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A particle carrying 45.3 uC of charge and enters a field of strength 1.99 T while...

A particle carrying 45.3 uC of charge and enters a field of strength 1.99 T while traveling at speed 150 km/s. The angle between velocity and field is 60 Degrees.

What is the Magnitude Force on the Charge?

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Answer #1

\vec F = q (\vec v \times \vec B) \\ \\ F = qv B sin(\theta)

here theta 60 degrees

so magnitude of force = 11.71044 N

please rate it up thanks :)

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