Question

Figure a shows an element of length $\mathbf{ds}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu m}$in a very long straight wire carrying current. The current in that element sets up a differential magnetic field at points in the surrounding space. Figure b gives the magnitude$\mathbf{d}\overrightarrow{\mathbf{B}}$of the field for points$\mathbf{2}\mathbf{.}\mathbf{5}\mathit{c}\mathit{m}$from the element, as a function of angle u between the wire and a straight line to the point. The vertical scale is set by${\mathbf{dB}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{pT}$. What is the magnitude of the magnetic field set up by the entire wire at perpendicular distance$\mathbf{2}\mathbf{.}\mathbf{5}\mathit{c}\mathit{m}$from the wire?

Short answer

The value of the magnetic field is$3.0\mu T$.

Step-by-step solution

Understanding the concept

Use the Biot- savart law to calculate the current through the wire. Then, using this current value and the equation for magnetic field due to an infinitely long wire carrying current, find the magnetic field.

$\mathbf{dB}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{}\frac{\mathbf{ids}\mathbf{}\mathbf{sin}\left(\mathbf{\theta }\right)}{{\mathbf{r}}^{\mathbf{2}}}$

Calculating the magnitude of the magnetic field set

Let us find the current through the wire first

For$\theta ={90}^0$

$dB=\frac{{\mu }_0}{4\pi }\frac{ids}{r^2}$

Hence,$i=0.375A$

Now, to find magnetic field $\left(B\right)$:

Magnetic field due to infinite wire carrying current$\left(i\right)$:$B=\frac{{\mu }_{0}i}{2\pi R}$

Substitute the values and solve as:

$$\begin{gathered} B=\frac{(4\mathrm{\pi }\times {10}^{-7})\times 0.375}{2\mathrm{\pi }\times 0.025} \\ B=3\times {10}^6 \\ B=3.0\mu T \end{gathered}$$