Question

Two long thin parallel conductors of the shape shown in figure above carry direct currents $I_1$ and $I_2$. The separation between the conductors is $a$, the width of the right-hand conductor is equal to $b$. With both conductors lying in one plane, the magnetic interaction force between them reduced to a unit of their length is given as $F_1=\frac{{\mu }_0{I}_1{I}_2}{x\pi b}\ln \frac{a+b}{a}$. Find $x$.

Answer 1

We Know that Ampere's force per unit length on a wire element in a magnetic field is given by
$d{\overrightarrow{F}}_n=i(\hat{n}\times \overrightarrow{B})$ where $\hat{n}$ is the unit vector along the direction of current. (1)
Now, let us take an element of the conductor $i_2$, as shown in figure below. This wire element is in the magnetic field, produced by the current $i_1$, which is directed normally into the sheet of the paper and its magnitude is given by,
$|\overrightarrow{B}|=\frac{{\mu }_0{I}_1}{2\pi r}$ ......(2)
From equations (1) and (2)
$d{\overrightarrow{F}}_n=\frac{I_2}{b}dr(\hat{n}\times \overrightarrow{B})$, (because the current through the element equals $\frac{I_2}{b}dr$)
So, $d{\overrightarrow{F}}_n=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{b}\frac{dr}{r}$, towards left (as $\hat{n}\perp \overrightarrow{B}$).
Hence the magnetic force on the conductor,
${\overrightarrow{F}}_n=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{b}{\int }_a^{a+b}\frac{dr}{r}$ (towardsleft) $=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{b}\ln \frac{a+b}{a}$ (towards left).
Then according to the Newton's third law the magnitude of sought magnetic interaction force
$=\frac{{\mu }_0{I}_1{I}_2}{2\pi b}\ln \frac{a+b}{a}$
$⟹x=2$