3. A velocity selector consists of electric and magnetic fields that are perpend

Question

3. A velocity selector consists of electric and magnetic fields that are perpendicular to one another. If in such an arrangement a B-field of 12.0 T is used, then what should be the value of the electric filed so that an electron with velocity of 0.2c passes the velocity selector undeflected? a) b) Now we turn off the electric field. Assuming the electron is moving to the right and magnetic field is horizontal and points into the paper, what is the magnitude and direction of the force acting on the electron at that instance? Ignore the effect of gravity. c What is the radius of curvature of the electron in part b

Explanation / Answer

given

B = 12 T

v = 0.2*c

= 0.2*3*10^8 m/s

= 6*10^7 m/s

a) For no deflection, Fe = Fb

q*E = q*v*B

E = v*B

= 6*10^7*12

= 7.2*10^8 m/s

b) FB = q*v*B*sin(theta)

= 1.6*10^-19*6*10^7*12*sin(90)

= 1.15*10^-10 N

direction : Dowbward(towards -y axis)

c) we know, r = m*v/(B*q)

= 9.1*10^-31*6*10^7/(12*1.6*10^-19)

= 2.84*10^-5 m

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