91Ó°ÊÓ

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$i_B=10\mathrm{A}$

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:

$\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{π}\mathbf{R}}$

Net magnetic field at the origin:

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

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}R}$

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 $-\hat{k}$.

$$\begin{gathered} \stackrel{→}{B_A}=-\frac{{\mathrm{μ}}_0{i}_A}{2\mathrm{π}{r}_A}\hat{\mathrm{k}} \\ =-\frac{\left(1.26×{10}^{-7}\frac{\mathrm{T}.\mathrm{m}}{\mathrm{A}}\right)\left(6\mathrm{A}\right)}{2\mathrm{π}×0.100\mathrm{m}}\hat{\mathrm{k}} \\ =-1.2×{10}^{-6}\mathrm{T}\hat{\mathrm{k}} \end{gathered}$$

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

${\stackrel{→}{B}}_{net}={\stackrel{→}{B}}_A+{\stackrel{→}{B}}_B$

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

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

Value of$Y$where$\stackrel{→}{B}$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" $\frac{{\mathrm{μ}}_0{i}_A}{2\mathrm{π}\left(r_A-y\right)}=\frac{{\mathrm{μ}}_0{i}_B}{2\mathrm{π}(y-r_B)}$

$2\mathrm{π}{\mathrm{μ}}_0{i}_A(y-r_B)=2\mathrm{π}{\mathrm{μ}}_0{i}_B\left(r_A-y\right)$

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

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

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

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

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

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.

$\frac{{\mathrm{μ}}_0{i}_A}{2\mathrm{π}\left(y-r_A\right)}=\frac{{\mathrm{μ}}_0{i}_B}{2\mathrm{π}(y-r_B)}$

Solving for y we get,

$2\mathrm{π}{\mathrm{μ}}_0{i}_A(y-r_B)=2\mathrm{π}{\mathrm{μ}}_0{i}_B\left(y-r_A\right)$

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

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

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

$y=0.0175\text{m}$

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