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:
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Solution and Explanation
Given:
- The current \( i = 5 \, \text{A} \). - The side length of the square loop is \( d = \frac{1}{2\sqrt{2}} \, \text{m} \).
Step 1: Magnetic Field Due to a Single Side of the Loop
The magnetic field \( B \) due to a single side of the square loop is given by 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 distance between the point and the wire, - \( \theta_1 \) and \( \theta_2 \) are the angles made by the magnetic field lines with respect to the wire.
Step 2: Calculation of the Magnetic Field for a Single Side
Substituting the given values, we get: \[ B = \frac{10^{-7} \times 5 \times 2}{\frac{1}{2\sqrt{2}}} = 2 \times 10^{-6} \, \text{T}. \]
Step 3: Net Magnetic Field at the Centre of the Loop
Since there are 4 sides to the square loop, and the magnetic field due to each side contributes equally at the center of the loop, 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: The Value of \( P \)
From the given data, we conclude that \( P = 8 \).
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