Question

Figure shows a cross section of a long thin ribbon of width $\mathbf{w}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{91}\mathbf{cm}$that is carrying a uniformlydistributed total currentlocalid="1663150167158" $\mathit{i}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{16}\mathit{m}\mathit{A}$into the page. In unit-vector notation,what is the magnetic field at a point P in the plane of the ribbon at adistance localid="1663150194995" $\mathit{d}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{16}\mathit{c}\mathit{m}$from its edge? (Hint: Imaginethe ribbon as being constructed from many long, thin, parallel wires.)

Short answer

The magnetic field at a point in the plane of the ribbon at $d=2.16cm$is$B_P=\left(2.23\times {10}^{-11}T\right)\hat{j}$

Step-by-step solution

Given

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

Understanding the concept

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

$d{B}_P=\frac{{\mu }_{0}di}{2\pi x}$

  Step 3: Calculate the magnetic field at point P in the plane of the ribbon at d=2.16cm

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

The current it carries is

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

And its contribution to the magnetic field$B_p$is

$d{B}_P=\frac{{\mu }_{0}di}{2\pi x}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

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

$d{B}_P=\frac{{\mu }_{0}idx}{2\pi xw}$

Now integrating the above equation, we get

$$\begin{gathered} B_p=\int d{B}_p={\int }_d^{d+w}\frac{{\mu }_{0}idx}{2\mathrm{\pi xw}} \\ {B}_p=\frac{{\mu }_{0}i}{2\mathrm{\pi w}}{\int }_d^{d+w}\frac{dx}{x} \\ {B}_p=\frac{{\mu }_{0}i}{2\mathrm{\pi w}}\ln \left(1+\frac{w}{d}\right) \end{gathered}$$$B_P=\frac{\left(4\pi \times {10}^{-7}\right)\times \left(4.61\times {10}^{-6}A\right)}{2\pi \times 0.0491}In\left(1+\frac{0.0491}{0.0216}\right)$

$B_P=2.23\times {10}^{-11}T$

And $B_p$points upward.

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