91影视

A moving charge produces the magnetism. To compute the magnetism at a specific location, use the Blot Savart law. The transfer function becomes multiplied because of radial and path length dimensions are always perpendicular. The magnetic force due to a given direction in the form is formed.

${\stackrel{鈫�}{B}}_{\text{point charge}}=\frac{{渭}_o}{4蟺}\frac{qv\sin 鈦�胃}{r^2}$

This same angles between of area's axis and also the charge's rate is zero

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

$=\left(\frac{4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}}{4蟺}\right)\left(\frac{\left(鈭�1.6脳{10}^{鈭�19}\mathrm{C}\right)\left(2脳{10}^7\mathrm{m}/\mathrm{s}\right)\sin 鈦�{90}^{鈭�}}{(0.01\mathrm{m}{)}^2}\right)$

$=鈭�3.2脳{10}^{鈭�15}\mathrm{T}$

The torque generated either by destination's direction and the mount's velocity is $胃=0$

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

$=\frac{{渭}_o}{4蟺}\frac{qv\sin 鈦�0^{鈭�}}{r^2}$

$=0\mathrm{T}$

The torque generated by the destination's axis or the mount's velocity is$胃={45}^{鈭�}$. The range again from original really does have a significance of

$r=\sqrt{(1\mathrm{cm}{)}^2+(1\mathrm{cm}{)}^2}=1.41\mathrm{cm}$

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

$=\left(\frac{4蟺脳{10}^{鈭�7}\mathrm{T}鈰�\mathrm{m}/\mathrm{A}}{4蟺}\right)\left(\frac{\left(鈭�1.6脳{10}^{鈭�19}\mathrm{C}\right)\left(2脳{10}^7\mathrm{m}/\mathrm{s}\right)\sin 鈦�\left({45}^{鈭�}\right)}{(0.0141\mathrm{m}{)}^2}\right)$

$=鈭�1.14脳{10}^{鈭�15}\mathrm{T}$