question_answer

A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is\[\overset{\to }{\mathop{\mathbf{F}}}\,\], the net force on the remaining three arms of the loop is

A) \[3\overset{\to }{\mathop{\mathbf{F}}}\,\]

B)                                        \[-\overset{\to }{\mathop{\mathbf{F}}}\,\]

C) \[-3\overset{\to }{\mathop{\mathbf{F}}}\,\]                     

D)        \[\overset{\to }{\mathop{\mathbf{F}}}\,\]

Correct Answer: B

Solution :

When a current carrying loop is placed in a magnetic field, the coil experiences a torque given by \[\tau =NB\,iA\sin \theta \]. Torque is maximum when \[\theta ={{90}^{o}},\]i.e., the plane of the coil is parallel to the field \[{{\tau }_{\max }}=NBiA\] Forces\[{{\overset{\to }{\mathop{\mathbf{F}}}\,}_{1}}\]and\[{{\overset{\to }{\mathop{\mathbf{F}}}\,}_{2}}\]acting on the coil are equal in magnitude and opposite in direction. As the forces \[{{\overset{\to }{\mathop{\mathbf{F}}}\,}_{1}}\] and \[{{\overset{\to }{\mathop{\mathbf{F}}}\,}_{2}}\] have the same line of action their resultant effect on the coil is zero. The two forces \[{{\overset{\to }{\mathop{\mathbf{F}}}\,}_{3}}\] and \[{{\overset{\to }{\mathop{\mathbf{F}}}\,}_{4}}\] are equal in magnitude and opposite in direction. As the two forces have different lines of action, they constitute a torque. Thus, if the force on one arc of the loop is \[\overset{\to }{\mathop{\mathbf{F}}}\,\], the net force on the remaining three arms of the loop is - F.