Question

In Fig. 29-44, point ${\mathbf{P}}_{\mathbf{2}}$is at perpendicular distance$\mathbf{R}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{cm}$ from one end of a straight wire of length $\mathbf{L}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{cm}$carrying current $\mathbf{i}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{693}\mathbf{}\mathbf{A}\mathbf{.}$(Note that the wire is notlong.) What is the magnitude of the magnetic field at ${\mathbf{P}}_{\mathbf{2}}$?

Short answer

Magnitude of the magnetic field at$P_2$ is $132\text{nT}$.

Step-by-step solution

Determine the concept

Using Biot – Savart’s law, we can find the magnitude of the magnetic field at${\mathbf{P}}_{\mathbf{2}}$due to small length segment. For the magnetic field due to complete wire, we can integrate the magnetic field of small segment.

Formula:

$B=\frac{{\mu }_0}{4\pi }\times \left(\frac{Idl\sin \theta }{r^2}\right)$

Calculate the magnitude of the magnetic field at P2

Consider the diagram for the condition as follows:

From the above diagram:

$\sin \theta =\frac{R}{r}$

According to Pythagoras theorem:

$r^2=L^2+R^2$

$r=\sqrt{L^2+R^2}$

$\sin \theta =\frac{R}{\sqrt{L^2+R^2}}$

Consider the formula for magnetic field as:

$B=\frac{{\mu }_0}{4\pi }\times \left(\frac{Idl\sin \theta }{r^2}\right)$

Here, L changes from $0$to $0.136\mathrm{m}$.

$B=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{0.136}\frac{\sin \theta }{r^2}dI$

$B=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{0.136}\frac{\left({\displaystyle \frac{R}{\sqrt{L^2+R^2}}}\right)}{L^2+R^2}dI$

$B=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{0.136}\frac{R}{{\left(L^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}dI$

role="math" localid="1663004095204" $B=\frac{{\mu }_{0}IR}{4\pi }{\int }_0^{0.136}\frac{dI}{{\left(L^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}$

Consider the integral as:

$\int \frac{dx}{{\left(x^2+a^2\right)}^{\frac{3}{2}}}=\frac{x}{a^2{\left(x^2+a^2\right)}^{\frac{1}{2}}}$

Solve further as:

$B=\frac{{\mu }_{0}IR}{4\pi }{\left[\frac{L}{R^2{\left(L^2+R^2\right)}^{\frac{1}{2}}}\right]}_0^{0.136}$

Substitute the values and solve as:

$B=\frac{{\mu }_{0}I}{4\pi R}{\left[\frac{0.136}{{\left({0.136}^2+{0.251}^2\right)}^{\frac{1}{2}}}\right]}_0^{0.136}$

$B=\frac{4\pi \times {10}^{-7}\times 0.693}{4\pi \times 0.251}\times \left(0.48\right)$

$$\begin{gathered} B=1.32\times {10}^{-7}\text{T} \\ B=132\text{nT} \end{gathered}$$

Thus, the magnitude of the magnetic field at $P_2$is $132\text{nT}$.