The current passing through a conducting loop in the form of equilateral triangle of side $4 \sqrt{3} cm$ is 2 A. The magnetic field at its centroid is $α \times 10^{- 5} T$ The value of $α$ is ______.

$(Given: μ_{0} = 4 π \times 10^{- 7} SI units)$ [2026]

Question

The current passing through a conducting loop in the form of equilateral triangle of side $4 \sqrt{3} cm$ is 2 A. The magnetic field at its centroid is $α \times 10^{- 5} T$ The value of $α$ is ______.

$(Given: μ_{0} = 4 π \times 10^{- 7} SI units)$ [2026]

Smart Achievers · Accepted Answer

(3)

$B = \frac{μ_{0}}{4 π} \times \frac{I}{d} [\sin 60 ° + \sin 60 °] \times 3$

$B = 10^{- 7} \times \frac{2}{2 \times 10^{- 2}} (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) \times 3$

$= \sqrt{3} \times 10^{- 5} \times 3 = 3 \sqrt{3} \times 10^{- 5}$

Solution

Q.
The current passing through a conducting loop in the form of equilateral triangle of side $4 \sqrt{3} cm$ is 2 A. The magnetic field at its centroid is $α \times 10^{- 5} T$ The value of $α$ is ______.

$(Given: μ_{0} = 4 π \times 10^{- 7} SI units)$ [2026]

1 $\frac{\sqrt{3}}{2}$

2 $\sqrt{3}$

3 $3 \sqrt{3}$

4 $2 \sqrt{3}$

Ans.
(3)

$B = \frac{μ_{0}}{4 π} \times \frac{I}{d} [\sin 60 ° + \sin 60 °] \times 3$

$B = 10^{- 7} \times \frac{2}{2 \times 10^{- 2}} (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) \times 3$

$= \sqrt{3} \times 10^{- 5} \times 3 = 3 \sqrt{3} \times 10^{- 5}$

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Solution

Q. The current passing through a conducting loop in the form of equilateral triangle of side 43 cm is 2 A. The magnetic field at its centroid is α×10-5 T The value of α is ______. (Given: μ0=4π×10-7 SI units) [2026]

1 32

2 3

3 33

4 23

Ans. (3) B=μ04π×Id[sin60°+sin60°]×3 B=10-7×22×10-2(32+32)×3 =3×10-5×3=33×10-5