Question

In Fig. 29-83, two infinitely long wires carry equal currents i. Each follows a $\mathbf{90}\mathbf{°}$arc on the circumference of the same circle of radius R. Show that the magnetic field $\overrightarrow{\mathbf{B}}$at the center of the circle is the same as the field $\overrightarrow{\mathbf{B}}$a distance R below an infinite straight wire carrying a current Ito the left.

Short answer

Proved.

Step-by-step solution

Given

Figure 29-83.

Understanding the concept

In this case, we take the vector sum of the magnetic fields and then use the relation for magnetic field in terms of current, distance, and angle subtended to get the required result. We will use the usual convention as $\otimes$which will represent the direction of the field going into the page and $\odot$will represent the direction of the field coming out of page

Formula:

Magnetic field due to the circular arc at the center of arc with current (i).

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

direction of B is determined by right hand thumb rule

 Step 3: Show that the magnetic field B→  at the center of the circle is the same as the field B→ a distance R below an infinite straight wire carrying a current I to the left

Let’s find the magnetic field due to the first and second wire separately. Adding them, we will get the net field due to both wires at the center of the arc using equation 29-9

$B_1=\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{2\pi R}·\frac{\pi }{2}+\frac{{\mu }_{0}i}{4\pi R}$

This field is pointing out of the page.

For wire 2

$B_2=0+\frac{{\mu }_{0}i}{2\pi R}·\frac{\pi }{2}+0$

This field is pointing into the page.

So the net field at the center is

$$\begin{gathered} B_c=B_1+B_2=\frac{{\mu }_{0}i}{4\pi R}+\frac{{\mu }_{0}i}{2\pi R}·\frac{\pi }{2}+\frac{{\mu }_{0}i}{4\pi R}-\frac{{\mu }_{0}i}{2\pi R}·\frac{\pi }{2} \\ {B}_c=\frac{{\mu }_{0}i}{2\pi R} \end{gathered}$$

We can see that this expression is exactly the same as the equation for the field by a single infinite straight wire (equation 29-4).

Hence, it is proved.