“I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent.”

– Mahatma Gandhi

The following question which appeared in IIT JEE 2010 question paper is worth noting:

A thin flexible wire of length *L* is connected to two adjacent fixed points and carries a current *I* in the clockwise direction, as shown in the figure.

When the system is put in a uniform magnetic field of strength *B* going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

(a) *iBL *

(b) *iBL /*π

(d) *iBL /*4π

Considering an elemental length *dL* of the wire, the magnetic force *F* acting on the element is given by

*F *= *idLB = iRdθB*, as is clear from the figure.

The tension *T * developed in the wire is resolved into rectangular components *T* sin(*dθ/*2) and *T* cos(*dθ/*2) as indicated in the figure.

The magnetic force *F * is related to the tension *T* as

*F *= *iRdθB = *2*T* sin(*dθ/*2)

Since *dθ* is a very small angle, sin(*dθ/*2) ≈ *dθ/*2

Therefore, the above equation gives *T* =* iRB*

But *R* = *L /*2π

Therefore, *T =* *iBL /*2π

Two straight, infinitely long parallel wires P and Q carry equal currents (*I*) in opposite directions (fig.). A square loop of side *a* carrying current* i* is arranged symmetrically in between the straight wires so that the plane of the loop is in the plane containing the wires P and Q. What is the torque acting on the current loop?

(a) Zero

(b) *μ*_{0}*Iia*^{2}/2π*d*

(c) (*μ*_{0}*Iia*^{2}/2π)[1/(*d*+*a*)]

(d) (*μ*_{0}*Iia*^{2}/2π)[1/*d* + 1/(*d*+*a*)]

(e) (*μ*_{0}*Iia*^{2}/2π)[1/*d* – 1/(*d*+*a*)]

The resultant magnetic forces acting on the four sides of the current loop act at their centres. The forces on the opposite sides of the loop are oppositely directed as you can easily verify using Fleming’s left hand rule. Since the lines of action of the resultant forces on opposite sides of the loop coincide, there is no lever arm to produce a torque and hence the torque is zero.

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