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A straight wire carrying a current of \(12\;A\)  is bent into a semi-circular arc of radius \(\text{2.0 cm}\)  as shown in Fig.\((a).\) Consider the magnetic field \(B\)  at the Centre of the arc.
In what way the contribution to \(B\)  from the semicircle differs from that of a circular loop and in what way does it resemble?
Question: A straight wire carrying a current of  12;A  is bent into a semi-circular arc of radius  text{2.0 cm}  as shown in Fig.(a). Consider the magnetic field  B  at the Centre of the arc. In what way the contribution to  B  from the semicircle differs from that of a circular loop and in what way does it resemble?

Answer

For all segments of the semicircular arc, \(\mathit{dl}\times r\) are all parallel to each other \((\)into the plane of the paper\().\) All such contributions add up in magnitude. Hence direction of \(B\) for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop.
Thus \(B=\frac{\mu_0I}{2R}\)
Here,
\(R=\) radius \(=2\ cm\ =\ 0.02\ m\)
\(I\ =\ \)current \(=\ 12\ A\)
\(\mu_0=4\pi\times10^{-7}\ Tm/A\)
\(\Rightarrow B=\frac{4\pi\times10^{-7}\times12}{2\times0.02}=1.9\times 10^{-4}\; \text T\) normal to the plane of the paper going into it.

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