Question

A long, straight wire lies along the y-axis and carries a current I = 8 A in the -y-direction. In addition to the magnetic field due to the current in the wire, a uniform magnetic field with magnitude is in the +x-direction. What is the total field (magnitude and direction) at the following points in the xz-plane: (a)$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{z}\mathbf{=}\mathbf{1}\mathit{m}$ ; (b)$\mathit{x}\mathbf{=}\mathbf{1}\mathit{m}\mathbf{,}\mathit{z}\mathbf{=}\mathbf{0}$ ; (c)$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{z}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{25}\mathit{m}$ ?

Short answer

a)$\overrightarrow{B}=-{10}^{-7}T\widehat{i}$ , along negative x-axis

b) $B=2.19\times {10}^{-6}\mathrm{T},\theta ={46.8}^0$, from x to z

c) $\overrightarrow{B}=\left(7.9\times {10}^{-6}\mathrm{T}\right)\widehat{i}$, along positive x-axis

Step-by-step solution

The scenario in the problem

Consider a long, straight wire which lies on the y-axis and carries a current in the -y direction. In addition, there is a magnetic field of$\overrightarrow{B_0}=\left(1.5\times {10}^{-6}\mathrm{T}\right)\widehat{i}$ .

Find the resultant magnetic field at the first point

The magnetic field due to a long wire is given by$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{\pi r}}$ .

This field points towards -x axis at$x=0,y=0,z=1m$ . So, total magnetic field is

$\begin{array}{r}\overrightarrow{B}=\overrightarrow{B_0}-\frac{{\mu }_{0}I}{2\pi r}\widehat{i}\\ \overrightarrow{B}=\left(1.5\times {10}^{-6}\mathrm{T}\right)\widehat{i}-\frac{4\pi \times {10}^{-7}\times 8}{2\pi \times 1}\\ \overrightarrow{B}=-\left({10}^{-7}\mathrm{T}\right)\widehat{i}\end{array}$

Since, the total magnetic field is negative, then it is along the negative x-axis

Find the resultant magnitude of magnetic field at the second point

Now, find the magnetic field at$x=1m,y=0,z=0$ , the magnetic field at this point is towards +z axis.

$\begin{array}{r}\overrightarrow{B}=\overrightarrow{B_0}+\frac{{\mu }_{0}I}{2\pi r}\widehat{k}\\ \overrightarrow{B}=\left(1.5\times {10}^{-6}T\right)\widehat{i}+\frac{4\pi \times {10}^{-7}\times 8}{2\pi \times 1}\widehat{k}\\ \overrightarrow{B}=\left(1.5\times {10}^{-6}\mathrm{T}\right)\widehat{i}+\left(1.6\times {10}^{-6}\mathrm{T}\right)\widehat{k}\end{array}$

The magnitude of the magnetic field is

$\begin{array}{r}B=\sqrt{B_x^2+B_y^2+B_z^2}\\ B=\sqrt{{\left(1.5\times {10}^{-6}\right)}^2+{\left(1.6\times {10}^{-6}\right)}^2}\\ B=2.19\times {10}^{-6}\mathrm{T}\end{array}$

The field makes an angle with the positive x-axis given by

$\begin{array}{r}\theta ={\tan }^{-1}\left(\frac{B_z}{B_x}\right)\\ \theta ={\tan }^{-1}\left(\frac{1.6}{1.5}\right)\\ \theta ={46.8}^{\circ }\end{array}$

Find the resultant magnitude of magnetic field at the third point

We need to find the magnetic field at$x=0,y=0,z=1m$ , the magnetic field at these points towards +x-axis.

$\begin{array}{r}\overrightarrow{B}=\overrightarrow{B_0}+\frac{{\mu }_{0}I}{2\pi r}\widehat{i}\\ \overrightarrow{B}=\left(1.5\times {10}^{-6}\right)\widehat{i}+\frac{4\pi \times {10}^{-7}\times 8}{2\pi \times 1}\widehat{i}\\ \overrightarrow{B}=\left(7.9\times {10}^{-6}T\right)\widehat{i}\end{array}$

Since, the total magnetic field is positive, then it is along the positive x-axis.