Q149 / 160 · TS EAMCET 2024 (10 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
$1=20 \mathrm{~cm}, \mathrm{I}=\frac{3}{\pi^2} \mathrm{~A}$
For circle, $1=2 \pi R \Rightarrow 20 \times 10^{-2}=2 \pi R$ $\therefore \quad \mathrm{R}=\frac{10^{-1}}{\pi} \mathrm{~m}$ $\therefore \quad$ Magnetic field at the centre of circle, $\begin{aligned} & B=\frac{\mu_0 I}{2 R}=\frac{4 \pi \times 10^{-7} \times \frac{3}{\pi^2}}{2 \times \frac{10^{-1}}{\pi}} \\ & =6 \times 10^{-6} \mathrm{~T} \end{aligned}$
For circle, $1=2 \pi R \Rightarrow 20 \times 10^{-2}=2 \pi R$ $\therefore \quad \mathrm{R}=\frac{10^{-1}}{\pi} \mathrm{~m}$ $\therefore \quad$ Magnetic field at the centre of circle, $\begin{aligned} & B=\frac{\mu_0 I}{2 R}=\frac{4 \pi \times 10^{-7} \times \frac{3}{\pi^2}}{2 \times \frac{10^{-1}}{\pi}} \\ & =6 \times 10^{-6} \mathrm{~T} \end{aligned}$
Hence, the correct answer is option: 'D'.