Question

Find the magnetic field at point $O$ due to a conductor of length $(\sqrt{3}+1)a$ carrying current $I$ at a distance of $a$ from the wire as shown in figure.

• A. $(\sqrt{2}+\sqrt{3})\frac{{\mu }_0}{4\pi }\frac{I}{a}$
• B. $(\sqrt{2}+\sqrt{3})\frac{{\mu }_0}{8\pi }\frac{I}{a}$
• C. $\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{a}$
• D. $\sqrt{3}\frac{{\mu }_0}{4\pi }\frac{I}{a}$

Answer 1

The correct option is B $(\sqrt{2}+\sqrt{3})\frac{{\mu }_0}{8\pi }\frac{I}{a}$


From above diagram, in $\Delta PSO$

$\tan {\theta }_1=\frac{PS}{SO}=\frac{a}{a}=1$

${\theta }_1={\tan }^{-1}\left(1\right)={45}^{\circ }$

From above diagram, in $\Delta PSO$

$\tan {\theta }_2=\frac{SQ}{SO}=\frac{\sqrt{3}a}{a}=\sqrt{3}$

${\theta }_2={\tan }^{-1}\left(\sqrt{3}\right)={60}^{\circ }$

Magnetic field due to wire $PQ$ at point $O$ is given by

$B=\frac{{\mu }_0}{4\pi }\frac{I}{OS}[\sin {\theta }_1+\sin {\theta }_2]$

$B=\frac{{\mu }_0}{4\pi }\frac{I}{\left(a\right)}[\sin {45}^{\circ }+\sin {60}^{\circ }]$

$B=(\sqrt{2}+\sqrt{3})\frac{{\mu }_0}{8\pi }\frac{I}{a}$

From right hand thumb rule, direction of magnetic field at point $O$ will be outside the plane of paper and perpendicular to it.

Hence, option $\left(b\right)$ is correct.