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torisob [31]
3 years ago
5

An electric field of 1.27 kV/m and a magnetic field of 0.490 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:
lesantik [10]3 years ago
8 0

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s

So, the speed of the electron is 2591.83 m/s.

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What is dark energy?
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Iron combines with oxygen and water from the air to form rust. If an iron nail were
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8 0
3 years ago
A coil is wrapped with 332 turns of wire on the perimeter of a circular frame (of radius 30 cm). Each turn has the same area, eq
almond37 [142]

Answer:

E=84.5V

Explanation:

From the question we are told that:

Number of Turns N=332turns

Radius r= 30cm

Field Change B=56mt-29mt=27mt

Time t=63ms

Generally the equation for Magnetic Field is mathematically given by

 \frac{dB}{dt}=\frac{27*10^{-3}}{29*10^{-3}}

 \frac{dB}{dt}=0.9T/s

Generally the Flux at 332 turns is  mathematically given by

 \phi=N*A*B

Generally the equation for Area of coil is mathematically given by

 A=\pi*r^2

 A=\pi*(r*10^{-2})^2

Since

 \phi=332*\pi*(\theta*10^{-2})^2*B

Therefore

 \frac{d \phi}{dt}=332*\pi*(900*10^{-4}*\frac{dB}{dt}

Generally the equation for emf Magnitude is mathematically given by

 E=\frac{d\phi}{dt}

 E=332*\pi*(900*10^{-4}*0.9

 E=84.5V

4 0
2 years ago
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