A proton moving in a uniform magnetic field with ${\overrightarrow{v}}_1=$$1.00\times {10}^6\hat{ı}\mathrm{m}/\mathrm{s}$experiences force ${\overrightarrow{F}}_1=1.20\times {10}^{-16}\hat{k}\mathrm{N}$. A second proton with ${\overrightarrow{v}}_2=2.00\times {10}^6\hat{j}\mathrm{m}/\mathrm{s}$experiences ${\overrightarrow{F}}_2=$$-4.16\times {10}^{-16}\hat{k}N$in the same field. What is $\overrightarrow{B}$? Give your answer as a magnitude and an angle measured $cCW$ from the$+x$- axis.

Depending on whether the currents run within the same or opposing directions, two parallel wires carrying electric currents repel or attract one another, in line with Ampère.

An equation describes the magnetic force:

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$

Where $q$is that the charge. Rearrange equation for $\overrightarrow{B}$to be within the form

$\overrightarrow{B}=\frac{\overrightarrow{F}}{q\overrightarrow{v}}$

The force field $B1$for the primary proton is obtained by

${\overrightarrow{B}}_1=\frac{{\overrightarrow{F}}_1}{q{\overrightarrow{v}}_1}$

$=\frac{1.2\times {10}^{-16}\hat{k}\mathrm{N}}{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(1\times {10}^6\hat{i}\mathrm{m}/\mathrm{s}\right)}$

$=7.5\times {10}^{-4}\hat{j}$

The force field $B2$for the second proton is

${\overrightarrow{B}}_2=\frac{{\overrightarrow{F}}_2}{q{\overrightarrow{v}}_2}$

$=\frac{-4.16\times {10}^{-16}\widehat{kN}}{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(2\times {10}^6\hat{j}\mathrm{m}/\mathrm{s}\right)}$

$=-13\times {10}^{-4}\hat{i}$

The magnitude of the magnetic flux is obtained by

$B=\sqrt{{\overrightarrow{B}}_1^2+{\overrightarrow{B}}_2^2}$

$=\sqrt{{\left(7.5\times {10}^{-4}\hat{j}\right)}^2+{\left(-13\times {10}^{-4}\hat{i}\right)}^2}$

$=1.5\times {10}^{-3}\mathrm{T}$

$=1.5\mathrm{mT}$

The angle formed by the $2$elements is

$\theta ={\tan }^{-1}\left(\frac{7.5\times {10}^{-4}}{13\times {10}^{-4}}\right)=30°$