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Physics Question on Magnetic Field Due To A Current Element, Biot-Savart Law

A current of 2 A2\,A is flowing in the sides of an equilateral triangle of side 9 cm9\, cm. The magnetic field at the centroid of the triangle is

A

1.66×10−5T1.66 \times 10^{-5} T

B

1.22×10−4T1.22 \times 10^{-4} T

C

1.33×10−5T1.33 \times 10^{-5} T

D

1.44×10−4T1.44 \times 10^{-4} T

Answer

1.33×10−5T1.33 \times 10^{-5} T

Explanation

Solution

The magnetic field at the centre O due to the current through side AB is
B1=μ0I4πa[sin⁡ θ1+sin⁡ θ2]B_1 = \frac{\mu_0 I}{4 \pi a} [\sin \, \theta_1 + \sin \, \theta_2]

As the magnetic field due to each of the three sides is the same in magnitude and direction. So, the total magnetic field atO is sum of all the fields.
i.e. B=3B1=3μ0I4πa[sin⁡θ1+sin⁡θ2]B = 3B_{1} = \frac{3\mu_{0}I}{4\pi a}\left[\sin \theta_{1} + \sin \theta_{2}\right]
Here, tan⁡θ1=ADOD\tan \theta_{1} = \frac{AD}{OD}
⇒tan⁡60∘=l2a\Rightarrow \tan60^{\circ} = \frac{\frac{l}{2}}{a}
⇒a=l23=9×10−223\Rightarrow a = \frac{l}{2\sqrt{3}} = \frac{9\times10^{-2}}{2\sqrt{3}}
Now B =3×4π×10−7×24π×9×10−223    [sin⁡ 60∘+sin⁡60∘] = 3 \times\frac{4\pi\times10^{-7} \times2 }{4 \pi\times\frac{9 \times10^{-2}}{2\sqrt{3}}} \, \, \, \, [\sin \, 60^{\circ} + \sin 60^{\circ} ]
=439×10−5[32+32]= \frac{4\sqrt{3}}{9} \times10^{-5} \left[ \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}\right]
=43×10−5=\frac{4}{3} \times 10^{-5}
=1.33×10−5T= 1.33 \times10^{-5} T