Question

The magnetic force between wires, as shown in figure is

• A. $\frac{{\mu }_{0}iI}{2\pi }\ln \left(\frac{x+L}{2x}\right)$
• B. $\frac{{\mu }_{0}iI}{2\pi }\ln \left(\frac{2x+L}{2x}\right)$
• C. $\frac{{\mu }_{0}iI}{2\pi }\ln \left(\frac{x+L}{x}\right)$
• D. $\frac{{\mu }_{0}iI}{2\pi }\ln \left(\frac{x-L}{x}\right)$

Answer 1

The correct option is C $\frac{{\mu }_{0}iI}{2\pi }\ln \left(\frac{x+L}{x}\right)$

Consider a small element of length $dr$ on the wire $AB$ at distance $r$ from the long wire.


Magnetic field at distance $r$ from a long current carrying wire,

${\overrightarrow{B}}_r=\frac{{\mu }_{0}I}{2\pi r}\odot \left(\text{From Right hand thumb rule}\right)$

Force on small current element at distance $r$,

$d\overrightarrow{F}=i(\overrightarrow{dr}\times \overrightarrow{B_r})$

$\Rightarrow dF=\frac{{\mu }_{0}iIdr\sin {90}^{\circ }}{2\pi r}$ (towards right from Fleming's right-hand rule)

Total magnetic force on current element of length $L$,

$\int dF=\frac{{\mu }_{0}iI}{2\pi r}\underset{x}{\overset{x+L}{\int }}\frac{dr}{r}$

$F=\frac{{\mu }_{0}iI}{2\pi }\ln \left(\frac{x+L}{x}\right)$ towards right.

Hence, option (c) is the correct answer.