A) 0.2T
B) 0.7T
C) 0.4T
D) 0.9T
Correct Answer: C
Solution :
As, magnetic moment \[M=niA=50\times 2\times 1.25\times {{10}^{-3}}=0.125A{{m}^{2}}\] If the normal to face of the coil makes an angle \[\theta \] with the magnetic induction B, then in 1st case, torque \[=MB\sin \theta =0.03\] \[\therefore \]\[MB=\sqrt{{{(0.04)}^{2}}+{{(0.03)}^{2}}}=0.05\] \[\Rightarrow \]\[B=\frac{0.05}{M}=\frac{0.05}{0.125}=0.4T\]You need to login to perform this action.
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