# Directions : Question are based on the following paragraph.A current loop $\dpi{100} ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $\dpi{100} BC(radius=b)$ and $\dpi{100} DA(radius=a)$ of the loop are joined by two straight wires $\dpi{100} AB$ and $\dpi{100} CD$ . A steady current $\dpi{100} I$ is flowing in the loop. Angle made by $\dpi{100} AB$ and $\dpi{100} CD$ at the origin $\dpi{100} O$ is $\dpi{100} 30^{\circ}$ . Another straight thin wire with steady current $\dpi{100} I_{1}$   flowing out of the plane of the paper is kept at the origin.Question : Due to the presence of the current $\dpi{100} I_{1}$ at the origin Option 1) the forces on $\dpi{100} AB\; and\; DC$ are zero Option 2) the forces on $\dpi{100} AD\; and\; BC$ are zero Option 3) the magnitude of the net force on the loop is given by $\dpi{100} \frac{I_{1}I}{4\pi }\mu _{0}\left [ 2(b-a)+\frac{\pi }{3}(a+b) \right ]\;$   Option 4) the magnitude of the net force on the loop is given by $\dpi{100} \frac{\mu _{0}II_{1}}{24ab}(b-a)$

As we learnt in

Magnetic Field due to Circular Current at the centre -

If a coil of radius r is carrying current i then magnetic feild at a distnace X from its centre is:

$B_{centre}= \frac{\mu_{0} }{4\pi } \frac{2\pi Ni}{r} = \frac{\mu_{0} Ni}{2r}$

- wherein

The force on AD and BC due to current $I_1$ is zero.

This is because the directions of current element $I\vec{dI}$ and magnetic field $\vec{B}$ are parallel.

Option 1)

the forces on $\dpi{100} AB\; and\; DC$ are zero

Incorrect

Option 2)

the forces on $\dpi{100} AD\; and\; BC$ are zero

Correct

Option 3)

the magnitude of the net force on the loop is given by

$\dpi{100} \frac{I_{1}I}{4\pi }\mu _{0}\left [ 2(b-a)+\frac{\pi }{3}(a+b) \right ]\;$

Incorrect

Option 4)

the magnitude of the net force on the loop is given by $\dpi{100} \frac{\mu _{0}II_{1}}{24ab}(b-a)$

Incorrect

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