91影视

It produces a magnetic field by only a charged particle. Use the Blot Savart law to calculate estimated magnetism at a specific location. Also because lengths and path distance directions are already perpendicular, the wave form is multiplication. T he field strength attributed to one flowing fluid is created.

That slope here between place's orientation and indeed the companion's velocity is zero.

$\begin{array}{r}\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\frac{qv\sin 鈦�胃}{r^2}\\ =\frac{{渭}_o}{4蟺}\frac{qv\sin 鈦�0^{鈭�}}{r^2}\end{array}$

$=0\mathrm{T}$

That viewpoint made mostly by mount's pace and even the path of the place. is owing to the reasons that the placement is diagonal to$\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\frac{qv\sin 鈦�胃}{r^2}$

$=\left(\frac{4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}}{4蟺}\right)\left(\frac{\left(1.6脳{10}^{鈭�19}\mathrm{C}\right)\left(1脳{10}^7\mathrm{m}/\mathrm{s}\right)\sin 鈦�{90}^{鈭�}}{(0.02\mathrm{m}{)}^2}\right)$

$=1.6脳{10}^{鈭�15}\mathrm{T}$

Even as area is diagonal to the movement of the energy, the distance between it line of such place and or the pace of the control is$\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\frac{qv\sin 鈦�胃}{r^2}$

$=\left(\frac{4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}}{4蟺}\right)\left(\frac{\left(1.6脳{10}^{鈭�19}\mathrm{C}\right)\left(1脳{10}^7\mathrm{m}/\mathrm{s}\right)\sin 鈦�\left(鈭�{90}^{鈭�}\right)}{(0.02\mathrm{m}{)}^2}\right)$

$=鈭�4脳{10}^{鈭�16}\mathrm{T}$