Part A A beam of electrons is moving due East at a speed of 5.0×10 6 m/s in the

Question

Part A

   A beam of electrons is moving due East at a speed of 5.0×106 m/s in the presence of a uniform electric field of 7400 V/m that points vertically upward. What is the magnitude of the uniform magnetic field that must be applied to prevent any deflection of the electron beam?  

B = ____________ ______

Part B

What is the proper direction for this magnetic field? (a or b)

a. The magnetic field points due North.

b. The magnetic field points due South.

Part C

   If the electric field is switched off, with what frequency will the electrons move around their circular orbits? Express your answer to two significant figures in Hz.

f = ___________ ____

Explanation / Answer

here,
velocity of electrons = 5 * 10^6 m/s
E = 7400 V/M

Part A:

as,Electron is moving in straight line,Under the influence of Electric field and magantic Field therefore

Electrostatic force = Electromagnetic force
qE = Bqv

B = E / V = 7400 /(5 * 10^6)

B = .00148 T

the magnitude of the uniform magnetic field that must be applied to prevent any deflection of the electron beam is .00148 T

Part B:

The magnetic field point due South.

Part C:

Force due to Circular motion = Electromagnetic force

M*V^/R = Q*V*B

R = m * V / Q *B

R = (9.1 * 10^-31 * 5 * 10^6) / (1.6*10^-19 *.00148 )

R = 0.0192 m

as

Frequency = V / Time = V / 2* pi * R

f = 5*10^6 / (2*3.14*0.0192)

f = 4.146 * 10^7 HZ

The frequency 4.146 * 10^7 HZ with which the electrons move around their circular orbits

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