In Figure, point P is at perpendicular distance $\mathit{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{cm}}$from a very long straight wire carrying a current. The magnetic field $\overrightarrow{\mathbf{B}}$set up at point Pis due to contributions from all the identical current length elements $\mathit{i}\mathbf{}\mathit{d}\overrightarrow{\mathbf{s}}$along the wire. What is the distanceto the element making (a) The greatest contribution to field $\mathbf{}\overrightarrow{\mathbf{B}}$and (b) 10.0% of the greatest contribution?

1. Distance of point P from the wire, $R=2.00\text{cm}$.
2. Figure of the long straight wire.

Consider the magnetic field due to a straight current carrying wire. Write the equation for magnetic field due to a small element of the wire. Find the distance between the small element and point P. Then, determine the maximum value.

$\mathbf{dB}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{}\frac{\mathbf{i}\left(\overrightarrow{\mathbf{ds}}\mathbf{\times }\hat{\mathbf{r}}\right)}{{\mathbf{r}}^{\mathbf{2}}}$

Distance s to the element making the greatest contribution to field $\overrightarrow{B}$:

We can write vector $\overrightarrow{r}$pointing towards P from the current element.

$\overrightarrow{r}=-s\hat{i}+R\hat{j}$

Small element we can write

$\overrightarrow{ds}=ds\hat{i}$

We can find the cross product of $\overrightarrow{r}$and $\overrightarrow{ds}$we get

$\left|\overrightarrow{ds}\times \overrightarrow{r}\right|=ds\hat{i}\times \left(-s\hat{i}+R\hat{j}\right)=Rds\hat{k}$

Also, we can find the magnitude of $\overrightarrow{r}$

$r=\sqrt{s^2+R^2}$

Using the equation,

$dB=\frac{{\mu }_0}{4\pi }\frac{i\left|\overrightarrow{ds}\times \hat{r}\right|}{r^2}$

We know, $\hat{r}=\frac{\overrightarrow{r}}{r}$we can write,

$dB=\frac{{\mu }_0}{4\pi }\frac{i\left|\overrightarrow{ds}\times \overrightarrow{r}\right|}{r^2\left(r\right)}$

We can plug the values of modulus and r,

$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{i\left(Rds\right)\hat{k}}{\left(s^2+R^2\right){\left(s^2+R^2\right)}^{\frac{1}{2}}}$

Taking magnitude,

$dB=\frac{{\mu }_0}{4\pi }\frac{i\left(Rds\right)}{{\left(s^2+R^2\right)}^{\frac{3}{2}}}$

For maximum value of magnetic field, the distance s should be zero as at this value, the denominator will become minimum resulting in the large value for dB. So for$s=0$. we get

$d{B}_{max}=\frac{{\mu }_0}{4\pi }\frac{ids}{R^2}$

This will give the maximum value of the magnetic field.

Distance s to the element making$10.0\%$ofgreatest contribution to field $\overrightarrow{B}$:

Write the equation as:

$dB=d{B}_{max}10\%$

Using above equations and solve as:

$$\begin{gathered} \frac{{\mu }_0}{4\pi }\frac{i\left(Rds\right)}{{\left(s^2+R^2\right)}^{\frac{3}{2}}}=\frac{{\mu }_0}{4\pi }\frac{ids}{R^2}\left(10\%\right) \\ \frac{{\mu }_0}{4\pi }\frac{i\left(Rds\right)}{{\left(s^2+R^2\right)}^{\frac{3}{2}}}=\frac{{\mu }_0}{4\pi }\frac{ids}{R^2}\left(0.10\right) \end{gathered}$$

Rewrote the equation as:

$$\begin{gathered} \frac{R}{{\left(s^2+R^2\right)}^{\frac{3}{2}}}=\frac{1}{R^2}\left(0.10\right) \\ {R}^3=0.10{\left(s^2+R^2\right)}^{\frac{3}{2}} \end{gathered}$$

Substitute the value and solve as:

$$\begin{gathered} 2.{00}^3=0.10{\left(s^2+{2.00}^2\right)}^{\frac{3}{2}} \\ \frac{8}{0.10}={\left(s^2+{2.00}^2\right)}^{\frac{3}{2}} \\ 80={\left(s^2+{2.00}^2\right)}^{\frac{3}{2}} \end{gathered}$$

Squaring on both sides and solve as:

$6400={\left(s^2+{2.00}^2\right)}^3$

Taking cube root we get

$$\begin{gathered} 18.566=s^2+4 \\ {s}^2=18.566-4 \\ {s}^2=14.566 \\ s=3.82\text{cm} \end{gathered}$$

Hence the distance (s) is, $3.82\text{cm}$