91影视

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$\stackrel{鈫�}{B_0}=\left(1.5脳{10}^{鈭�6}\mathrm{T}\right)\widehat{i}$ .

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

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

$\begin{array}{r}\stackrel{鈫�}{B}=\stackrel{鈫�}{B_0}鈭�\frac{{渭}_{0}I}{2蟺r}\widehat{i}\\ \stackrel{鈫�}{B}=\left(1.5脳{10}^{鈭�6}\mathrm{T}\right)\widehat{i}鈭�\frac{4蟺脳{10}^{鈭�7}脳8}{2蟺脳1}\\ \stackrel{鈫�}{B}=鈭�\left({10}^{鈭�7}\mathrm{T}\right)\widehat{i}\end{array}$

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

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

$\begin{array}{r}\stackrel{鈫�}{B}=\stackrel{鈫�}{B_0}+\frac{{渭}_{0}I}{2蟺r}\widehat{k}\\ \stackrel{鈫�}{B}=\left(1.5脳{10}^{鈭�6}T\right)\widehat{i}+\frac{4蟺脳{10}^{鈭�7}脳8}{2蟺脳1}\widehat{k}\\ \stackrel{鈫�}{B}=\left(1.5脳{10}^{鈭�6}\mathrm{T}\right)\widehat{i}+\left(1.6脳{10}^{鈭�6}\mathrm{T}\right)\widehat{k}\end{array}$

The magnitude of the magnetic field is

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

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

$\begin{array}{r}胃={\tan }^{鈭�1}鈦�\left(\frac{B_z}{B_x}\right)\\ 胃={\tan }^{鈭�1}鈦�\left(\frac{1.6}{1.5}\right)\\ 胃={46.8}^{鈭�}\end{array}$

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

$\begin{array}{r}\stackrel{鈫�}{B}=\stackrel{鈫�}{B_0}+\frac{{渭}_{0}I}{2蟺r}\widehat{i}\\ \stackrel{鈫�}{B}=\left(1.5脳{10}^{鈭�6}\right)\widehat{i}+\frac{4蟺脳{10}^{鈭�7}脳8}{2蟺脳1}\widehat{i}\\ \stackrel{鈫�}{B}=\left(7.9脳{10}^{鈭�6}T\right)\widehat{i}\end{array}$

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