A current of 5 A exists in a square loop of side \frac1\sqrt2 ~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 \underline\hspace2cm[. Take \mu_0=4 \pi \times 10^-7 ~T ~mA^-1 ].

Magnetics PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q72 with Solution

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Correct answer: 8

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 loop, for each side:

r=\frac{a}{2}=\frac{1}{2\sqrt{2}}\,\text{m}

From the centre, each side subtends equal angles, so

\theta_1=\theta_2=45^\circ

Thus magnetic field due to one side is

B_1=\frac{\mu_0 I}{4\pi r}(\sin45^\circ+\sin45^\circ)

B_1=\frac{\mu_0 I}{4\pi r}(2\cdot \frac{1}{\sqrt{2}}) =\frac{\mu_0 I}{4\pi r}(\sqrt{2})

Now substitute $r=\dfrac{1}{2\sqrt{2}}$ :

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

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

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

Substitute $\mu_0=4\pi\times10^{-7}$ and $I=5$ A:

B=4\cdot \frac{4\pi\times10^{-7}\cdot 5}{\pi}

B=4\cdot 20\times10^{-7}=80\times10^{-7}

B=8\times10^{-6}\,\text{T}

So,

p=8

Derived answer: $8$

Stored correct answer: $8$

They match.