Calculate the magnetic field induction at the center of the coil in the form of a square of side `10cm`, carrying a current of `10A`.

Explanation

To calculate the magnetic field induction at the center of a square coil carrying a current, we can follow these steps:

### Step 1: Understand the Geometry of the Coil
The coil is in the shape of a square with each side measuring 10 cm. The center of the coil is equidistant from all four sides.

### Step 2: Convert Units
Convert the side length from centimeters to meters for standard SI units:
\[ \text{Side length} = 10 \text{ cm} = 0.1 \text{ m} \]

### Step 3: Calculate the Distance from the Center to a Side
The distance from the center of the square to the midpoint of any side can be calculated as:
\[ r = \frac{\text{Side length}}{2} = \frac{0.1 \text{ m}}{2} = 0.05 \text{ m} \]

### Step 4: Use the Formula for Magnetic Field Due to a Straight Conductor
The magnetic field \( B \) at a distance \( r \) from a long straight conductor carrying current \( I \) is given by:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where \( \mu_0 = 4\pi \times 10^{-7} \text{ T m/A} \) (permeability of free space).

### Step 5: Calculate the Magnetic Field Due to One Side of the Square
Substituting the values into the formula:
\[ B_{\text{one side}} = \frac{(4\pi \times 10^{-7} \text{ T m/A}) \times (10 \text{ A})}{2 \pi (0.05 \text{ m})} \]
\[ B_{\text{one side}} = \frac{(4 \times 10^{-7}) \times 10}{2 \times 0.05} \]
\[ B_{\text{one side}} = \frac{4 \times 10^{-6}}{0.1} = 4 \times 10^{-5} \text{ T} \]

### Step 6: Calculate the Total Magnetic Field at the Center
Since there are four sides to the square, and the magnetic fields due to each side at the center will add up (they are all in the same direction at the center):
\[ B_{\text{total}} = 4 \times B_{\text{one side}} \]
\[ B_{\text{total}} = 4 \times (4 \times 10^{-5} \text{ T}) = 16 \times 10^{-5} \text{ T} \]
\[ B_{\text{total}} = 1.6 \times 10^{-4} \text{ T} \]

### Final Answer
The magnetic field induction at the center of the square coil is:
\[ B = 1.6 \times 10^{-4} \text{ T} \]

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