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salantis [7]
3 years ago
12

An electron (e = 1.6 x 10-19 C) is traveling at 4.00 x 107 m/s due North in a horizontal plane through a point where the earth’s

magnetic field has a component to the North of 2.00 x 10-5 T and a downward (in to the earth) of 5.00 x 10-5 T. Calculate the magnetic force (magnitude and direction) on the electron and its acceleration. Me = 9.11 x 10-31 kg
Physics
1 answer:
Irina18 [472]3 years ago
4 0

Answer:

Explanation:

Force on the electron due to magnetic field 's north component will be zero because both velocity of electron and direction of magnetic field are same .

Force due to downward component of magnetic field

= B q v where B is magnetic field , q is charge moving and v is velocity of charge

F = 5 .00 x 10⁻⁵ x 1.6 x 10⁻¹⁹ x 4 x 10⁷

= 32 x 10⁻¹⁷ N

acceleration = F / m where m is mass of electron

= 32 x 10⁻¹⁷ / 9.11 x 10⁻³¹

= 3.5 x 10¹⁴ m/s²

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This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

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