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Expert-verified Found in: Page 860 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Figure 29-67 shows a cross section across a diameter of a long cylindrical conductor of radius ${\mathbit{a}}{\mathbf{=}}{\mathbf{}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbit{c}}{\mathbit{m}}$ carrying uniform current ${\mathbf{170}}{\mathbf{}}{\mathbit{A}}$. What is the magnitude of the current’s magnetic field at radial distance (a) 0 (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm? (a) The magnitude of the current’s magnetic field at a radial distance 0 cm is zero.

(b)The magnitude of the current’s magnetic field at a radial distance 1.00cm is $8.50×{10}^{-4}T$

(c) The magnitude of the current’s magnetic field at a radial distance 2.00 cm is $1.70×{10}^{-3}T$

(d) The magnitude of the current’s magnetic field at a radial distance 4.00 cm is $8.50×{10}^{-4}T$

See the step by step solution

## Step 1: Listing the given quantities

• The radius of the conductor is $a=2.00cm=0.0200m$
• Current through the conductor is $i=170A$

## Step 2: Understanding the concept of the magnetic field at a point

Ampere’s law states that,

${\mathbf{\oint }}\stackrel{\mathbf{\to }}{\mathbf{B}}{\mathbf{·}}{\mathbit{d}}\stackrel{\mathbf{\to }}{\mathbf{s}}{\mathbf{=}}{{\mathbit{\mu }}}_{{\mathbf{0}}}{\mathbit{i}}$

The line integral in this equation is evaluated around a closed-loop called an Amperian loop. The current ion the right side is the net current encircled by the loop.

Using Ampere’s law, we can derive the magnetic field inside the wire ${\mathbf{\left(}}{\mathbit{r}}{\mathbf{}}{\mathbf{<}}{\mathbf{}}{\mathbit{a}}{\mathbf{\right)}}$ as,

${\mathbit{B}}{\mathbf{=}}\left(\frac{{\mu }_{0}i}{2\pi {a}^{2}}\right){\mathbit{r}}$ (i)

Here, ${\mathbit{i}}$ is the current, ${\mathbit{a}}$ is area of cross-section of the wire, ${\mathbit{r}}$ is the radius.

For the first two conditions ${\mathbit{r}}{\mathbf{}}{\mathbf{<}}{\mathbf{}}{\mathbit{a}}$ , therefore, using, we can find the magnitude of the current’s magnetic field at the corresponding radii. We can also use the same equation for the third condition ${\mathbit{a}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbit{r}}$. For the fourth condition, ${\mathbit{r}}{\mathbf{}}{\mathbf{>}}{\mathbf{}}{\mathbit{a}}$ , therefore, we can use equation (i) to find the corresponding magnitude of the current’s magnetic field.

The magnetic field outside the wire ${\mathbf{\left(}}{\mathbit{r}}{\mathbf{}}{\mathbf{>}}{\mathbf{}}{\mathbit{a}}{\mathbf{\right)}}$ is given by,

${\mathbit{B}}{\mathbf{=}}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$ (ii)

## Step 3: (a) Calculations of the magnitude of the current’s magnetic field at a radial distance 0 cm

For $r=0$,

Since $a>r$, therefore, the magnetic field is given by,

$B=\left(\frac{{\mu }_{0}i}{2\pi {a}^{2}}\right)r\phantom{\rule{0ex}{0ex}}=\left(\frac{{\mu }_{0}i}{2\pi {a}^{2}}\right)×\left(0\right)\phantom{\rule{0ex}{0ex}}=0$

The magnitude of the current’s magnetic field at a radial distance 0 cm is zero.

## Step 4: (b) Calculations of the magnitude of the current’s magnetic field at a radial distance 1.00 cm

For $r=0.0100m$

Since,$a>r$ , therefore, from equation (i), the magnetic field is given by,

$B=\left(\frac{{\mu }_{0}i}{2\pi {a}^{2}}\right)r\phantom{\rule{0ex}{0ex}}=\left(\frac{\left(4\pi ×{10}^{-7}T.m/A\right)\left(170A\right)}{2\pi {\left(0.0200m\right)}^{2}}\right)\left(0.0100m\right)\phantom{\rule{0ex}{0ex}}=8.50×{10}^{-4}T$

The magnitude of the current’s magnetic field at a radial distance 1.00 cm is $8.50×{10}^{-4}T$.

## Step 5: (c) Calculations of the magnitude of the current’s magnetic field at a radial distance 2.00 cm

For $r=0.0200m$

Since,$a=r$ , therefore, from equation (ii), the magnetic field is given by,

$B=\left(\frac{{\mu }_{0}i}{2\pi {a}^{2}}\right)r\phantom{\rule{0ex}{0ex}}=\left(\frac{\left(4\pi ×{10}^{-7}T.m/A\right)\left(170A\right)}{2\pi {\left(0.0200m\right)}^{2}}\right)\left(0.0200m\right)\phantom{\rule{0ex}{0ex}}=1.70×{10}^{-3}T$

The magnitude of the current’s magnetic field at a radial distance 2.00 cm is $1.70×{10}^{-3}T$.

## Step 6: (d) Calculations of the magnitude of the current’s magnetic field at a radial distance 4.00 cm

For $r=0.0400m$,

Since,$a , therefore, from equation (ii), the magnetic field is given by

$B=\left(\frac{{\mu }_{0}i}{2\pi r}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{\left(4\pi ×{10}^{-7}T.m/A\right)\left(170A\right)}{2\pi \left(0.0400m\right)}\right)\phantom{\rule{0ex}{0ex}}=8.50×{10}^{-4}T$

The magnitude of the current’s magnetic field at a radial distance 4.00 cm is $8.50×{10}^{-4}T$. ### Want to see more solutions like these? 