Question

A wire is bent into the form of semicircle of radius R and carry current i. Magnetic field at point P as shown in figure is $B=\frac{{\mu }_{0}i}{a\pi R}ln(b+c)$

• A. $a=5$
• B. $b=\sqrt{2}$
• C. $C=0$
• D. $a+c=5$

Answer 1

Answer: (B), (D)

$a+c=5$

$dB=\frac{{\mu }_0}{4\pi }\frac{iRd\theta sin\alpha }{r^2}$
$\alpha +\beta =\frac{\pi }{2}\&2\beta +\theta =\frac{\pi }{2}$
$sin\alpha =cos\beta \Rightarrow cos\beta =\frac{r}{2R}$
$dB=\frac{{\mu }_0}{4\pi }\frac{iRcos\beta d\theta }{4R^{2}co{s}^2\beta }=\frac{{\mu }_{0}i}{16\pi R}sec(\frac{\pi }{4}-\frac{\theta }{2})d\theta$
$B=\int dB=\frac{{\mu }_{0}i}{16\pi R}{\int }_0^{\pi }sec(\frac{\pi }{4}-\frac{\theta }{2})d\theta =\frac{{\mu }_{0}i}{16\pi R}4In(\sqrt{2}+1)$
$B=\frac{{\mu }_0}{4\pi R}ln(\sqrt{2}+1)$