Question

Figure 29-50ashows, in cross section, two long, parallel wires carrying current and separated by distance L. The ratio i₁/i₂ of their currents is$\mathbf{4}\mathbf{.}\mathbf{00}$; the directions of the currents are not indicated. Figure 29-50bshows the ycomponent B_yof their net magnetic field along the xaxis to the right of wire 2. The vertical scale is set by ${\mathbf{B}}_{\mathbf{ys}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{nT}$ , and the horizontal scale is set by ${\mathbf{x}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ . (a) At what value of $\mathit{x}\mathbf{}\mathbf{}\mathbf{0}$ is B_ymaximum?(b) If $\mathbf{}{\mathbf{i}}_{\mathbf{2}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{}\mathbf{mA}$, what is the value of that maximum? What is the direction (into or out of the page) of (c) i₁ and (d) i₂?

Short answer

1. The value of $x(>0)$when $B_y$is maximum is $30\text{cm}$.
2. If $i=3\text{mA}$the value of $B_y$is $2\text{nT}$.
3. Direction of$i_1$is out of the plane of paper.
4. Direction of $i_2$is into the plane of paper.

Step-by-step solution

Given

1. Permeability of free space, ${\mu }_0=4\pi \times {10}^{-7}\frac{\text{Tm}}{\text{A}}$.
2. Ratio of currents $\frac{i_1}{i_2}=4$.

Understanding the concept

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:

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

(a) Calculate at what value of   x>0 By is maximum

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:

$B_1=\frac{{\mu }_0{i}_1}{2\pi \left(L+x\right)}$

Magnetic field of 2^nd wire is:

$B_2=\frac{{\mu }_0{i}_2}{2\pi x}$

Hence,

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

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

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

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

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

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

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

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:

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

Solve further as:

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

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

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

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$x=30\text{cm}$.

(b) Calculate if, i2=3 mA, the value of By

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:

$B_2=\frac{{\mu }_0{i}_2}{2\pi x}$

$B_{max}=\frac{4\pi \times {10}^{-7}\times 0.003}{2\pi \times 0.3}$

$B_{max}=2\times {10}^{-9}T$

$B_{max}=2nT$

(c) Calculate direction of i1

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.

(d) Calculate direction of i2

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