Question

Figure shows a long straight current carrying wire with current $I_1=0.5\text{A}$, and a thin strip of width $b=10\text{cm}$ and length $l=1.5\text{m}$ placed parallel to it with a current $I_2=2\text{A}$ as shown. The distance between wire and strip is $r=1\text{m}$. Find the magnetic force of interaction between current $I_1$ and $I_2$.

• A. $5\times {10}^{-6}\text{N}$
• B. $3\ln \left(11\right)\times {10}^{-6}\text{N}$
• C. $5\ln \left(11\right)\times {10}^{-6}\text{N}$
• D. $\ln \left(11\right)\times {10}^{-6}\text{N}$

Answer 1

The correct option is B $3\ln \left(11\right)\times {10}^{-6}\text{N}$

Consider an elemental strip of width $dx$ in the strip at a distance $x$ from the wire as shown in the figure.

The current in the elemental strip is given by

$dI=\frac{I_2}{b}dx$

So, the force experienced by elemtary strip due tio the wire is given by,

$dF=BIL=(\frac{{\mu }_0{I}_1}{2\pi x})\left(dI\right)\left(l\right)$

So the force interaction between $I$ and $dI$ is given as

$dF=\frac{{\mu }_0{I}_{1}dIl}{2\pi x}$

Total force on the strip can be calculated by integrating the above force within limits from $x=r$ to $x=r+b$

$\therefore F=\int dF={\int }_r^{r+b}\frac{{\mu }_0{I}_{1}dIl}{2\pi x}$

$F=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi b}{\int }_r^{r+b}\frac{dx}{x}(\because dI=I_{2}dx/b)$

$F=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi b}\ln \left(\frac{r+b}{b}\right)$

Here, $I_1=0.5\text{A},I_2=2\text{A},l=1.5\text{m},$
$b=0.1\text{m}$ and $r=1\text{m}$

$F=\frac{4\pi \times {10}^{-7}\times 0.5\times 2\times 1.5}{2\pi \times 0.1}\times \ln \left(\frac{0.1+1}{0.1}\right)$

$\therefore F=3\ln \left(11\right)\times {10}^{-6}\text{N}$

Hence, option $\left(\text{B}\right)$ is correct.