In Figure, a long circular pipe with outside radius $\mathit{R}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{}\mathit{c}\mathit{m}$carries a (uniformly distributed) current $\mathit{i}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathit{A}$into the page. A wire runs parallel to the pipe at a distance of $\mathbf{3}\mathbf{.}\mathbf{00}\mathit{R}$from center to center. (a) Find the magnitude and (b) Find the direction (into or out of the page) of the current in the wire such that the net magnetic field at point Phas the same magnitude as the net magnetic field at the center of the pipe but is in the opposite direction.

• The outside radius of the pipe is$R=2.6cm=0.026m$
• Current in the pipe is$i_p=8mA$
• Distance between pipe and wire$=3R$
  

The magnetic field Bdue to the current-carrying conductor with current${\mathit{i}}_{\mathbf{w}}$at a perpendicular distance Ris,

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{i}}_{\mathbf{w}}}{\mathbf{2}\mathbf{\pi }\mathbf{R}}$

By using the expression for the magnetic field due to the current-carrying conductor to wire and pipe and applying the given conditions, we can find the magnitude and the direction of the current in the wire.

The center of the pipe be at a point C.

Suppose the magnetic field due to wire at point P is$B_{Pw}$ , the magnetic field due to wire at point C is $B_{Cw}$, the magnetic field due to pipe at point P is$B_{Pp,}$ and the magnetic field due to pipe at point C is$B_{Cp}=0$ since the electric field inside the pipe is zero, which leads to zero magnetic field.

The magnetic field due to wire at point P is$B_{Pw}=\frac{{\mu }_0{i}_w}{2\pi R}$

The magnetic field due to wire at point C is$B_{Cw}=\frac{{\mu }_0{i}_w}{2\pi \left(3R\right)}$

Thus, we can conclude that$B_{Pw}>B_{Cw}$

As, $B_{Cp}=0$the magnetic field at point C will be due to the wire alone, i.e.,

$$\begin{gathered} B_C=B_{Cw} \\ =\frac{{\mu }_0{i}_w}{2\pi \left(3R\right)} \end{gathered}$$ (i)

Since it is given that the current through the pipe$i_p$is into the page.

Forthewire, we have $B_{Pw}>B_{Cw}$. Thus, for $B_P=B_C=B_{Cw}$, the current through the wire$i_w$must also be into the page.

Thus,

$$\begin{gathered} B_P=B_{Pw}-B_{Pp} \\ =\frac{{\mu }_0{i}_w}{2\pi R}-\frac{{\mu }_0{i}_p}{2\pi \left(2R\right)} \\ =\frac{{\mu }_0}{2\pi R}\left[i_w-\frac{i_p}{2}\right] \end{gathered}$$

(ii)

But it is given that magnetic field at point is equal to magnetic field at point C but in opposite direction. So, setting $B_C=-B_P$, from equation (i) and (ii), we get

role="math" localid="1663114023064" $$\begin{gathered} \frac{{\mu }_0}{2\pi R}\left[i_w-\frac{i_p}{2}\right]=-\frac{{\mu }_0{i}_w}{2\pi \left(3R\right)} \\ \left[i_w-\frac{i_p}{2}\right]=-\frac{i_w}{3} \\ \left[i_w+\frac{i_w}{3}\right]=\frac{i_p}{2} \\ {i}_w\left[1+\frac{1}{3}\right]=\frac{i_p}{2} \\ \frac{4}{3}{i}_w=\frac{i_p}{2} \\ {i}_w=\frac{3i_p}{8} \end{gathered}$$

Substituting the given value,

$$\begin{gathered} i_w=\frac{3\left(8mA\right)}{8} \\ =3mA \end{gathered}$$

The current through the wire is$3mA$ .

As explained in part (a), the direction of the current is into the page.