Taliyah Spencer

2022-05-03

Calculate magnetic field from magnetic force on a charged particle

The magnetic force, charge, and velocity of a particle are given. I am asked to find the magnetic field.

$q=-2\phantom{\rule{thinmathspace}{0ex}}\mu \mathrm{C}$

$\overrightarrow{v}=(-\hat{i}+3\hat{j})\times {10}^{6}\mathrm{m}/\mathrm{s}$

$\overrightarrow{F}=(3\hat{i}+\hat{j}+2\hat{k})\mathrm{N}$

${B}_{x}=0$

The magnetic force, charge, and velocity of a particle are given. I am asked to find the magnetic field.

$q=-2\phantom{\rule{thinmathspace}{0ex}}\mu \mathrm{C}$

$\overrightarrow{v}=(-\hat{i}+3\hat{j})\times {10}^{6}\mathrm{m}/\mathrm{s}$

$\overrightarrow{F}=(3\hat{i}+\hat{j}+2\hat{k})\mathrm{N}$

${B}_{x}=0$

Alexzander Guerrero

Beginner2022-05-04Added 7 answers

Use the equation of Lorentz force to calculate the field vector.

In your case, there is no electric field. So calculate accordingly.

Given: ${B}_{x}=0$

Take ${B}_{y}$ and ${B}_{z}$ to be the other components. Calculate the cross product and then compare the coefficients of the unit vectors, to get the answer.

Hope it helps.

In your case, there is no electric field. So calculate accordingly.

Given: ${B}_{x}=0$

Take ${B}_{y}$ and ${B}_{z}$ to be the other components. Calculate the cross product and then compare the coefficients of the unit vectors, to get the answer.

Hope it helps.

timbreoizy

Beginner2022-05-05Added 15 answers

Let $\overrightarrow{B}={B}_{y}\hat{j}+{B}_{z}\hat{k}$ (since, ${B}_{x}=0$) be the magnetic field then the magnetic force (${F}_{m}$) acting on the charge, is given as

${F}_{m}=q(\overrightarrow{v}\times \overrightarrow{B})$

${F}_{m}=q(\overrightarrow{v}\times \overrightarrow{B})$

setting the corresponding values of $\overrightarrow{F}$, $q$, $\overrightarrow{v}$ & $\overrightarrow{B}$,

$3\hat{i}+\hat{j}+2\hat{k}=2\times {10}^{-6}\times {10}^{6}((-\hat{i}+3\hat{j}))({B}_{y}\hat{j}+{B}_{z}\hat{k})$

$3\hat{i}+\hat{j}+2\hat{k}=2(3{B}_{z}\hat{i}+{B}_{z}\hat{k}-{B}_{y}\hat{k})$

$3\hat{i}+\hat{j}+2\hat{k}=6{B}_{z}\hat{i}+2{B}_{z}\hat{k}-2{B}_{y}\hat{k}$

Now, compare the corresponding coefficients on both the sides, to find ${B}_{y}$ & ${B}_{z}$ & then the magnetic field $\overrightarrow{B}$

${F}_{m}=q(\overrightarrow{v}\times \overrightarrow{B})$

${F}_{m}=q(\overrightarrow{v}\times \overrightarrow{B})$

setting the corresponding values of $\overrightarrow{F}$, $q$, $\overrightarrow{v}$ & $\overrightarrow{B}$,

$3\hat{i}+\hat{j}+2\hat{k}=2\times {10}^{-6}\times {10}^{6}((-\hat{i}+3\hat{j}))({B}_{y}\hat{j}+{B}_{z}\hat{k})$

$3\hat{i}+\hat{j}+2\hat{k}=2(3{B}_{z}\hat{i}+{B}_{z}\hat{k}-{B}_{y}\hat{k})$

$3\hat{i}+\hat{j}+2\hat{k}=6{B}_{z}\hat{i}+2{B}_{z}\hat{k}-2{B}_{y}\hat{k}$

Now, compare the corresponding coefficients on both the sides, to find ${B}_{y}$ & ${B}_{z}$ & then the magnetic field $\overrightarrow{B}$

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