Anelectron( q = 1.60×1019 C)istraveling through the Earth\'s magnetic field at a

Question

Anelectron(q = 1.60×1019 C)istraveling through the Earth's magnetic field at a location ontheEarth's surface where the field has a strength ofexactly5.00×105 T, directed due north. At aparticularinstant, a magnetic force of6.44×1021 N,directed straight down, acts on theelectron. If the electron istraveling in a cardinal compassdirection, what are
(a) the magnitude
(b) the direction of the electron's velocity at that instant?
a) m/s b) Answer in North, South, East or West
(a) the magnitude
(b) the direction of the electron's velocity at that instant?
a) m/s b) Answer in North, South, East or West
a) m/s b) Answer in North, South, East or West

Explanation / Answer

we know that                F = B q v ==>      v = F / B q                = 6.44 x 10-24 / 5.00 x 10-5 * (-1.6 x10-19 )                = - 0.805 m/s towards east.

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