MHT CET · Physics · Magnetic Effects of Current
A long straight conductor is bent into shape as shown. If it carries \(1 \mathrm{~A}\) and its radius is \(R\), then magnetic ficld \(B\) at the centre of circular coil

- A \(\infty\)
- B Zero
- C \(\frac{\mu_{0} i(\pi+1)}{2 \pi R}\)
- D \(\frac{\mu_{0} i(\pi-1)}{2 \pi R}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mu_{0} i(\pi-1)}{2 \pi R}\)
Step-by-step Solution
Detailed explanation
Magnetic field at centre \(O=\) magnetic field due to reaction \(A B+\) magnetic field due to section \(P Q R\)
\(
B_{0}=B_{A B}+B_{P Q R}
\)

but \(B_{A B}=\frac{\mu_{0} i}{2 \pi R}\) (upside the plane of paper)
and \(B_{P Q R}=\frac{\mu_{0} i}{2 R}\) (inside the plane of paper)
Hence \(B_{0}=\frac{\mu_{0} i}{2 R}-\frac{\mu_{0} i}{2 \pi R}\)
\(
B_{0}=\frac{\mu_{0} i(\pi-1)}{2 \pi R}
\)
\(
B_{0}=B_{A B}+B_{P Q R}
\)

but \(B_{A B}=\frac{\mu_{0} i}{2 \pi R}\) (upside the plane of paper)
and \(B_{P Q R}=\frac{\mu_{0} i}{2 R}\) (inside the plane of paper)
Hence \(B_{0}=\frac{\mu_{0} i}{2 R}-\frac{\mu_{0} i}{2 \pi R}\)
\(
B_{0}=\frac{\mu_{0} i(\pi-1)}{2 \pi R}
\)
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