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3 years ago

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A 6.41 $\mu C$ particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive $x$

direction, and a magnetic field of magnitude 1.28 T points in the positive $z$ direction. If the net force acting on the particle is 6.40E-3 N in the positive $x$ direction, calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the $x$-$y$ plane.

1 answer:

Marat540 [252]3 years ago

5 0

Answer:

The particle's velocity is 212.15 m/s.

Explanation:

Given that,

Charge of particle, $q=6.41\ \mu C=6.41\times 10^{-6}\ C$

The magnitude of electric field, E = 1270 N/C

The magnitude of magnetic field, B = 1.28 T

Net force, $F=6.4\times 10^{-3}\ N$

We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :

$F=qE+qvB\\\\v=\dfrac{F-qE}{qB}\\\\v=\dfrac{6.4\times 10^{-3}-6.41\times 10^{-6}\times 1270}{6.41\times 10^{-6}\times 1.28}\\\\v=-212.15\ m/s$

So, the particle's velocity is 212.15 m/s.

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Answer:

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