Question

The magnetic field $B$ at point $P$ at a distance $R$ due to the wire of length $L$ carrying current $I$ , when ${\varphi }_1={\varphi }_2$ is

• A. $\frac{{\mu }_o}{4\pi }\frac{I}{R}\frac{L}{\sqrt{R^2+L^2}}$
• B. $\frac{{\mu }_o}{2\pi }\frac{I}{R}\frac{L}{\sqrt{R^2+L^2}}$
• C. $\frac{{\mu }_o}{2\pi }\frac{I}{R}\frac{L}{\sqrt{2R^2+L^2}}$
• D. $\frac{{\mu }_o}{2\pi }\frac{I}{R}\frac{L}{\sqrt{4R^2+L^2}}$

Answer 1

The correct option is D $\frac{{\mu }_o}{2\pi }\frac{I}{R}\frac{L}{\sqrt{4R^2+L^2}}$

For the infinitesimal portion, magnetic field
$dB=\frac{{\mu }_0}{4\pi }\frac{I\overrightarrow{d}l\times \overrightarrow{r}}{r^3}=\frac{{\mu }_0}{4\pi }\frac{Idl\sin (180-\theta )}{r^2}$

Since angle between $Idl$ and $r$ is $(180-\theta )$
Now, $EG=EF\sin \theta =dl\sin \theta$
and $EG=EP\sin d\varphi =r\sin d\varphi$

Therefore, comparing two values of $EG$ we have:

$dlsin\theta =rd\varphi$
$dB=\frac{{\mu }_0}{4\pi }\frac{Id\varphi }{r}$

Again from triangle $EQP$
$r=\frac{R}{\cos \varphi }$
$dB=\frac{{\mu }_0}{4\pi }\frac{I\cos \varphi d\varphi }{R}$

Total magnetic field at point P due the whole conductor is obtained by integrating $dB$ from $-{\varphi }_1$ to ${\varphi }_2$.
Therefore,
$B=\frac{{\mu }_{0}I}{4\pi R}(\sin {\varphi }_1+\sin {\varphi }_2)$

When ${\varphi }_1$ = ${\varphi }_2$
$B=\frac{{\mu }_{0}I}{4\pi R}\left(2\sin {\varphi }_1\right)$

$\sin {\varphi }_1=\frac{l}{2\sqrt{R^2+\frac{l^2}{4}}}=\frac{l}{\sqrt{4R^2+l^2}}$

$B=\frac{{\mu }_{0}I}{2\pi R}\frac{l}{\sqrt{4R^2+l^2}}$