A wire shown in figure carries a current of 10 A. The magnitude of the magnetic field at the centre O is: (Given: radius of the bent coil is 3 cm.)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.57 × 10−4 T
The wire has two kinds of parts. The two straight parts point along radii: extended, they pass through O (the dashed lines). For a current element directed towards or away from a point, \( d\vec{l} \times \vec{r} = 0 \), so by the Biot–Savart law the straight parts give no field at O.
Readings from the figure:
\( \text{angle not covered by the arc} = 90^\circ \)
\( \text{arc} = 360^\circ - 90^\circ = 270^\circ \)
\( R = 3\,\mathrm{cm} = 0.03\,\mathrm{m}, \; I = 10\,\mathrm{A} \)
\( \displaystyle \frac{270^\circ}{360^\circ} = \frac{3}{4}\text{ of a circle} \)
The field at the centre of a full loop is \( \mu_0 I/2R \), and an arc gives the same fraction of it as its fraction of the circle:
\( \displaystyle B = \frac{3}{4}\,\frac{\mu_0 I}{2R} \)
\( \displaystyle B = \frac{3}{4} \times \frac{4\pi\times10^{-7} \times 10}{2 \times 0.03}\,\mathrm{T} \)
\( \displaystyle B = \frac{3}{4} \times 2.09\times10^{-4}\,\mathrm{T} \approx 1.57\times10^{-4}\,\mathrm{T} \)