question_answer

 A current loop of area 0.01 m2 and carrying acurrent of 10 A is held perpendicular to a magnetic field of intensity 0.1 T. The torque acting on the loop (in N-m) is

A)  1.1                                        

B)  0.8

C)  0.001                                   

D)  0.01

Correct Answer: D

Solution :

                 Torque acting on loop is given by                 \[\tau =IAB\,\sin \theta \] where \[I\] is current, A the area and B the magnetic field intensity.                 Given,   \[I=10\,A\],        \[A=0.01\,\,{{m}^{2}}\],                 \[B=0.1\,T\],      \[\theta ={{90}^{o}}\] \[\therefore \]  \[\tau =10\times 0.01\times 0.1\times \sin \,{{90}^{o}}\] \[\Rightarrow \]               \[\tau =0.01\,\,N-m\]