A wire carrying current \( I \) is bent in the form of a quadrant of a circle of radius \( r \). The magnetic field at the center of the quadrant loop is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\mu_{0}I}{8r} \)
By the Biot–Savart law, a circular arc of radius r subtending an angle \( \theta \) at the centre produces a field that is the fraction \( \theta/2\pi \) of the field of a full loop. The two straight parts of the quadrant lie along radii, so \( d\vec{l} \) is parallel to \( \vec{r} \) and they give no field at the centre.
\( \displaystyle B = \frac{\mu_0 I}{2r}\times\frac{\theta}{2\pi} \)
\( \displaystyle B = \frac{\mu_0 I}{2r}\times\frac{1}{4} \)
\( \displaystyle B = \frac{\mu_0 I}{8r} \)