Q149 / 160 · TS EAMCET 2024 (11 May Shift 1) · Physics
A straight wire of length 20 cm carrying a current of $\frac{3}{\pi^2}$ A is bent in the form of a circle. The magnetic field at the centre of the circle is
- A$8 \times 10^{-6} \mathrm{~T}$
- B$3 \times 10^{-6} \mathrm{~T}$
- C$12 \times 10^{-6} \mathrm{~T}$
- Correct answerD$6 \times 10^{-6} \mathrm{~T}$
Solution
$\mathrm{L}=20 \mathrm{~cm}, \mathrm{I}=\frac{3}{\pi^2} \mathrm{~A}$
For circle, $\mathrm{L}=2 \pi \mathrm{R} \Rightarrow 20 \times 10^{-2}=2 \pi \mathrm{R}$ $\therefore \mathrm{R}=\frac{10^{-1}}{\pi} \mathrm{~m}$ $\therefore \quad$ Magnetic field at the centre of circle, $\begin{aligned} & \mathrm{B}=\frac{\mu_{\mathrm{0}} \mathrm{I}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times \frac{3}{\pi^2}}{2 \times \frac{10^{-1}}{\pi}} \\ & =6 \times 10^{-6} \mathrm{~T} \end{aligned}$
For circle, $\mathrm{L}=2 \pi \mathrm{R} \Rightarrow 20 \times 10^{-2}=2 \pi \mathrm{R}$ $\therefore \mathrm{R}=\frac{10^{-1}}{\pi} \mathrm{~m}$ $\therefore \quad$ Magnetic field at the centre of circle, $\begin{aligned} & \mathrm{B}=\frac{\mu_{\mathrm{0}} \mathrm{I}}{2 \mathrm{R}}=\frac{4 \pi \times 10^{-7} \times \frac{3}{\pi^2}}{2 \times \frac{10^{-1}}{\pi}} \\ & =6 \times 10^{-6} \mathrm{~T} \end{aligned}$
Hence, the correct answer is option: 'D'.