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 \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\left( \text{g} \right)\text{=}{{\text{p}}_{\text{1}}}\] \[{{H}_{2}}\left( g \right)={{p}_{2}}mm\] \[\therefore \] \[{{p}_{1}}+{{p}_{2}}=60mm\] \[{{C}_{6}}{{H}_{6}}\left( g \right)=0\]You need to login to perform this action.
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