question_answer

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

A) \[1.66\times {{10}^{-5}}T\]         

B) \[1.22\times {{10}^{-4}}T\]

C) \[1.33\times {{10}^{-5}}T\]         

D) \[1.44\times {{10}^{-4}}T\]

Correct Answer: C

Solution :

The magnetic field at the centre O due to the current through side AB is                \[\Delta {{I}_{b}}=15\mu A=15\times {{10}^{-6}}A\]                As the magnetic field due to each of the three sides is the same in magnitude and direction So, the total magnetic field at O is sum of all the fields. i.e.   \[{{R}_{L}}=5k\Omega =5\times {{10}^{3}}\Omega \] Here,    \[\tan {{\theta }_{1}}=\frac{AD}{OD}\] \[{{A}_{r}}=\frac{\Delta {{I}_{C}}\times {{R}_{L}}}{\Delta {{I}_{b}}\times {{R}_{i}}}\]    \[\tan {{60}^{o}}=\frac{\frac{l}{2}}{a}\] \[=\frac{100000}{99}\approx 1010\]   \[a=\frac{l}{2\sqrt{3}}=\frac{9\times {{10}^{-2}}}{2\sqrt{3}}\] Now   \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( \text{g} \right)\text{=}{{\text{p}}_{\text{1}}}\]                 \[{{H}_{2}}\left( g \right)={{p}_{2}}mm\]                 \[\therefore \] \[{{p}_{1}}+{{p}_{2}}=60mm\]          \[{{C}_{6}}{{H}_{6}}\left( g \right)=0\]