A current of 5 A exists in a square loop of side . Then the magnitude of the magnetic field B at the centre of the square loop will be . Where, value of p is ______. [Take ]. [2025]

Question

A current of 5 A 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 ______. [Take $μ_{0} = 4 π \times 10^{- 7} T {mA}^{- 1}$ ]. [2025]

Smart Achievers · Accepted Answer

(8)

Let B be the magnetic field due to single side then

$p \times 10^{- 6} T$

$B = \frac{10^{- 7} \times 5 \times 2}{\frac{1}{2 \sqrt{2}}} \times \frac{1}{\sqrt{2}} = 2 \times 10^{- 6}$

$∴ B_{net}$ at centre O = 4 B

$B_{net} = 8 \times 10^{- 6} T$

$∴ p = 8$

Solution

Q.
A current of 5 A 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 ______. [Take $μ_{0} = 4 π \times 10^{- 7} T {mA}^{- 1}$ ]. [2025]

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Solution

Q. A current of 5 A exists in a square loop of side 12m. 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]. [2025]

Ans. (8) Let B be the magnetic field due to single side then B=μ0i4πd(sin θ1+sin θ2) B=10–7×5×2122×12=2×10–6 ∴ Bnet at centre O = 4 B Bnet=8×10–6 T ∴ p=8

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Q.
A current of 5 A 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 ______. [Take $μ_{0} = 4 π \times 10^{- 7} T {mA}^{- 1}$ ]. [2025]