A) 31T
B) 33.34 T
C) 32T
D) 40T
Correct Answer: B
Solution :
The magnetic moment of the current loop will be \[M=NiA=500\times 0.4\times 3\times {{10}^{-4}}=6\times {{10}^{-2}}A-{{m}^{2}}\] Also torque \[\tau =M\times B\] \[\Rightarrow \] \[|\tau |=MBsin\theta \] Here \[\theta =90{}^\circ ,\tau =2N-m\] So \[B=\frac{\tau }{M}=\frac{2}{6\times {{10}^{-2}}}\] \[B=33.34T\]You need to login to perform this action.
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