A current carrying wire is in form of square loop of side length 10 cm. current in wire is 5A. Find out magnetic field at centre of loop.

Options

A. 2nB
B. `2n^(2)B`
C. nB
D. `n^(2)B`

Answer

To find the magnetic field at the center of a square loop carrying current, we can use the formula for the magnetic field due to a straight current-carrying wire and apply it to each side of the square loop.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The square loop has a side length \( a = 10 \, \text{cm} = 0.1 \, \text{m} \).
- The current \( I = 5 \, \text{A} \) flows through the wire.

2. **Magnetic Field due to a Straight Wire**:
- The magnetic field \( B \) at a distance \( r \) from a long straight wire carrying current \( I \) is given by the formula:
\[ B = \frac{\mu_0 I}{2\pi r} \]
- Here, \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Distance from the Center to the Wire**:
- For a square loop, the distance from the center of the loop to the midpoint of each side is:
\[ r = \frac{a}{2\sqrt{2}} = \frac{0.1}{2\sqrt{2}} = \frac{0.1}{2 \times 1.414} \approx 0.0354 \, \text{m} \]

4. **Calculating the Magnetic Field from One Side**:
- For one side of the square loop, the magnetic field at the center due to that side is:
\[ B_{\text{one side}} = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.0354} \]
- Simplifying this gives:
\[ B_{\text{one side}} = \frac{2 \times 10^{-7} \times 5}{0.0354} \approx 2.83 \times 10^{-5} \, \text{T} \]

5. **Total Magnetic Field from All Four Sides**:
- Since there are four sides contributing to the magnetic field at the center and they all contribute equally, the total magnetic field \( B_{\text{total}} \) is:
\[ B_{\text{total}} = 4 \times B_{\text{one side}} = 4 \times 2.83 \times 10^{-5} \approx 1.13 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnetic field at the center of the square loop is approximately \( 1.13 \times 10^{-4} \, \text{T} \).

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