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Q43P

Expert-verifiedFound in: Page 860

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 ${\mathit{a}}{\mathbf{=}}{\mathbf{}}{\mathbf{2}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathit{c}}{\mathit{m}}$**** carrying uniform current ${\mathbf{170}}{\mathbf{}}{\mathit{A}}$****. 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 $8.50\times {10}^{-4}T$

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

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

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

**Ampere’s law states that,**

** **

${\mathbf{\oint}}\overrightarrow{\mathbf{B}}{\mathbf{\xb7}}{\mathit{d}}\overrightarrow{\mathbf{s}}{\mathbf{=}}{{\mathit{\mu}}}_{{\mathbf{0}}}{\mathit{i}}$

** **

**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 magn****etic field ****inside the wire ${\mathbf{(}}{\mathit{r}}{\mathbf{}}{\mathbf{<}}{\mathbf{}}{\mathit{a}}{\mathbf{)}}$**** as,**

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

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

** **

For the first two conditions ${\mathit{r}}{\mathbf{}}{\mathbf{<}}{\mathbf{}}{\mathit{a}}$ , 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 ${\mathit{a}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathit{r}}$. For the fourth condition, ${\mathit{r}}{\mathbf{}}{\mathbf{>}}{\mathbf{}}{\mathit{a}}$ , therefore, we can use equation (i) to find the corresponding magnitude of the current’s magnetic field.

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

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

For $r=0$,

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

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

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

For $r=0.0100m$

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

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

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

For $r=0.0200m$

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

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

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

For $r=0.0400m$,

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

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

The magnitude of the current’s magnetic field at a radial distance 4.00 cm is $8.50\times {10}^{-4}T$.

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