A proton moves along the $x\text{-axis}$with $v_x=1.0\times {10}^7\mathrm{m}/\mathrm{s}$. As it passes the origin, what are the strength and direction of the magnetic field at the $(x,y,z)$positions (a) (1 cm, 0 cm, 0 cm) (b) (0 cm, 1 cm, 0 cm) (c) (0 cm, -2 cm, 0 cm)

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.

$\begin{array}{r}\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{qv\sin \theta }{r^2}\\ =\frac{{\mu }_o}{4\pi }\frac{qv\sin 0^{\circ }}{r^2}\end{array}$

$=0\mathrm{T}$

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$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{qv\sin \theta }{r^2}$

$=\left(\frac{4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}}{4\pi }\right)\left(\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(1\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin {90}^{\circ }}{(0.02\mathrm{m}{)}^2}\right)$

$=1.6\times {10}^{-15}\mathrm{T}$

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$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{qv\sin \theta }{r^2}$

$=\left(\frac{4\pi \times {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}}{4\pi }\right)\left(\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(1\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin \left(-{90}^{\circ }\right)}{(0.02\mathrm{m}{)}^2}\right)$

$=-4\times {10}^{-16}\mathrm{T}$