Q.7 Assuming an equation of magnetic force due to arbitrarily shaped wire, obtain an expression for force on a closed circuit in a uniform magnetic field.
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Answer: The net magnetic force on a closed current loop in a uniform magnetic field is zero.
The force on a small element \(d\vec{l}\) of a wire carrying current \(I\) in a magnetic field \(\vec{B}\) is \(d\vec{F} = I\,d\vec{l}\times\vec{B}\). For an arbitrarily shaped wire between points \(a\) and \(b\):
\[\vec{F} = I\int_a^b d\vec{l}\times\vec{B}\]If \(\vec{B}\) is uniform it can be taken out of the integral: \(\vec{F} = I\left(\int_a^b d\vec{l}\right)\times\vec{B} = I\,\vec{L}\times\vec{B}\), where \(\vec{L}\) is the straight vector from \(a\) to \(b\).
For a closed circuit the starting and end points coincide, so the vector sum of all elements is zero: \(\oint d\vec{l} = 0\). Therefore
\[\vec{F} = I\left(\oint d\vec{l}\right)\times\vec{B} = 0\](The loop may still experience a torque.)
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