91Ó°ÊÓ

1. Width of the long thin ribbon is$w=4.91cm=0.0491m$
2. The total current is$i=4.61\mathrm{μ}A=4.61×{10}^{-6}A$.
3. The point P is at distance$d=2.16cm=0.0216m$.
4. The ribbon is being constructed from many long, thin, parallel wires.

By using the equation for thecurrent carriedby the section of the ribbon of thickness$\mathbf{dx}$in its contribution toand integrating it with respect to$\mathit{x}$, we can find themagnetic field at a point P in the plane of the ribbon at$\mathit{d}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{16}\mathit{c}\mathit{m}$.

Formulas:

The current carried by the section of the ribbon of thickness $dx$is

$di=\frac{idx}{w}$

The contribution ofto magnetic field$B_P$is

$d{B}_P=\frac{{\mathrm{μ}}_{0}di}{2\mathrm{π}x}$

Let’s consider a section of the ribbon of thickness$dx$located at distance$x$away from point P.

The current it carries is

$di=\frac{idx}{w}…………………………………………………………………\left(1\right)$

And its contribution to the magnetic field$B_p$is

$d{B}_P=\frac{{\mathrm{μ}}_{0}di}{2\mathrm{π}x}………………………………………………………………\left(2\right)$

From equations $\left(1\right)$and$\left(2\right)$, we get

$d{B}_P=\frac{{\mathrm{μ}}_{0}idx}{2\mathrm{π}xw}$

Now integrating the above equation, we get

$$\begin{gathered} B_p=∫d{B}_p={∫}_d^{d+w}\frac{{\mathrm{μ}}_{0}idx}{2\mathrm{Ï€³\ae ·É}} \\ {B}_p=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€·É}}{∫}_d^{d+w}\frac{dx}{x} \\ {B}_p=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€·É}}\ln \left(1+\frac{w}{d}\right) \end{gathered}$$$B_P=\frac{\left(4\mathrm{Ï€}×{10}^{-7}\right)×\left(4.61×{10}^{-6}A\right)}{2\mathrm{Ï€}×0.0491}In\left(1+\frac{0.0491}{0.0216}\right)$

$B_P=2.23×{10}^{-11}T$

And $B_p$points upward.

Therefore, in unit vector notation$B_P=\left(2.23×{10}^{-11}T\right)\hat{j}$