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Question: The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m whic...

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is: (take μ0=4π×107NA - 2{\mu _0} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{A}}^{{\text{ - 2}}}})
(A) 18μT18\mu T
(B) 4μT4\mu T
(C) 1μT1\mu T
(D) 9μT9\mu T

Explanation

Solution

The angle between the sides of an equilateral triangle is 60 degrees. The magnetic field of a triangular loop is inversely proportional to the length of its sides, directly proportional to the current. The constant of the proportion can be given as the product of the permeability of free space, the number 3, the sine and tangent of the angle in the triangle and the inverse of the number pi.
Formula used: In this solution we will be using the following formulae;
B=3μ0IπasinθtanθB = \dfrac{{3{\mu _0}I}}{{\pi a}}\sin \theta \tan \theta where BB is the magnetic field due to an equilateral triangular loop, μ0{\mu _0} is the permeability of free space, II is the current, aa is the length of the sides and θ\theta is the angle between two lengths.

Complete Step-by-Step Solution:
We are told to find the magnetic field at the centre of an equilateral triangular loop. This loop has a side 1 m, and carries a current of 10 A.
A triangular loop, like circular loops (one turn solenoid) and square loop (like in motors and generators) is one of the famous loops used in electromagnetic. The formula of the magnetic field generated at the centre can be given as
B=3μ0IπasinθtanθB = \dfrac{{3{\mu _0}I}}{{\pi a}}\sin \theta \tan \theta where BB is the magnetic field due to an equilateral triangular loop, μ0{\mu _0} is the permeability of free space, II is the current, aa is the length of the sides and θ\theta is the angle between two lengths. The angle between the sides of an equilateral triangle is 60 degrees.
Hence, by inserting all known and given values, we have,
B=3(4π×107)(10)π(1)sin60tan60B = \dfrac{{3\left( {4\pi \times {{10}^{ - 7}}} \right)\left( {10} \right)}}{{\pi \left( 1 \right)}}\sin 60^\circ \tan 60^\circ
By computation, we shall have that
B=1.8×105TB = 1.8 \times {10^{ - 5}}T which in the unit given to us in the option, can be written as
B=18μTB = 18\mu T

Hence, the correct option is A

Note: For clarity, an equilateral triangle has angle 60 degree between its sides because, as we know from mathematics, the total angle in a triangle is 180 degrees, and there are three angles in a triangle, hence, the angle of one would be 1803=60\dfrac{{180^\circ }}{3} = 60^\circ