A long straight wire carries a current of 5 A. In its middle, the wire is bent into a semicircular arc of radius \(\pi\) cm. The straight parts on both sides lie along the diameter of the arc. The magnetic field at the centre of the arc is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(5.0\times10^{-5}\ \mathrm{T}\)
For every small part of the straight segments, \( d\vec{l} \) and \( \vec{r} \) are parallel, so \( d\vec{l}\times\vec{r}=0 \). The straight parts give no field at the centre. All parts of the semicircle give fields in the same direction. So the arc gives half the field of a full loop, \( B=\mu_0I/4R \).
Here \( I=5\ \mathrm{A} \) and the radius is \( \pi \) cm, so \( R=\pi\times10^{-2}\ \mathrm{m} \).
\( \begin{aligned} B &= \frac{\mu_0 I}{4R} \\ &= \frac{4\pi\times10^{-7}\times5}{4\times\pi\times10^{-2}} \\ &= 5.0\times10^{-5}\ \mathrm{T} \end{aligned} \)
NCERT (2026-27):
Magnetic Field on the Axis of a Circular Current Loop – “… As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain,”
\( B_0=\frac{\mu_0 I}{2R} \)
Example 4.5 – “(a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|. (b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. …”