91Ó°ÊÓ

It produces a magnetic field by one charged particle.

The Biot-Savart law is used to calculate the force of attraction at a specific position.

Since this radial and route lengths components also are perpendicular, then transfer function becomes multiplication.

It can be used to express the magnetism related with a constant force.

${\stackrel{→}{B}}_{\text{point charge}}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{q\stackrel{→}{v}×\widehat{r}}{r^3}$

The idea is at, although the protons is at. As a response, the proton's length from the Pinpoint point is $\stackrel{→}{r}=(−1\widehat{i}\mathrm{cm},−1\widehat{j}\mathrm{cm})−(1\widehat{i}\mathrm{cm},1\widehat{j}\mathrm{cm})=(−2\widehat{i}\mathrm{cm},−2\widehat{j}\mathrm{cm})$

The distance is provided by the magnitude which is,

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

Since the proton comes in the progressive direction. Then we'll enter up the amount for

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\frac{q\stackrel{→}{v}×\widehat{r}}{r^2}$

$=\left(\frac{4\mathrm{π}×{10}^{−7}\mathrm{T}×\mathrm{m}/\mathrm{A}}{4\mathrm{π}}\right)\left(1.6×{10}^{−19}\mathrm{C}\right)\frac{\left(2×{10}^7\mathrm{m}/\mathrm{s}\widehat{j}\right)(−0.02\widehat{i}\mathrm{m},−0.02\widehat{j}\mathrm{m})}{{\left(2\sqrt{2}×{10}^{−2}\mathrm{m}\right)}^3}$

$=\left(\frac{4\mathrm{π}×{10}^{−7}\mathrm{T}×\mathrm{m}/\mathrm{A}}{4\mathrm{π}}\right)\left(1.6×{10}^{−19}\mathrm{C}\right)\frac{\left(0.04×{10}^7\widehat{k}\right)}{{\left(2\sqrt{2}×{10}^{−2}\mathrm{m}\right)}^3}$

$\stackrel{â‡\P Ä}{B}=2.8×{10}^{−16}\widehat{k}$