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

The magnetic field at the center of a square loop carrying a current can be calculated using the formula: \[ B = \frac{\mu_0 I}{2a} \cdot \sqrt{2}, \] where: - \( I \) is the current flowing through the loop, - \( a \) is the side length of the square loop, - \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4 \pi \times 10^{-7} \, {T m A}^{-1} \)). 
Step 1: Substitute the given values into the formula: - \( I = 5 \, {A} \), - \( a = \frac{1}{\sqrt{2}} \, {m} \). Thus, the magnetic field at the center is: \[ B = \frac{4 \pi \times 10^{-7} \times 5}{2 \times \frac{1}{\sqrt{2}}} \cdot \sqrt{2}. \] Step 2: Simplifying the expression: \[ B = \frac{4 \pi \times 10^{-7} \times 5}{2 \times \frac{1}{\sqrt{2}}} \cdot \sqrt{2} = \frac{4 \pi \times 10^{-7} \times 5 \times \sqrt{2}}{2 \times \frac{1}{\sqrt{2}}}. \] \[ B = \frac{4 \pi \times 10^{-7} \times 5 \times 2}{2}. \] Step 3: Now we calculate the value: \[ B = \frac{4 \pi \times 10^{-7} \times 10}{2} = 2 \times 10^{-6} \, {T}. \] Thus, the value of \( p \) is 8. The correct answer is \( \boxed{8} \).