clovnerie0q
2022-10-02
Answered

A particle with a charge of $5\mu C$ is moving at $3\cdot {10}^{6}\text{}m/s$ perpendicularly through a magnetic field with a strength of 0.06T. What is the magnitude of the force on the particle?

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Salma Baird

Answered 2022-10-03
Author has **8** answers

Charge of the particle (q) = $5\times {10}^{-6}\text{}C$

Velocity of the particle (v) = $3\times {10}^{6}\text{}m/s$

Magnetic field ( B ) = 0.06 T

Magnitude of force

$F=q(v\times B)\phantom{\rule{0ex}{0ex}}F=qvB\mathrm{sin}{90}^{0}\phantom{\rule{0ex}{0ex}}F=5\times {10}^{-6}\times 3\times {10}^{6}\times 0.06\phantom{\rule{0ex}{0ex}}F=0.9\text{}Newton$

So,magnitude of force = 0.9 Newton

Velocity of the particle (v) = $3\times {10}^{6}\text{}m/s$

Magnetic field ( B ) = 0.06 T

Magnitude of force

$F=q(v\times B)\phantom{\rule{0ex}{0ex}}F=qvB\mathrm{sin}{90}^{0}\phantom{\rule{0ex}{0ex}}F=5\times {10}^{-6}\times 3\times {10}^{6}\times 0.06\phantom{\rule{0ex}{0ex}}F=0.9\text{}Newton$

So,magnitude of force = 0.9 Newton

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I want to derive Gauss's law for magnetism,

$\mathrm{\nabla}\cdot \overrightarrow{B}=0\phantom{\rule{thinmathspace}{0ex}}.$

The derivation in Griffiths Introduction to elecrodynamics uses

$\mathrm{\nabla}\cdot \overrightarrow{B}\text{}=\text{}\frac{{\mu}_{0}}{4\pi}\int \mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})\phantom{\rule{thinmathspace}{0ex}},$

but to use this, I would need $\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}}$ and $\mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})$ to be continuous.

Since $\overrightarrow{J}=\rho \overrightarrow{v}$ , then if $\rho $ (charge density) has a continuous derivative, I'm done; but, I'm not sure why can I affirm this because clearly, I can construct charge distributions with jump discontinuities.

$\mathrm{\nabla}\cdot \overrightarrow{B}=0\phantom{\rule{thinmathspace}{0ex}}.$

The derivation in Griffiths Introduction to elecrodynamics uses

$\mathrm{\nabla}\cdot \overrightarrow{B}\text{}=\text{}\frac{{\mu}_{0}}{4\pi}\int \mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})\phantom{\rule{thinmathspace}{0ex}},$

but to use this, I would need $\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}}$ and $\mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})$ to be continuous.

Since $\overrightarrow{J}=\rho \overrightarrow{v}$ , then if $\rho $ (charge density) has a continuous derivative, I'm done; but, I'm not sure why can I affirm this because clearly, I can construct charge distributions with jump discontinuities.