91Ó°ÊÓ

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

Ampere’s law states that,

$\mathbf{âˆ\circledR }\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{·}\mathit{d}\stackrel{\mathbf{→}}{\mathbf{s}}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}\mathit{i}$

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 magnetic fieldinside the wire$\mathbf{(}\mathit{r}\mathbf{}\mathbf{<}\mathbf{}\mathit{a}\mathbf{)}$as,

$\mathit{B}\mathbf{=}\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}{a}^2}\right)\mathit{r}$ (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,

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

For$r=0$,

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

$$\begin{gathered} B=\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}{a}^2}\right)r \\ =\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}{a}^2}\right)×\left(0\right) \\ =0 \end{gathered}$$

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

For$r=0.0100m$

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

$$\begin{gathered} B=\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}{a}^2}\right)r \\ =\left(\frac{\left(4\mathrm{π}×{10}^{-7}T.m/A\right)\left(170A\right)}{2\mathrm{π}{\left(0.0200m\right)}^2}\right)\left(0.0100m\right) \\ =8.50×{10}^{-4}T \end{gathered}$$

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

For$r=0.0200m$

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

$$\begin{gathered} B=\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}{a}^2}\right)r \\ =\left(\frac{\left(4\mathrm{π}×{10}^{-7}T.m/A\right)\left(170A\right)}{2\mathrm{π}{\left(0.0200m\right)}^2}\right)\left(0.0200m\right) \\ =1.70×{10}^{-3}T \end{gathered}$$

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

For$r=0.0400m$,

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

$$\begin{gathered} B=\left(\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}\right) \\ =\left(\frac{\left(4\mathrm{π}×{10}^{-7}T.m/A\right)\left(170A\right)}{2\mathrm{π}\left(0.0400m\right)}\right) \\ =8.50×{10}^{-4}T \end{gathered}$$

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