Question

What is the magnetic field at the position of the dot in FIGURE EX29.5? Give your answer as a vector.

Short answer

The magnetic field at a point $\overrightarrow{B}=2.8\times {10}^{-16}\widehat{k}$.

Step-by-step solution

Blot- Savart Law

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.

${\overrightarrow{B}}_{\text{point charge}}=\frac{{\mu }_o}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^3}$

Proton and Dot point

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

The distance is provided by the magnitude which is,

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

Proton Travels

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

$\overrightarrow{B}=\frac{{\mu }_o}{4\pi }\frac{q\overrightarrow{v}\times \widehat{r}}{r^2}$

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

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

$\stackrel{\rightharpoonup }{B}=2.8\times {10}^{-16}\widehat{k}$