Question

What current is needed in the top wire in Figure 22.58(a) to produce a field of zero at the point equidistant from the wires if the currents in the bottom two wires are both 10.0 A into the page?

Short answer

The top wire requires 10.0A directed into the page to produce a field of zero at the point of equidistance from the wires.

Step-by-step solution

Definition of magnetic field

A magnetic field is defined as a position in space near a magnet or an electric current where a physical field is formed by a moving electric charge applying force on another moving electric charge.

Given information and Formula to use

Current through the wire B is,${\mathrm{I}}_{\mathrm{B}}=10.0\mathrm{A}$

Current through the wire C is,${\mathrm{I}}_{\mathrm{c}}=10.0\mathrm{A}$

The magnetic field created by a long straight wire can be calculated using the following formula:

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}$.........(1)

Where ${\mathrm{\mu }}_0$absolute permeability.

Step 3: The current required in the top wire to produce a field of zero at the point of equidistance from the wires

Take a look at the diagram below, which shows the point O equidistance from each of the three wires.

From the above diagram

$$\begin{gathered} \mathrm{d}=0.05\tan 30° \\ {\mathrm{=2.887\times 10}}^{-2}\mathrm{m} \end{gathered}$$

The distance r can be calculated as,

$$\begin{gathered} r=\sqrt{0.{05}^5+{\left(2.887\times {10}^{-2}\right)}^2} \\ =5.77\times {10}^{-2}\mathrm{m} \end{gathered}$$

Substituting the values from the figure for wire B in equation (1), we get

$$\begin{gathered} B_B=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(10.0A\right)}{2\mathrm{\pi }\left(5.77\times {10}^{-2}\mathrm{m}\right)} \\ =3.466\times {10}^{-5}\mathrm{T} \end{gathered}$$

Substituting the values from the figure for wire C in equation (1), we get

$$\begin{gathered} B_C=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(10.0A\right)}{2\mathrm{\pi }\left(5.77\times {10}^{-2}\mathrm{m}\right)} \\ =3.466\times {10}^{-5}T \end{gathered}$$

The magnetic field vector diagram from the right-hand thumb rule at point O is,

The component of the net magnetic field will be zero, as shown in the vector diagram. The net magnetic field is made up of all of the x components. As a result, the net magnetic field generated by wires B and C have positive x-direction.

The magnitude of the net magnetic field can be calculated as,

$$\begin{gathered} B_{net}=\left(B_B+B_C\right)\cos 60° \\ =\left(3.466+3.466\right)\times {10}^{-5}\times \cos 60° \\ =3.466\times {10}^{-5}\mathrm{T} \end{gathered}$$

To calculate the magnetic field at point O, the magnetic field due to wire A must be equal to B in the negative x-direction.

The current in the wire A can be calculated as,

$$\begin{gathered} B_A=\frac{{\mu }_0{I}_A}{2\mathrm{\pi r}} \\ {\mathrm{I}}_{\mathrm{A}}=\frac{2{\mathrm{\pi B}}_{\mathrm{A}}}{{\mathrm{\mu }}_0} \\ {\mathrm{I}}_{\mathrm{A}}=\frac{2\mathrm{\pi }\times 5.77\times {10}^{-2}\times 3.466\times {10}^{-5}}{4\mathrm{\pi }\times {10}^{-7}} \\ {\mathrm{I}}_{\mathrm{A}}=10.0\mathrm{A} \end{gathered}$$

From the right-hand thumb rule, the direction of the current must be into the page.

Hence, the current required in the top wire to produce a field of zero at the point equidistance from the wires is 10.0A directed into the page.