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

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

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

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