TS EAMCET · Physics · Magnetic Effects of Current
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 \(12 \times 10^{-6} \mathrm{~T}\)
- D \(6 \times 10^{-6} \mathrm{~T}\)
Answer & Solution
Correct Answer
(D) \(6 \times 10^{-6} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
\(1=20 \mathrm{~cm}, \mathrm{I}=\frac{3}{\pi^2} \mathrm{~A}\) 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,…
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