91Ó°ÊÓ

We have given,

the diagram shown the current carrying wires.

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

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

$\mathrm{B}=\frac{{\mathrm{μ}}_0\mathrm{I}}{2\mathrm{Ï€¸é}}$

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,

$$\begin{gathered} {\mathrm{B}}_{\mathrm{a}1}=\frac{{\mathrm{μ}}_0\mathrm{I}}{2\mathrm{Ï€¸é}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{a}1}=\frac{4\mathrm{Ï€}×{10}^{-7}×10}{2\mathrm{Ï€}×0.01×\sqrt{2}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{a}1}={10}^{-4}(\mathrm{i}+\mathrm{j})\mathrm{T} \end{gathered}$$

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

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

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

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

then Total magnetic field at b

localid="1649340438824" $$\begin{gathered} {\mathrm{B}}_{\mathrm{b}}=2×\frac{{\mathrm{μ}}_0\mathrm{i}}{2\mathrm{Ï€°ù}} \\ {\mathrm{B}}_{\mathrm{b}}=\frac{2×4\mathrm{Ï€}×{10}^{-7}×10}{2\mathrm{Ï€}(0.01)} \\ {\mathrm{B}}_{\mathrm{b}}=4×{10}^{-4}\mathrm{T} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{B}}_{\mathrm{c}1}=\frac{{\mathrm{μ}}_0\mathrm{I}}{2\mathrm{Ï€¸é}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{c}1}=\frac{4\mathrm{Ï€}×{10}^{-7}×10}{2\mathrm{Ï€}×0.01×\sqrt{2}}(\cos {45}^0\mathrm{i}+\sin {45}^0\mathrm{j}) \\ {\mathrm{B}}_{\mathrm{c}1}={10}^{-4}(\mathrm{i}+\mathrm{j})\mathrm{T} \end{gathered}$$

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

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