A) 0.32 Nm tending to rotate the side AD out of the page
B) 0.32 Nm tending to rotate the side AD into the page
C) 0.0032 Nm tending to rotate the side AD out of the page
D) 0.0032 Nm tending to rotate the side AD into the page
Correct Answer: A
Solution :
\[\tau =NBiA=100\times 0.2\times 2\times \left( 0.08\times 0.1 \right)=0.32N\times m\] Direction can be found by Fleming's left hand rule.
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