Deduce the expression for the torque 
acting on a planar loop of area 
and carrying current I placed in a uniform magnetic field 
. If the loop is free to rotate, what would be its orientation in stable equilibrium?
Let there be a rectangular coil PQRS, placed in a magnetic field 
such that its area vector 
is perpendicular to direction of magnetic field 
. Let θ be the angle between 
and 
.
Length of side PQ = b and QR = a
The side PQ experiences a normal inward force equal to Ib
while the side RS experiences an equal normal outward force. These two forces form a couple which exerts a torque given by
= Force × perpendicular distance
=(Ib
)×
= I
)
|τ| = IBAsinθ
The direction of the torque τ is such that it rotates the loop clockwise about the axis of suspension and will be given by Fleming’s left hand rule.
In stable equilibrium, torque will vanish. ⇒ θ = 0. That is, the angle between the area vector 
and the magnetic field 
will be zero. Therefore, the loop will be perpendicular to the direction of the magnetic field.
Couldn't generate an explanation.
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