magnetic Moment (M)
The magnetic moment is given as:
\[ M = 5 \, \text{A m}^2 \]

### Step 4: Calculate the Induction (B)
Now we can substitute the values into the formula for magnetic induction:
\[ B = \mu_0 \left( H + \frac{M}{V} \right) \]
Substituting the known values:
\[ B = 4\pi \times 10^{-7} \left( 5000 + \frac{5}{10^{-6}} \right) \]

Calculating \( \frac{M}{V} \):
\[ \frac{M}{V} = \frac{5}{10^{-6}} = 5 \times 10^6 \]

Now substituting this back into the equation:
\[ B = 4\pi \times 10^{-7} \left( 5000 + 5 \times 10^6 \right) \]
\[ B = 4\pi \times 10^{-7} \left( 5.005 \times 10^6 \right) \]

Calculating:
\[ B \approx 4\pi \times 10^{-7} \times 5.005 \times 10^6 \]
\[ B \approx 4 \times 3.14 \times 10^{-7} \times 5.005 \times 10^6 \]
\[ B \approx 4 \times 3.14 \times 5.005 \]
\[ B \approx 62.52 \times 10^{-1} \]
\[ B \approx 6.252 \, \text{Wb/m}^2 \]

### Final Answer:
The magnetic induction in the rod is approximately \( 6.252 \, \text{Wb/m}^2 \).

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