clovnerie0q

2022-10-02

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?

Salma Baird

Beginner2022-10-03Added 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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