Question

A current of 5A exists in a square loop of side \( \frac{1}{\sqrt{2}} \, {m} \). Then the magnitude of the magnetic field \( B \) at the centre of the square loop will be \( p \times 10^{-6} \, {T} \). Where, value of \( p \) is:

Hint:

The magnetic field at the center of a square loop is proportional to the current and inversely proportional to the side length of the loop. Remember to account for geometric factors like \( \sqrt{2} \) when deriving the formula for the field.

Answer 1

Given:

- Current \( i = 5 \, \text{A} \). - Square loop side length \( d = \frac{1}{2\sqrt{2}} \, \text{m} \).

Step 1: Magnetic Field from a Single Loop Side

The magnetic field \( B \) produced by one side of the square loop is calculated using the formula: \[ B = \frac{\mu_0 i}{4 \pi d} \left( \sin \theta_1 + \sin \theta_2 \right), \] where: - \( \mu_0 \) is the permeability of free space, - \( i \) is the current, - \( d \) is the perpendicular distance from the point to the wire, - \( \theta_1 \) and \( \theta_2 \) are the angles subtended by the ends of the wire at the point.

Step 2: Calculate Magnetic Field for a Single Side

Using the provided values: \[ B = \frac{10^{-7} \times 5 \times 2}{\frac{1}{2\sqrt{2}}} = 2 \times 10^{-6} \, \text{T}. \]

Step 3: Total Magnetic Field at the Loop's Center

A square loop has 4 sides. Each side contributes equally to the magnetic field at the center. Therefore, the net magnetic field at the center is: \[ B_{\text{net}} = 4B = 4 \times 2 \times 10^{-6} = 8 \times 10^{-6} \, \text{T}. \]

Final Answer:

The net magnetic field at the center of the square loop is \( \boxed{8 \times 10^{-6} \, \text{T}} \).

Step 4: Determine the Value of \( P \)

Based on the data, we find that \( P = 8 \).