Question

What are the magnetic fields at points a to c in FIG URE EX29.12? Give your answers as vectors.

Short answer

At a. $2\times {10}^{-4}\mathrm{T}$

At b. $4\times {10}^{-4}\mathrm{T}$

AT C.$2\times {10}^{-4}\mathrm{T}.$

Step-by-step solution

Given information

We have given,

the diagram shown the current carrying wires.

We have to find the magnetic field at a and c.

Simplify

The magnetic field due to current carrying wire at a distance is

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi R}}$

Let us find the magnetic field due to upper wire at point a.

The distance between the point a and wires are,

$\mathrm{r}=\sqrt{1+1}=\sqrt{2}\mathrm{cm}$

then,

$$\begin{gathered} {\mathrm{B}}_{\mathrm{a}1}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi R}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{a}1}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 10}{2\mathrm{\pi }\times 0.01\times \sqrt{2}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{a}1}={10}^{-4}(\mathrm{i}+\mathrm{j})\mathrm{T} \end{gathered}$$

Since the distance of point a from both the wire is same. so magnetic fields will be same but the direction will opposite.

Then total magnetic field will be

${\mathrm{B}}_{\mathrm{a}}=2\times {10}^{-4}\mathrm{T}$

Simplify

For magnetic field at b, Since the distance is perpendicular.

$\mathrm{r}=0.01\mathrm{m}$

then Total magnetic field at b

localid="1649340438824" $$\begin{gathered} {\mathrm{B}}_{\mathrm{b}}=2\times \frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{\pi r}} \\ {\mathrm{B}}_{\mathrm{b}}=\frac{2\times 4\mathrm{\pi }\times {10}^{-7}\times 10}{2\mathrm{\pi }(0.01)} \\ {\mathrm{B}}_{\mathrm{b}}=4\times {10}^{-4}\mathrm{T} \end{gathered}$$

For magnetic field at c, we find it similarly as we find for at a.

$$\begin{gathered} {\mathrm{B}}_{\mathrm{c}1}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi R}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{c}1}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 10}{2\mathrm{\pi }\times 0.01\times \sqrt{2}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{c}1}={10}^{-4}(\mathrm{i}+\mathrm{j})\mathrm{T} \end{gathered}$$

Since the distance of point c from both the wire is same. so magnetic fields will be same but the direction will opposite.

Then total magnetic field will be

${\mathrm{B}}_{\mathrm{c}}=2\times {10}^{-4}\mathrm{T}$