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Katen [24]
3 years ago
6

An electric field of 1.32 kV/m and a magnetic field of 0.516 T act on a moving electron to produce no net force. If the fields a

re perpendicular to each other, what is the electron's speed?
Physics
1 answer:
Lapatulllka [165]3 years ago
3 0

Answer:

The speed of the electron is 2.55\times 10^3\ m/s.

Explanation:

Given that,

The magnitude of electric field, E=1.32\ kV/m=1.32\times 10^3\ V/m

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then,  the magnitude of electric field and magnetic field is balanced such that :

evB=eE\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.32\times 10^3}{0.516}\\\\v=2558.13\ m/s

or

v=2.55\times 10^3\ m/s

So, the speed of the electron is 2.55\times 10^3\ m/s. Hence, this is the required solution.

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The magnitude of the induced emf is given by:

ℰ = |Δφ/Δt|

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