A) 0.2 T
B) 0.3 T
C) 0.4 AT
D) 0.1T
E) 0.05 T
Correct Answer: E
Solution :
Given \[A=2\times {{10}^{-2}}{{m}^{2}}\] \[N=100\] \[I=5A\] When the coil is held with its plane in North-South direction then its torque \[{{\tau }_{1}}=0.3Nm\] \[{{\tau }_{1}}=MB\sin \theta \] When the plane is in East-West direction then its torque \[{{\tau }_{2}}=0.4\,Nm\] \[{{\tau }_{2}}=MB\sin ({{90}^{o}}-\theta )\] \[{{\tau }_{2}}=MB\cos \theta \] \[Ratio=\frac{MB\,\sin \theta }{MB\cos \theta }=\frac{{{\tau }_{1}}}{{{\tau }_{2}}}\] \[\tan \theta =\frac{3}{4}\] Or \[\theta ={{\tan }^{-1}}\left( \frac{3}{4} \right)\] then \[sin\text{ }\theta =0.6\] \[\therefore \] \[B=\frac{{{\tau }_{1}}}{M\,\sin \theta }\] \[\Rightarrow \] \[B=\frac{0.3}{NIA\sin \theta }\] \[\Rightarrow \] \[B=\frac{0.3}{100\times 5\times 2\times {{10}^{-2}}\times 0.6}\] \[\Rightarrow \] \[B=0.05\,T\]
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