An ion with charge of Q = + 3.2 x 10-19 C is in a region where a uniform electric field of magnitude E = 5.0 x 105 V/m is perpendicular to a uniform magnetic field B = 0.80 T. If the ion's acceleration is zero then what is its speed v? a) v = 1.6 x 10-5 m/s b) v = 4.0 x 105 m/s c) v = 6.3 x 10 m/s d) V = 0 m/s

Question

Physics

Allison Washington

6 November, 08:01

An ion with charge of Q = + 3.2 x 10-19 C is in a region where a uniform electric field of magnitude E = 5.0 x 105 V/m is perpendicular to a uniform magnetic field B = 0.80 T. If the ion's acceleration is zero then what is its speed v? a) v = 1.6 x 10-5 m/s b) v = 4.0 x 105 m/s c) v = 6.3 x 10 m/s d) V = 0 m/s

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Answers (1)

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Thumper · Answer 1

speed is 6.3 * 10^5 m/s

Explanation:

given data

charge Q = + 3.2 x 10^-19 C

uniform electric field E = 5.0 x 10^5 V/m

uniform magnetic field B = 0.80 T

acceleration = 0

to find out

speed

solution

we know here that acceleration = 0

so that force on uniform electric field E and uniform magnetic field B on charge particle = 0

so here force of magnitude will be equal

so we can say that

q*E = q*v*B ... 1

so v = E / B

put here value

v = 5 * 10^5 / 0.80

v = 6.3 * 10^5

so speed is 6.3 * 10^5 m/s