Question

A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point $P$ which lies on the centre of the semicircle?

• A. $\frac{{\mu }_{0}I}{4\pi r}(2-\pi )$
  
• B. $\frac{{\mu }_{0}I}{4\pi r}(2+\pi )$
  
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• C. $\frac{{\mu }_{0}I}{2\pi r}(2+\pi )$
  
• D. $\frac{{\mu }_{0}I}{2\pi r}(2-\pi )$
  

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{4\pi r}(2+\pi )$


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Magnetic field due to each of the semi-infinite straight wire $=\frac{{\mu }_{0}I}{4\pi r}$ (into the paper)

Magnetic field due to semicircular arc $=\frac{{\mu }_{0}I}{2r}\left(\frac{\pi }{2\pi }\right)=\frac{{\mu }_{0}I}{4r}$ (into the paper)

Thus, total magnetic field at point $P$:
$B=\frac{{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{4\pi r}+\frac{{\mu }_{0}I}{4r}$

$B=\frac{{\mu }_{0}I}{2\pi r}+\frac{{\mu }_{0}I}{4r}$

$B=\frac{{\mu }_{0}I}{4\pi r}(2+\pi )$