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E- C Secure I https:// edugen.wileyplus.com/edugen/student/mainfr.un Wiley PLUS: MyWileyPLUS I Help I Contact Us I Log Out WileyPLUS Halliday, Fundamentals of Physics, 10e FUNDAMENTAL PHYSICS (PHYS 225/226/227) Home Read, Study & Practice Gradebook ORION Assignment Assignment Open Assignment MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK NEXT ASSIGNMENT RESOURCES Chapter 28, Problem 009 etic Fields Chapter 28, Problem In the figure, an electron accelerated from rest through potential difference V 1.12 kV enters the gap between two parallel plates having separation d a 29.6 mm and potential difference V 28, Prob field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? 009 Chapter 28 Problem Chapter 28, Problem Chapter 28 Problem 027 Cha er 28, Problem Number Units Review Results by Study objective LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI LECTURE LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE MATH HELP License Agreement l Privacy Policy I 00 lohn Wiley s Sons Inc All Rights Reserved. A Division of lohn Wile Sons Inc. Version 4.22.2.9Explanation / Answer
kinetic energy gained by he electron when it is accelerated = work done by the potential differnce on it
= e*1.12 kV = 1.792*(10^-16)
where e is the magnitude of charge on the electron.
so, (1/2)m*(v*v) = 1.792*(10^-16)
=> v = 1.984* (10^7)
Now, between the plates, potential difference = 66.4 V
separation = 29.6 mm = 0.0296 m
So, electric field between the plates = V/d = 66.4/0.0296
= 2243.24 N/C
Now, the magnetic field is applied perpendicular to the velocity of the electron, let its magnitude be B.
Then, magnetic force on the electron = e*v*B = electric force = e*2243.24
So, B= 2243.24 /v = 2243.24/1.984* (10^7) = 1.13*10^(-4) j Tesla
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