A current of 5 A exists in a square loop of side 1/√(2) m. Then the magnitude of the magnetic field B at the centre of the square loop will be p × 10⁻⁶ T. where, value of p is ________ [ Take μ₀ = 4π × 10⁻⁷ T mA⁻¹ ]. Numerical answer type. Work it out, then open the answer.

Answer: 8. Square loop of side a: B = (2√(2)μ₀ I)/(π a). B = 2√(2) × 4 × 10⁻⁷ × 51/√(2) = 2√(2) × √(2) × 20 × 10⁻⁷ B = 80 × 10⁻⁷ = 8 × 10⁻⁶ T So p = 8.

A current of 5 A exists in a square loop of side 1/√(2) m. Then the… | JEE Main 2025 (24 Jan, Shift 1)

Q26JEE Main 2025 · 24 JAN · SHIFT 1Medium

Magnetic Field of Straight Wires

A current of 5 A exists in a square loop of side $\frac{1}{\sqrt{2}}\text{ m}$ . Then the magnitude of the magnetic field $B$ at the centre of the square loop will be $p \times 10^{-6}\text{ T}$ . where, value of p is ________ $[$ Take $\mu_0 = 4\pi \times 10^{-7}\text{ T mA}^{-1}$ ].

Numerical answer type. Work it out, then open the answer.

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Answer: 8

Square loop of side $a$ : $B = \frac{2\sqrt{2}\mu_0 I}{\pi a}$ .
$B = \frac{2\sqrt{2} \times 4 \times 10^{-7} \times 5}{1/\sqrt{2}} = 2\sqrt{2} \times \sqrt{2} \times 20 \times 10^{-7}$
$B = 80 \times 10^{-7} = 8 \times 10^{-6}$ T
So $p = 8$ .

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