Question

Two long straight wires are parallel and $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$apart .They are to carry equal currents such that the magnetic field at a point halfway between them has magnitude $\mathbf{}\mathbf{300}\mathbf{}\mathbf{\text{μT}}$. (a) Should the currents be in the same or opposite directions? (b) How much current is needed?

Short answer

1. The direction of current in the wire should be opposite each other
2. The amount of current needed is$I=30\text{A}$

Step-by-step solution

Given

1. The distance between two wires is$d=8.0\text{cm}=8.0\times {10}^{-2}\text{m}$
2. Magnetic field is$B=300\text{μT}=300\times {10}^{-6}\text{T}$

Determine the concept

Use the concept of magnetic field at the center of circular arc. Using the equation, find the magnetic field at the given point. Using right hand rule determine the direction of the magnetic field.

Formulae:

$\mathit{B}\mathbf{=}\frac{{\mathit{\mu }}_{\mathbf{0}}\mathit{i}\mathit{\varphi }}{\mathbf{4}\mathit{\pi }\mathit{R}}$

(a) Figure out if the currents should be in the same or opposite directions

If current in the wires is equal in magnitude and opposite in direction, then in the region half way between the wires, the magnetic field due to current carrying wires is equal, and they are in the same direction. So they will be added.

If the currents in the wires are in the same direction, then the region at halfway between the wires’ magnetic fields due to current-carrying wires is equal in magnitude and opposite in direction. Thus, net magnetic field will become zero.

But here, the magnetic field in the region at half way between the wires is not zero. So, the direction of current in the wire should be opposite each other.

(b) Calculate the current needed

Consider the diagram for the condition as follows:

The magnitude of magnetic field due to first long wire carrying currentI at perpendicular distance$R$is given as follows

$B_1=\frac{\mu I}{2\pi R}$

The magnitude of the magnetic field due to the second long wire carrying current I at perpendicular distance $R$is given as follows

$B_2=\frac{\mu I}{2\pi R}$

So, net magnetic field is addition of these two as follows

$$\begin{gathered} B=B+B_2 \\ B=\frac{\mu I}{2\pi R}+\frac{\mu I}{2\pi R} \\ B=\frac{\mu I}{\pi R} \\ I=\frac{B\times \pi \times R}{\mu } \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} I=\frac{\left(300\times {10}^{-6}\text{T}\right)\times \pi \times 4\times {10}^{-2}\text{m}}{4\pi \times {10}^{-7}{\text{N/A}}^{\text{2}}} \\ I=30\text{A} \end{gathered}$$

Hence the current is, $30\text{A}$.