Question

Question: In Fig. 29-77, a closed loop carries current $\mathbf{200}\mathbf{}\mathbf{mA}$. The loop consists of two radial straight wires and two concentric circular arcs of radii $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$and $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$. The angle is role="math" localid="1662809179609" $\mathit{\theta }\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{4}}\mathit{r}\mathit{a}\mathit{d}$. What are the (a) magnitude and (b) direction (into or out of the page) of the net magnetic field at the center of curvature P?

Short answer

1. Magnitude of net magnetic field at P is $2.75\times {10}^{-8}\mathrm{T}$.
2. Direction of magnetic field is into the page

Step-by-step solution

Given Data

1. Loop carries a current of $200\mathrm{mA}$
2. Radius$r_1=2.00\mathrm{m}$
3. Radius $r_2=4.00\mathrm{m}$
4. The angle is $\theta =\frac{\mathrm{\pi }}{4}\mathrm{rad}$

Understanding the concept

We use the formula for magnetic field at the center of a circular arc, and it depends on current, radius of circular arc, and central angle.

Formula:

$\mathit{B}\mathbf{=}\frac{\mathbf{\mu }\mathbf{i}\mathbf{\Phi }}{\mathbf{4}\mathbf{\pi R}}$

(a) Calculate the magnitude of net magnetic field at P

The central angle is as follows-

$$\begin{gathered} \varnothing =2\mathrm{\pi }-\frac{\mathrm{\pi }}{4} \\ =\frac{7\mathrm{\pi }}{4}\mathrm{rad}. \end{gathered}$$

Now magnetic field due to the outer wire is as follows

$$\begin{gathered} B_1=\frac{\mu i\varnothing }{4\mathrm{\pi R}} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(200\times {10}^{-3}\mathrm{A}\right)\left({\displaystyle \frac{7\mathrm{\pi }}{4}}\right)}{4\mathrm{\pi }\times 4} \\ =2.74\times {10}^{-8}\mathrm{T} \end{gathered}$$

From the right hand rule, this field is directed out of page.

Now magnetic field due to inner wire is as follows

$$\begin{gathered} B_2=\frac{\mu i\varnothing }{4\mathrm{\pi R}} \\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \raisebox{1ex}{$\mathrm{H}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{m}$}\right.}\right)\left(200\times {10}^{-3}\mathrm{A}\right)\left({\displaystyle \frac{7\mathrm{\pi }}{4}}\right)}{4\mathrm{\pi }\times 2} \\ =5.49\times {10}^{-8}\mathrm{T} \end{gathered}$$

From the right hand rule, this field is directed into the page.

Net field is as follows

role="math" localid="1662811190294" $$\begin{gathered} B=B_2-B_1 \\ =5.49\times {10}^{-8}\mathrm{T}-2.74\times {10}^{-8}\mathrm{T} \\ =2.75\times {10}^{-8}\mathrm{T} \end{gathered}$$

(b) Calculate direction of magnetic field

Net field is directed into the page because field by the inner arc is greater than field by the outer arc.