Problem a) What value of magnetic field would make a beam of electrons, travelin
ID: 2035090 • Letter: P
Question
Problem a) What value of magnetic field would make a beam of electrons, traveling West at a speed of 2010 ms, go undeflected through a region where there is a uniform electric field of 3.0x10impointing North? Suomit AnswerTries 0/6 b) What is the direction of the magnetic field assuming it is not pointing East? West O North O Into the Page O Out of the page O South Submit AnswerTries 0/6 c) What is the frequency of the circular orbit if the electric field is turned off? Submit AnserTries 0/6 Due Friday April 06 11:59 am (EDT) Problem2 A 1000 turn loop (radius-0.039 m) of wire is connected to a 18.0 ? resistor as shown in the figure. A magnetic field is directed perpendicular to the plane of the loop. The field points into the paper and has a magnitude that varies in time as B-gt, where g-0.12 Ts. Neglect the resistance of the wire 1000-turn loop a) What is the magnitude of the potential difference between points a and b? Submit AnseTries 06 b) What is the electrical energy dissipated in the resistor in 30.70 s? Suomit AnswTries 06 c) Which direction does the current flow? (Maximum one try!) O Counter clockwise O Clockwise ONo direction Submit AngTries 0/1 Due Friday April 06 11:59 am (EDT)Explanation / Answer
a)
for the electron to go undeflected
the net force = 0
electric force = magnetic force
Fe = Fb
E*q = q*v*B
B = E/v = 3*10^4/(2*10^6) = 0.015 T
(b)
the electric force is along south ( -j)
the magnetic force should be along north ( j)
Fb = q*(v X B)
Fyj = -e*(-vxi X (Bxi + Byj + Bzk) ) = e*v*By k - e*vy*Bzj
By = 0
Bz is negative ( into the page)
Into the Page
=======================
(c)
In magentic field
magnetic force Fb = centripetal force
q*v*B = m*v^2/r
r = m*v/(qB)
time period T = 2*pi*r/v = 2*pi*m*v/(q*B*v) = 2*pi*m/(qB)
frequency f = 1/T = q*B/(2*pi*m)
frequency f = 1.6*10^-19*0.015/(2*pi*9.11*10^-31) = 4.2*10^8 Hz
===================================
problem 2
magnetic flux = N*B*A
N = number of turns = 1000
B = magnetic field = gt
A = pi*r^2
r = radius of loop
induced emf = rate of change in magnetic flux
emf = N*A*dB/dt
emf = N*A*g = 1000*pi*0.039^2*0.12 = 0.573 V
magnitude of potential difference dV = 0.573 V
(b)
electrical energy E = ((dV)^2/R)*t = 0.573^2*30.7/18 = 0.56 J
(c)
counter clockwise
DONE please check the answer. any doubts post in comment box
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.