can you please ans (a) (b) (c)? Problem 9: An electron beam is shot into a region where there is bothan electric field pointing upward of magnitude E = 9.6 N/C, and amagnetic field, B, as shown in the figure. While moving through theregion the electron beam stays in a straight line with speed v = 390 m/s.Randomized VariablesE = 9.6 N/Cv = 390 m/sPart (a) In which direction does the electric force act on the electrons?MultipleChoice:1) Upward.2) To the right.3) To the left.4) Downward.Part (b) In order to keep the electrons in a straight line, in which direction should the magnetic force act?MultipleChoice :1) Out of the screen.2) To the left.3) To the right.4) Downward.5) Into the screen.6) Upward.Part (c) What should the direction of the magnetic field be in order to produce the force in part (b)?MultipleChoice :1) Out of the screen.2) Into the screen.Part (d) Express the magnitude of the electric force using the electric field E and the elementary charge e.Expression :FE =Select from the variables below to write your expression. Note that all variables may not be required.a, B, 0, B, d, e, E, g, h, i, j, m, P, t, vPart (e) Calculate the numerical value of the magnitude of FE, in newtons.Numeric : A numeric value is expected and not an expression.FE =Part (f) Express the magnitude of the magnetic force in terms of the elementary charge e, speed v, and B, the magnitude ofthe magnetic field.Expression :FB =Select from the variables below to write your expression. Note that all variables may not be required.a, B, 0, B, d, e, E, g, h, i, j, m, P, t, vPart (g) If the electron is continuing in a straight line, express the magnitude of the magnetic field in terms of v and E.Expression :B =Select from the variables below to write your expression. Note that all variables may not be required.a, B, 0, B, d, e, E, g, h, i, j, m, P, t, vPart (h) Calculate the magnitude of the magnetic field if the electron continues in a straight line, in tesla.Numeric : A numeric value is expected and not an expression.B =

Answered: Image /qna-images/answer/cb207417-df29-446a-ae4b-33c6bd81b4d8.jpg

Problem 9: An electron beam is shot into a region where there is both an electric field pointing upward of magnitude E = 9.6 N/C, and a magnetic field, B, as shown in the figure. While moving through the region the electron beam stays in a straight line with speed v = 390 m/s. Randomized Variables E = 9.6 N/C v = 390 m/s Part (a) In which direction does the electric force act on the electrons? MultipleChoice: 1) Upward. 2) To the right. 3) To the left. 4) Downward. Part (b) In order to keep the electrons in a straight line, in which direction should the magnetic force act? MultipleChoice: 1) Out of the screen. 2) To the left. 3) To the right. 4) Downward. 5) Into the screen. 6) Upward. Part (c) What should the direction of the magnetic field be in order to produce the force in part (b)? MultipleChoice : 1) Out of the screen. 2) Into the screen. Part (d) Express the magnitude of the electric force using the electric field E and the elementary charge e. Expression : FE =

Problem 9: An electron beam is shot into a region where there is both an electric field pointing upward of magnitude E = 9.6 N/C, and a magnetic field, B, as shown in the figure. While moving through the region the electron beam stays in a straight line with speed v = 390 m/s. Randomized Variables E = 9.6 N/C v = 390 m/s Part (a) In which direction does the electric force act on the electrons? MultipleChoice: 1) Upward. 2) To the right. 3) To the left. 4) Downward. Part (b) In order to keep the electrons in a straight line, in which direction should the magnetic force act? MultipleChoice: 1) Out of the screen. 2) To the left. 3) To the right. 4) Downward. 5) Into the screen. 6) Upward. Part (c) What should the direction of the magnetic field be in order to produce the force in part (b)? MultipleChoice : 1) Out of the screen. 2) Into the screen. Part (d) Express the magnitude of the electric force using the electric field E and the elementary charge e. Expression : FE =

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter19: Electric Forces And Electric Fields

Section: Chapter Questions

Problem 6P

See similar textbooks

Similar questions

The electric field is zero everywhere except in the region 0≤x≤4.00 cm, where there is a uniform electric field of 110 N/C in the +y direction. A proton is moving in the +x direction with a speed of v = 1.00×10^6 m/s. When the proton passes through the region 0≤x≤4.00 cm, the electric field exerts a force on it. 1)When the x coordinate of the proton’s position is 4.00 cm, what is the y component of its velocity?(Express your answer to three significant figures.)
An electron in a TV CRT moves with a speed of 6.00×107 m/s , in a direction perpendicular to the Earth’s field, which has a strength of 5.00×10−5 T . (a) What strength electric field must be applied perpendicular to theEarth’s field to make the electron moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)
At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.5 * 10^5 m/s and vy =3.0 * 10^3 m/s. Suppose the electric field between the plates is uniform and given by E: = (120 N/C)jˆ . In unit-vector notation, what are (a) the electron’s acceleration in that field and (b) the electron’s velocity when its x coordinate has changed by 2.0 cm?
Consider two identical particles. The magnitude of their gravitational attraction is equal to the magnitude of their electrical repulsion. What is the ratio of q/m for these particles,given G=6.67×1.^-11Nm^2/kg^2 and Eo=8.854×10^-12C^2/N.m^2?
A proton moves at 4.50 x 105 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 9.60 x 103 N/C. Ignoring any gravitational effects, find (a) the time interval required for the proton to travel 5.00 cm horizontally, (b) its vertical displacement during the time interval in which it travels 5.00 cm horizontally, and (c) the horizontal and vertical components of its velocity after it has traveled 5.00 cm horizontally.
Why is the following situation impossible? Two identical dust particles of mass 1 μg arefloating in empty space, far from any external sources of large gravitational or electricfields, and at rest with respect to each other. Both particles carry electric charges that areidentical in magnitude and sign. The gravitational and electric forces between the particleshappen to have the same magnitude, so each particle experiences zero net force and thedistance between the particles remains constant.
At one instant, one electron that is moving through a uniform magnatic filed has velocity V= 40i + 35j (m/s) when it experience a force due to the magnatic field given by Fb=(-4.2i + 4.8j)×(10^-18 N). If Bx is equal to zero calculate the magnatic field. [Give your answer in unit vector notation.
A particle of charge q and mass m moves in the uniform fields E⃗ =E0k^ and B⃗ =B0k^. At t = 0, the particle has velocity v⃗ 0=v0i^. What is the particle's speed at a later time t?
A particle is launched from the origin on the ground with an initialvelocity? = (2.40 m/s) ? + (2.00 m/s) ?. The particle falls freely under earth’sgravitational field.
A proton, mass 1.67 x 10-27 kg, is projected horizontally midway between two parallel plates thatare separated by 0.10 cm, with an electrical field with magnitude 470,000 N/C between the plates.If the plates are 2.90 cm long, find minimum speed of the proton that just misses the lower plate asit emerges from the field.
I have a physics question as follows, A proton moves at 4.5 e5 m/s in the horizontal direction. It enters a uniform electruc field with magnitude of 9.6 e3 N/C. Ignoring any gravitational effects find (a) the time interval required for the proton to travel 5 cm, (b) its vertical displacement during the time interval in which it travels 5 cm horizontally, (c) the horizontal and vertical components of its velocity after it has traveled 5 cm horizontally.
An oil droplet of mass 1.00 × 10–14 kg loses an electron while it is in an electric field of 1.00 × 106 N/C. The resulting change in the acceleration of the oil droplet is approximately Group of answer choices 16.0 m/s2 1.76 × 1017 m/s2 1.60 m/s2 1.76 × 1018 m/s2 176 m/s2