The figure shows a current loop having two circular arcs joined by two radial lines. If radius OA is 'a' and OD is 'b', the magnetic field at the centre O is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \left(\frac{\theta}{2\pi}\right)\left(\frac{\mu_0 i}{2}\right)\left(\frac{b-a}{ab}\right) \)
By the Biot-Savart law, a current element gives no field at a point on its own line, so the radial lines AD and BC give no field at O. Only the two arcs contribute.
An arc of radius \( r \) subtending angle \( \theta \) at its centre gives the fraction \( \theta/2\pi \) of the field of a full loop:
\( \displaystyle B_{\text{arc}} = \left(\frac{\theta}{2\pi}\right)\frac{\mu_0 i}{2r} \)
In the figure the current goes round the outer arc (D to C) and the inner arc (B to A) in opposite senses, so their fields at O are opposite. The inner arc, being closer, gives the larger field.
\( \displaystyle B_O = \left(\frac{\theta}{2\pi}\right)\frac{\mu_0 i}{2a} - \left(\frac{\theta}{2\pi}\right)\frac{\mu_0 i}{2b} \)
\( \displaystyle B_O = \left(\frac{\theta}{2\pi}\right)\left(\frac{\mu_0 i}{2}\right)\left(\frac{1}{a} - \frac{1}{b}\right) = \left(\frac{\theta}{2\pi}\right)\left(\frac{\mu_0 i}{2}\right)\left(\frac{b-a}{ab}\right) \)