Q148 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Physics
A current carrying loop $\mathrm{ABCD}$ has two circular arcs $\mathrm{AD}$ and $\mathrm{BC}$ with radius $1 \mathrm{~cm}$ and $2 \mathrm{~cm}$ respectively as shown in the figure. The two arcs $\mathrm{AD}$ and $\mathrm{BC}$ subtend a common angle $30^{\circ}$ at the centre $\mathrm{O}$. If the current flowing in the loop is $\frac{1.2}{\pi} \mathrm{A}$, then the magnitude of net magnetic field at $O$ is (Given $\mu_0=4 \pi \times 10^{-7}$ )
- A$0.5 \mu \mathrm{T}$
- B$3 \mu \mathrm{T}$
- Correct answerC$1 \mu \mathrm{T}$
- D$1.5 \mu \mathrm{T}$
Solution
$\begin{aligned} &\mathrm{B}_0=\mathrm{B}_{\mathrm{AB}}+\mathrm{B}_{\mathrm{BC}}+\mathrm{B}_{\mathrm{CD}}+\mathrm{B}_{\mathrm{DA}} \\ & =\mathrm{O}+\frac{\mu_0 \mathrm{i}}{24 \times 0.02} \otimes+\mathrm{O}+\frac{\mu_0 \mathrm{i}}{24 \times 0.01} \odot \\ & =\frac{\mu_0 \mathrm{i}}{24}\left[\frac{1}{0.02}-\frac{1}{0.01}\right] \otimes \\ & =\frac{4 \pi \times 10^{-7}}{24} \times \frac{1.2}{\pi}[50-100] \otimes \\ & =\frac{4 \times 10^{-7} \times 1.2}{24} \times-50 \otimes=-1 \mu \mathrm{T} \otimes=1 \mu \mathrm{T} \odot\end{aligned}$
Hence, the correct answer is option: 'C'.