A current of \( 10\,\mathrm{A} \) is passing through a long wire which has a semicircular loop of the radius \( 20\,\mathrm{cm} \) as shown in the figure. The magnetic field produced at the centre \( P \) of the loop is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 5\pi\,\mu\mathrm{T} \)
The field at P is the sum of the contributions of the straight parts and the semicircle (Biot-Savart law). The straight parts lie on a line passing through P, so each of their elements \( d\vec{l} \) is parallel to \( \vec{r} \) and \( d\vec{l}\times\vec{r} = 0 \): they give no field at P. A semicircle gives half the field of a full circular loop at its centre.
\( \displaystyle B = \frac{1}{2}\times\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R} \)
\( R = 20\,\mathrm{cm} = 0.2\,\mathrm{m} \)
\( \displaystyle B = \frac{4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}\times10\,\mathrm{A}}{4\times0.2\,\mathrm{m}} \)
\( B = 5\pi\times10^{-6}\,\mathrm{T} = 5\pi\,\mu\mathrm{T} \)