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. $F =\frac{\mu_{0}}{4 \pi} \quad I _{1} I _{2} \quad \ell n 3(-\hat{ k })$
• B. $F =\frac{\mu_{0}}{4 \pi} 2 I _{1} I _{2} \quad \ell n 3(-\hat{ k })$
• C. $F =\frac{\mu_{0}}{4 \pi} \quad I _{1} I _{2} \quad \ell n 3(\hat{ k })$
• D. $F =\frac{\mu_{0}}{4 \pi} 2 I _{1} I _{2} \quad \ell n 3(\hat{ k })$

Answer 1

Answer: (A)

$\frac{x}{a}=\tan \theta $
$x=a \tan \theta $
$d x=\operatorname{asec}^2 \theta d \theta$
$ \mathrm{B}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1}{\mathrm{r}} $
$ \overrightarrow{d F}=\mathrm{I}_2 \mathrm{dx} \mathrm{B} \sin \theta(-\hat{\mathrm{k}}) \cdots(3) $
From (1),(2) and (3)
$ \overrightarrow{d F}=\mathrm{I}_2 \operatorname{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} 2 I_1 I_2 \quad \int_{-30^{\circ}}^{60^{\circ}} \tan \theta d \theta=\frac{\mu_o}{4 \pi} 2 I_1 I_2\left[-\log _e \cos \theta\right]_{-30^{\circ}}^{60^{\circ}}(-\hat{k}) \quad \vec{F}=\frac{\mu_o}{4 \pi} 2 I_1 I_2$$\left[-\log _e \cos 60^{\circ}+\log _e \cos \left(-30^{\circ}\right)\right]$=$\frac{\mu_0}{4 \pi} 2 I_1 I_2\left[\log _e \frac{\sqrt{3}}{2} \times 2\right](-\hat{k}) \quad \vec{F}=\frac{\mu_0}{4 \pi} I_1 I_2 \ell n 3(-\hat{k})$