Question

The magnetic field $B$ at point $P$ at a distance $R$ due to the wire of carrying current $I$ is

• A. $\frac{{\mu }_o}{4\pi }\frac{I}{R}(cos{\varphi }_1-cos{\varphi }_2)$
• B. $\frac{{\mu }_o}{4\pi }\frac{I}{R}(cos{\varphi }_1+cos{\varphi }_2)$
• C. $\frac{{\mu }_o}{4\pi }\frac{I}{R}(sin{\varphi }_1+sin{\varphi }_2)$
• D. $\frac{{\mu }_o}{4\pi }\frac{I}{R}(sin{\varphi }_1-sin{\varphi }_2)$

Answer 1

Answer: (C)

$\frac{{\mu }_o}{4\pi }\frac{I}{R}(sin{\varphi }_1+sin{\varphi }_2)$

$\therefore$ Magnetic field at P due to $\overrightarrow{dl}$ is

$dB=\frac{{\mu }_{o}idl}{4\pi }\frac{\sin (90-\theta )}{r^2}$

$dB=\frac{{-\mu }_{o}idl}{4\pi }\frac{\cos \theta }{r^2}$$(\therefore \sin (90-\theta )=-\cos \theta )$

From figure, let $EQ=l$

Integrating $dB$ over limits of $AB$

$\int dB=\frac{{\mu }_{o}i}{4\pi R}\int \frac{dlcos\theta }{r^2}$

$=\frac{{-\mu }_{o}i}{4\pi R}\times {\left[\sin \theta \right]}_{{\varphi }_1}^{{\varphi }_2}$

$B=\frac{{\mu }_{o}i}{4\pi R}(\sin {\varphi }_1+\sin {\varphi }_2)$