Question

Question: Two long wires lie in an xyplane, and each carries a current in the positive direction of the xaxis.Wire 1 is at $\mathit{y}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{cm}$and carries role="math" localid="1662817900403" ${\mathit{i}}_{\mathbf{A}}\mathbf{=}\mathbf{6}\mathbf{}\mathbf{A}$; wire 2 is at role="math" localid="1662817917709" $\mathit{y}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{cm}$and carries role="math" localid="1662817934093" ${\mathit{i}}_B\mathbf{=}\mathbf{10}\mathbf{}\mathbf{A}$. (a) In unitvector notation, what is the net magnetic field at the origin? (b) At what value of $\mathit{y}$doesrole="math" localid="1662818220108" $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$? (c) If the current in wire 1 is reversed, at what value of role="math" localid="1662818150179" $\mathit{y}$does$\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{0}$?

Short answer

1. In unit vector notation the net magnetic field at the origin is$\left(-5.2\times {10}^{-6}T\right)\hat{k}.$
2. Value ofwhere magnetic field is zero is$y=0.0813\text{m}.$
3. If the current in wire 1 is reversed, then the value of where is$y=0.0175\text{m}.$

Step-by-step solution

Given Data

1. Wire A distance $r_A=10\text{cm}$
2. Current in wire A$i_A=6\text{A}$
3. Wire B distance$r_B=5\text{cm}$
4. Current in wire B$i_B=10\mathrm{A}$

Understanding the concept

We use the formula of the magnetic field due to a long straight wire (infinitely long) at a perpendicular distance to calculate the magnetic field due to both wires. Adding the magnetic field of both wires, we get the net magnetic field.

Formulae:

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

(a) In unit vector notation the net magnetic field at the origin

Net magnetic field at the origin:

The magnitude of magnetic field at perpendicular distance from the long straight wire is given as

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

Suppose we have a wire $A$along $y$axis at a distance $y=r_A=0.100\text{m}$and wire $B$along $y=r_B=0.050\mathrm{m}.$

The magnetic field due to both wires is along $-Z$direction that is along $-\hat{k}$.

$$\begin{gathered} \overrightarrow{B_A}=-\frac{{\mu }_0{i}_A}{2\pi r_A}\hat{\mathrm{k}} \\ =-\frac{\left(1.26\times {10}^{-7}\frac{\mathrm{T}.\mathrm{m}}{\mathrm{A}}\right)\left(6\mathrm{A}\right)}{2\mathrm{\pi }\times 0.100\mathrm{m}}\hat{\mathrm{k}} \\ =-1.2\times {10}^{-6}\mathrm{T}\hat{\mathrm{k}} \end{gathered}$$

The net magnetic field at the origin is the sum of both these magnetic fields, and is given by

${\overrightarrow{B}}_{net}={\overrightarrow{B}}_A+{\overrightarrow{B}}_B$

role="math" localid="1662819683790" ${\overrightarrow{B}}_{net}=\left(-1.2\times {10}^{-6}\mathrm{T}-4.0\times {10}^{-6}\mathrm{T}\right)\hat{\mathrm{k}}$

role="math" localid="1662819885383" ${\overrightarrow{B}}_{net}=\left(-5.2\times {10}^{-6}\mathrm{T}\right)\hat{\mathrm{k}}$

Calculate value of  where magnetic field is zero.

Value of$Y$where$\overrightarrow{B}$is zero :

The magnetic field will be zero only in region$r_B<y<r_A$.

Because, in this region, the magnetic field of wire A cancels the magnetic field due to wire B.

role="math" localid="1662820464300" $\frac{{\mu }_0{i}_A}{2\pi \left(r_A-y\right)}=\frac{{\mu }_0{i}_B}{2\pi (y-r_B)}$

$2\pi {\mu }_0{i}_A(y-r_B)=2\pi {\mu }_0{i}_B\left(r_A-y\right)$

$y\left(i_A+i_B\right)=i_B{r}_A+i_A{r}_B$

role="math" localid="1662820237305" $y=\frac{i_B{r}_A+i_A{r}_B}{\left(i_A+i_B\right)}$

role="math" localid="1662820442193" $y=\frac{10\text{A}\times 0.100\text{m}+6\text{A}\times 0.050\text{m}}{\left(6\text{A}+10\text{A}\right)}$

role="math" localid="1662820451897" $y=0.0813\text{m}$

(c) Calculate value of  Y does B→=0 if the current in wire 1 is reversed 

If the current in wire 1 is reversed, then value of $Y$where $\overrightarrow{B}=0$:

We eliminate the$y<r_B$possibility due to wire B carrying the larger current. Hence, we expect a solution in the region $y>r_A$where magnetic field will be zero.

$\frac{{\mu }_0{i}_A}{2\pi \left(y-r_A\right)}=\frac{{\mu }_0{i}_B}{2\pi (y-r_B)}$

Solving for y we get,

$2\pi {\mu }_0{i}_A(y-r_B)=2\pi {\mu }_0{i}_B\left(y-r_A\right)$

$y\left(i_A-i_B\right)=i_A{r}_B-i_B{r}_A$

$y=\frac{i_A{r}_B-i_B{r}_A}{\left(i_A-i_B\right)}$

role="math" localid="1662821418906" $y=\frac{\left(6\text{A}\times 0.050\text{m}\right)-\left(10\text{A}\times 0.100\text{m}\right)}{\left(6\text{A}-10\text{A}\right)}$

$y=0.0175\text{m}$

On reversing the direction of current, $\overrightarrow{B}=0$at$y=0.0175\text{m}$.