Question

An infinite wire placed along z-axis, has current ${\text{I}}_1$ in positive z-direction. A conducting rod placed in xy plane parallel to y-axis has current ${\text{I}}_2$ in positive y-direction. The ends of the rod subtend angles of +30^o and -60^o at the origin with positive x-direction. The rod is at a distance 'a' from the origin. Find net force on the rod.

• A. $F=\frac{{\mu }_0}{4\pi }{I}_1{I}_2\ell n3(-\hat{k})$
• B. $F=\frac{{\mu }_0}{4\pi }2I_1{I}_2\ell n3(-\hat{k})$
• C. $F=\frac{{\mu }_0}{4\pi }{I}_1{I}_2\ell n3(\hat{k})$
• D. $F=\frac{{\mu }_0}{4\pi }2I_1{I}_2\ell n3(\hat{k})$

Answer 1

Answer: (A)

$\frac{x}{a}=\tan \theta$
$x=a\tan \theta$
$dx={\mathrm{asec}}^2\theta d\theta$
$\mathrm{B}=\frac{{\mu }_0}{4\pi }\frac{2{\mathrm{I}}_1}{\mathrm{r}}$
$\overrightarrow{dF}={\mathrm{I}}_2\mathrm{dx}\mathrm{B}\sin \theta (-\hat{\mathrm{k}})\cdots (3)$
From (1),(2) and (3)
$\overrightarrow{dF}={\mathrm{I}}_2{\mathrm{asec}}^2\theta \mathrm{d}\theta \cdot \frac{{\mu }_0}{4\pi }\frac{2{\mathrm{I}}_1}{\mathrm{a}\sec \theta }\cdot \sin \theta =\frac{{\mu }_0}{4\pi }2{\mathrm{I}}_1{\mathrm{I}}_2\sin \theta \sec \theta \mathrm{d}\theta (-\hat{k})$
$\vec{F}=\frac{{\mu }_o}{4\pi }2I_1{I}_2{\int }_{-30^{\circ }}^{60^{\circ }}\tan \theta d\theta =\frac{{\mu }_o}{4\pi }2I_1{I}_2{\left[-{\log }_e\cos \theta \right]}_{-30^{\circ }}^{60^{\circ }}(-\hat{k})\vec{F}=\frac{{\mu }_o}{4\pi }2I_1{I}_2$$\left[-{\log }_e\cos 60^{\circ }+{\log }_e\cos \left(-30^{\circ }\right)\right]$=$\frac{{\mu }_0}{4\pi }2I_1{I}_2\left[{\log }_e\frac{\sqrt{3}}{2}\times 2\right](-\hat{k})\vec{F}=\frac{{\mu }_0}{4\pi }{I}_1{I}_2\ell n3(-\hat{k})$