find Magnitude and direction A particle carrying an electrical charge q and movi

Question

find Magnitude and direction

A particle carrying an electrical charge q and moving with a velocity u through a magnetic field B would experience a force f given by f=qu x B as shown in the figure below. Here f ,u and B are vector quantities, and x is the symbol for vector product. If B = 0.5 T, u = 0.1 m/s, theta = 30degree and the charged particle is a single electron (q = -1.60times10-19 C), what is the magnitude and direction of the force that acts on the particle? Select the correct answer among the following:

Explanation / Answer

we have,

F = q(v x B)

= 1.6 x 10-19 ( 0.1 sin 30 i    x    0.5 j)

= 8.0 x 10-21 normal to the page, away from it....

:)

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