Question

In Fig. 29-44 point ${\mathit{P}}_{\mathbf{1}}$is at distance $\mathit{R}\mathbf{=}\mathbf{}\mathbf{13}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{\text{cm}}\mathbf{}$on the perpendicular bisector of a straight wire of length $\mathit{L}\mathbf{=}\mathbf{}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{cm}}$. carrying current. (Note that the wire is notlong.) What is the magnitude of the magnetic field at ${\mathit{P}}_{\mathbf{1}}$due to i?

Short answer

The magnitude of the magnetic field at $P_1$due to i is$5.03\times {10}^{-8}\text{T}$

Step-by-step solution

Given

1. Current $i=58.2\text{mA}$
2. Distance of point P on the perpendicular bisector of wire$R=13.1\text{cm}=0.131\text{m}$
3. Length of wire$L=18.0\text{cm}=0.18\text{m}$

Determine the formulas:

Use the formula of Biot-Savarts law to find the magnitude of the field at ${\mathit{P}}_{\mathbf{1}}$due to i

Formula:

$\mathit{d}\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}\mathbf{}\mathbf{d}\mathbf{l}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{{\mathbf{r}}^{\mathbf{2}}}$

Calculate the magnitude of the magnetic field at P1 due to i

According to Bio-Savarts law

$dB=\frac{{\mu }_{0}idl}{4\pi }\frac{\sin \theta }{r^2}$

If r makes an angle $\theta$withl then

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

And

$$\begin{gathered} \sin \theta =\frac{R}{R}=\frac{R}{\sqrt{l^2+R^2}} \\ l=\frac{L}{2} \\ =\frac{0.18\text{m}}{2} \\ =0.09\text{m} \end{gathered}$$

Integrating an equation of Bio-Savarts law

$$\begin{gathered} B={\int }_0^{0.09}dB \\ ={\int }_0^{0.09}\frac{{\mu }_{0}idl}{4\pi }\frac{\sin \theta }{r^2} \\ B=\frac{{\mu }_{0}i}{4\pi }{\int }_0^{0.09}\frac{\sin \theta dl}{r^2} \\ B=\frac{{\mu }_{0}i}{4\pi }{\int }_0^{0.09}\frac{R}{\sqrt{l^2+R^2}}\frac{dl}{{\left(\sqrt{l^2+R^2}\right)}^2} \\ B=\frac{{\mu }_{0}iR}{4\pi }{\int }_0^{0.09}\frac{dl}{{\left(l^2+R^2\right)}^{3/2}} \end{gathered}$$

Solve further as:

$$\begin{gathered} B=\frac{{\mu }_{0}iR}{4\pi }\frac{1}{R^2}{\left|\frac{l}{{\left(l^2+R^2\right)}^{1/2}}\right|}_0^{0.09} \\ B=\frac{{\mu }_{0}i}{2\pi R}\left[\frac{l}{{\left(l^2+R^2\right)}^{1/2}}\right] \\ B=\frac{\left(4\pi \times {10}^{-7}\frac{\text{Tm}}{\text{A}}\right)\left(0.0582\text{A}\right)}{2\pi \left(0.131\text{m}\right)}\left[\frac{0.09\text{m}}{{\left[{\left(0.09\text{m}\right)}^2+{\left(0.131\text{m}\right)}^2\right]}^{1/2}}\right] \\ B=5.03\times {10}^{-8}\text{T} \end{gathered}$$

Therefore, the magnitude of the magnetic field at $P_1$due to i is$5.03\times {10}^{-8}\text{T}$