Question

Find the force and torque on the wire AB due to the infinitely long wire. Neglect gravity.

• A. $\frac{{\mu }_0{i}^2}{2\pi },lo{g}_e\left(\frac{a+l}{a}\right),\frac{{\mu }_0{i}^{2}l}{2\pi }$
• B. $zero,\frac{{\mu }_0{i}^{2}l}{2\pi }$
• C. $\frac{{\mu }_0{i}^2}{2\pi }ln\left(\frac{l}{a}\right),\frac{{\mu }_0{i}^{2}l}{2\pi }$
• D. $\frac{{\mu }_0{i}^2}{2\pi }ln\left(\frac{l+a}{q}\right),zero$

Answer 1

The correct option is A $\frac{{\mu }_0{i}^2}{2\pi },lo{g}_e\left(\frac{a+l}{a}\right),\frac{{\mu }_0{i}^{2}l}{2\pi }$


$dF=BidxSin{90}^{\circ }$ $d\tau =df\times x$
$dF=\frac{{\mu }_{0}i}{2\pi x}idx$ $=\frac{{\mu }_0{i}^2}{2\pi x}\times xdx$
$dF=\frac{{\mu }_0{i}^2}{2\pi }\frac{dx}{x}$ $\tau =\frac{{\mu }_0{i}^2}{2\pi }{\int }_a^{a+1}dx$
$F=\frac{{\mu }_0{i}^2}{2\pi }{\int }_a^{a+l}\frac{dx}{x}$ $=\frac{{\mu }_0{i}^2}{2\pi }(a+l-a)$
$F=\frac{{\mu }_0{i}^2}{2\pi }lo{g}_e\left(\frac{a+l}{2}\right)$ $=\frac{{\mu }_0{i}^2}{2\pi }\times l$