Directions : Question are based on the following paragraph.

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC(radius=b) and DA(radius=a) of the loop are joined by two straight wires AB and CD . A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30^{\circ} . Another straight thin wire with steady current I_{1}   flowing out of the plane of the paper is kept at the origin.

Question : Due to the presence of the current I_{1} at the origin

  • Option 1)

    the forces on AB\; and\; DC are zero

  • Option 2)

    the forces on AD\; and\; BC are zero

  • Option 3)

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

    \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 \frac{\mu _{0}II_{1}}{24ab}(b-a)

 

Answers (1)

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 AB\; and\; DC are zero

Incorrect

Option 2)

the forces on AD\; and\; BC are zero

Correct

Option 3)

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

\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 \frac{\mu _{0}II_{1}}{24ab}(b-a)

Incorrect

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