Question

A wire carrying a 28 A current bends through a right angle. Consider two 2mm segments of wire, each 3 cm from the bend. Find the magnitude and direction of the magnetic field these two segments produce at point P, which is midway between them.

Short answer

The total magnitude of the magnetic field is $1.76\times {10}^{-5}\mathrm{T}$. According to the right-hand thumb rule, the direction of the magnetic field will be in the inward direction.

Step-by-step solution

The formula of magnetic field

The magnetic field due to a segment of length dl and carrying a current of I is given by Biot and Savart law

$\mathbf{dB}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{ldlsin\varphi }}{{\mathbf{r}}^{\mathbf{2}}}$

Where the angle$\mathbf{\varphi }$ is the angle between the position unit vector and the length of the line segment.

Calculate the magnitude and direction of the field

Here $\mathrm{\varphi }=45°,\mathrm{I}=28\mathrm{A}\mathrm{and}\mathrm{dl}=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m}$.

From the given figure:

$$\begin{gathered} \mathrm{r}=\frac{1}{2}\sqrt{3^2+3^2} \\ \mathrm{r}=2.121\mathrm{cm} \end{gathered}$$

The magnetic field can be calculated as:

$$\begin{gathered} {\mathrm{B}}_{\mathrm{tot}}=\frac{2\left(4\mathrm{\pi }\times {10}^{-7}\right)}{4\mathrm{\pi }}\times \frac{2\times {10}^{-3}\times 28\times \sin 45°}{{\left(2.121\times {10}^{-2}\right)}^2} \\ {\mathrm{B}}_{\mathrm{tot}}=1.76\times {10}^{-5}\mathrm{T} \end{gathered}$$

So, the total magnitude of the magnetic field is $1.76\times {10}^{-5}\mathrm{T}$. According to the right-hand thumb rule, the direction of the magnetic field will be in the inward direction.