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Question

Question: The current \(I\) is flowing in an equilateral triangle of side \(a\) as shown in figure. The magnet...

The current II is flowing in an equilateral triangle of side aa as shown in figure. The magnetic field at the centroid OO will be:

A) 9μ0I2πa\dfrac{{9{\mu _0}I}}{{2\pi a}}
B) 52μ0I3πa\dfrac{{5\sqrt 2 {\mu _0}I}}{{3\pi a}}
C) 3μ0I2πa\dfrac{{3{\mu _0}I}}{{2\pi a}}
D) μ0I33πa\dfrac{{{\mu _0}I}}{{3\sqrt 3 \pi a}}

Explanation

Solution

Use the Biot-savart’s law to calculate the magnetic field on one side of the triangle,
B=μ04π×Ir×(sinθ+sinθ)B = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{I}{r} \times (\sin \theta + \sin \theta)
BB is magnetic field intensity,
II is the current intensity,
μ0{\mu _0} is the permeability of free space.

Complete step by step solution:
The Biot – Savart Law is an equation that defines a steady electric current produced by the magnetic field. It refers to magnitude, direction, length and proximity of the electrical current in the magnetic field. The Biot-Savart’s law is important for magneto-statics, and plays a similar role in the electrostatics law of Coulomb. If magneto statics are not applicable, Jefimenko equations can replace the Biot-Savart’s rule. The rule shall be in line with both the circuit rules of Ampere and the magnet law of Gauss. Let’s consider that an angle θ\theta is made by each end of each side at centroid is600{60^0} and rr is the perpendicular distance of each side from centroid which is equal to3a6\dfrac {{\sqrt 3 a}} {6} .
Using Biot-savart's law, magnetic field at the centroid by each side can be written as,
B=μ04π×Ir×(sinθ+sinθ)B = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{I}{r} \times (\sin \theta + \sin \theta)
BB is magnetic field intensity,
II is the current intensity,
μ0{\mu _0} is the permeability of free space.

Putting all the values given in the question, we get,
B=μ04π×6I3a×(sin60+sin60)B = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{{6I}}{{\sqrt 3 a}} \times (\sin 60 + \sin 60)
B=μ04π×6I3a×3\Rightarrow B = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{{6I}}{{\sqrt 3 a}} \times \sqrt 3
B=μ04π×6Ia\Rightarrow B = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{{6I}}{a}
Total magnetic field on all the three sides of the equilateral triangle will be three times the calculated magnetic field,
B=3BB' = 3B
B=μ04π×6Ia×3\Rightarrow B' = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{{6I}}{a} \times 3
B=μ04π×9Ia\Rightarrow B' = \dfrac{{{\mu _0}}} {{4\pi}} \times \dfrac{{9I}}{a}
The magnetic field at the centroid OO will be 9μ0I2πa\dfrac {{9{\mu _0} I}} {{2\pi a}} .

Hence, the correct option is A.

Note: In a plane perpendicular to the element line and the location vector always the direction of the magnetic field. The left thumb rule is given where the thumb points towards the current position and the other fingers shows the direction of the magnetic field.