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: The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is

  • Option 1)

    zero

  • Option 2)

    \frac{\mu_{0} I(b-a)}{24ab}\;\; \;

  • Option 3)

    \; \frac{\mu _{0}I}{4\pi }\left [ \frac{b-a}{ab} \right ]\;

  • Option 4)

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

 

Answers (1)

As we learnt in 

Arc subtends angle theta at the centre -

B=\frac{\mu_{0}}{4\pi }.\frac{\theta i}{r}

- wherein

 

  B_1 =\frac{\mu_0I}{4\pi a}\times \frac{\pi }{6} (directed\:vertically\:upward)

B_2 =\frac{\mu_0I}{4\pi b}\times \frac{\pi }{6} (directed\:vertically\:downward)

B=B_1 -B_2 =\frac{\mu_0I}{24}\left [ \frac{1}{a}-\frac{1}{b} \right ]=\frac{\mu_0I}{24ab} (b-a)


Option 1)

zero

Incorrect

Option 2)

\frac{\mu_{0} I(b-a)}{24ab}\;\; \;

Correct

Option 3)

\; \frac{\mu _{0}I}{4\pi }\left [ \frac{b-a}{ab} \right ]\;

Incorrect

Option 4)

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

Incorrect

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