The current of 5 ~A flows in a square loop of sides 1 ~m is placed in air. The magnetic field at the centre of the loop is X \sqrt2 \times 10^-7 T . The value of X is \underline\hspace2cm .

Magnetics PYQ — JEE Main 2024 · 30 Jan · Shift 2 · Q83 with Solution

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Correct answer: 40

For a finite straight wire, magnetic field at a point at perpendicular distance $r$ is

B = \frac{\mu_0 I}{4\pi r}(\sin\theta_1 + \sin\theta_2)

At the centre of a square, for each side:

Perpendicular distance from centre to side: $r = \frac{a}{2} = \frac{1}{2}\,\text{m}$
The point is symmetrically placed, so $\theta_1 = \theta_2 = 45^\circ$

Thus field due to one side is

B_1 = \frac{\mu_0 I}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ)

B_1 = \frac{\mu_0 I}{2\pi a}\left(2 \cdot \frac{1}{\sqrt{2}}\right)

B_1 = \frac{\mu_0 I\sqrt{2}}{2\pi a}

All four sides produce magnetic field in the same direction at the centre, so

B = 4B_1 = 4 \cdot \frac{\mu_0 I\sqrt{2}}{2\pi a}

B = \frac{2\mu_0 I\sqrt{2}}{\pi a}

Now use

\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A}

So,

B = \frac{2(4\pi \times 10^{-7})I\sqrt{2}}{\pi a}

B = 8 \times 10^{-7} \cdot \frac{I\sqrt{2}}{a}

Substitute $I=5$ and $a=1$ :

B = 8 \times 10^{-7} \cdot 5\sqrt{2}

B = 40\sqrt{2} \times 10^{-7}\,\text{T}

Given:

B = X\sqrt{2} \times 10^{-7}\,\text{T}

Hence,

X = 40

Stored correct answer = $40$

This matches the derived answer.