Question

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

Short answer

The Magnetic field,$\overrightarrow{B}=1.13\times {10}^{-15}\mathrm{T}\hat{k}$

Step-by-step solution

Introduction

The source of a field is the moving charge. The Biot-Savart law is used to calculate the flow at a given degree. Because of radial and route length directions are always at right angles, the vector is doubled. Equation gives the magnetic flux due to the moving charge.

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

Using Pythagorean Theorem

The distance $r$between the proton and therefore the dot point by

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

The angle between the proton direction and therefore the real number is calculated by

$\theta ={\tan }^{-1}\frac{1\mathrm{cm}}{1\mathrm{cm}}=45°$

$\theta ={\tan }^{-1}\frac{1\mathrm{cm}}{1\mathrm{cm}}=45°$

Using the right-hand rule, locate the force field.

the value for $\theta ,q,v$and $r$into equation to urge$\overrightarrow{B}$by

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

$=\left(\frac{4\pi \times {10}^{-7}\mathrm{T}·\mathrm{m}/\mathrm{A}}{4\pi }\right)\left(\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(2\times {10}^7\mathrm{m}/\mathrm{s}\right)\sin 45°}{{\left(\sqrt{2}\times {10}^{-2}\mathrm{m}\right)}^2}\right)$

$=1.13\times {10}^{-15}\mathrm{T}$$\hat{k}$