91Ó°ÊÓ

1. Permeability of free space, ${\mathrm{μ}}_0=4\mathrm{π}×{10}^{-7}\frac{\text{Tm}}{\text{A}}$.
2. Ratio of currents $\frac{i_1}{i_2}=4$.

Using the formula for the magnetic field in terms of current and distance from the wire and given graph in the figure, find the condition when the field becomes maximum. Using this condition, find the maximum value of field. Also, using the given graph, find the directions of the currents.

Formula:

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

From the graph, we can say that $B_y=0$at $x=10\text{cm}$. So, the direction of current should be opposite.

So, $B_y=B_1-B_2$.

Here, the magnetic field of 1^st wire is:

$B_1=\frac{{\mathrm{μ}}_0{i}_1}{2\mathrm{π}\left(L+x\right)}$

Magnetic field of 2^nd wire is:

$B_2=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}x}$

Hence,

$B_y=\frac{{\mathrm{μ}}_0{i}_1}{2\mathrm{π}\left(L+x\right)}-\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}x}$

Since $i_1=4i_2$, rewrite the equation as:

$B_y=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}}\left(\frac{4}{L+x}-\frac{1}{x}\right)$ ….. (1)

From the given graph, we can see that $B_y=0$at $x=10\text{cm}$

$0=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}}\left(\frac{4}{L+0.1}-\frac{1}{0.1}\right)$

$∴\left(\frac{4}{L+0.1}-\frac{1}{0.1}\right)=0$

$\frac{4}{L+0.1}=\frac{1}{0.1}$

Solving this, we get$L=0.3\text{m}$

For the maximum value of localid="1663178138447" $B_{y,}$ the derivative of the above term should be equal to zero.

Differentiating this with respect to x and setting localid="1663165587828">dBydx=0as follows:

$0=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}}\left(-\frac{4}{{\left(L+x\right)}^2}+\frac{1}{x^2}\right)$

Solve further as:

$0=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}}\left(\frac{3x^2-2Lx-L^2}{{\left(L+x\right)}^2{x}^2}\right)$

$3x^2-2Lx-L^2=0$

$∴\left(3x+L\right)\left(x-L\right)=0$

Since in the problem it is mentioned that x is measured in positive direction, then:

$x=L$

So localid="1663178119536" $B_y$is maximum at $x=L$

Therefore, $B_y$ is maximum at$x=30\text{cm}$.

For $i_2=0.003A$, find value of maximum field by finding the value oflocalid="1663177725735" $B_1$or$B_2$at$x=30\text{cm}$(as both fields are equal in magnitude and opposite in direction), so we can use as:

$B_2=\frac{{\mathrm{μ}}_0{i}_2}{2\mathrm{π}x}$

$B_{max}=\frac{4\mathrm{π}×{10}^{-7}×0.003}{2\mathrm{π}×0.3}$

$B_{max}=2×{10}^{-9}T$

$B_{max}=2nT$

From the diagram, it is observed that as one approaches the wire 2, the field becomes more and more negative. Using right hand rule, conclude that the current in wire 2 is into the plane of the paper. From the direction of the current $i_{2,}$one can say that the direction of currentlocalid="1663172504809" $i_1$is out of the plane of the paper.

As mentioned in the previous step, current $i_2$ is into the plane of the paper.