Therefore we can write the solutions for the electrons velocity components as

# Therefore we can write the solutions for the

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Therefore, we can write the solutions for the electron’s velocity components as follows: or equivalently: In-Class Exercises Pb. 8.20 Plot the 3-D curve, with time as parameter, for the tip of the velocity vector of an electron with an initial velocity v = v 0 ê 1 , where v 0 = 10 5 m/s, enter- ing a region of space where a constant electric field and a constant magnetic ﬂux density are present and are described by: = E 0 ê 3 , where E 0 = –10 4 V/m, and = B 0 ê 3 , where B 0 = 10 –2 Wb/m 2 . The mass of the electron is m e = 9.1094 × 10 –31 kg, and the magnitude of the electron charge is e = 1.6022 × 10 –19 C. Pb. 8.21 Integrate the expression of the velocity vector in Pb. 8.20 to find the parametric equations of the electron position vector for the preceding prob- lem configuration, and plot its 3-D curve. Let the origin of the axis be fixed to where the electron enters the region of the electric and magnetic fields. Pb. 8.22 Find the parametric equations for the electron velocity if the electric field and the magnetic ﬂux density are still parallel, the magnetic ﬂux density is still constant, but the electric field is now described by = E 0 cos( ω t ) ê 3 . e t t t t t A = cos( ) sin( ) sin( ) cos( ) α α α α 0 0 0 0 1 v t v t v t t t t t v v v t 1 2 3 1 2 3 0 0 0 0 1 0 0 0 0 0 ( ) ( ) ( ) cos( ) sin( ) sin( ) cos( ) ( ) ( ) ( ) = + α α α α β v t v t v t v t v t v t v t v t 1 1 2 2 1 2 3 3 0 0 0 0 0 ( ) ( )cos( ) ( )sin( ) ( ) ( )sin( ) ( )cos( ) ( ) ( ) = + = − + = + α α α α β r E r B r E © 2001 by CRC Press LLC Example 8.14 Find the motion of an electron in the presence of a constant electric field and a constant magnetic ﬂux density perpendicular to it. Solution: Let the electric field and the magnetic ﬂux density be given by: The matrix A is given in this instance by: while the vector B is still given by: The matrix e A t is now given by: and the solution for the velocity vector is for this configuration given, using Eq. (8.40), by: leading to the following parametric representation for the velocity vector: r r E E ê B B ê = = 0 3 0 1 A = α 0 0 0 0 0 1 0 1 0 B = β 0 0 1 e t t t t t A = 1 0 0 0 0 cos( ) sin( ) sin( ) cos( ) α α α α v t v t v t t t t t v v v t t t t 1 2 3 1 2 3 1 0 0 0 0 0 0 0 1 0 0 0 0 ( ) ( ) ( ) cos( ) sin( ) sin( ) cos( ) ( ) ( ) ( ) cos[ ( )] sin[ ( )] sin[ ( )] cos[ ( = + + α α α α α τ α τ α τ α τ )] 0 0 0 t d β τ © 2001 by CRC Press LLC Homework Problems Pb. 8.23 Plot the 3-D curve, with time as parameter, for the tip of the veloc- ity vector of an electron with an initial velocity where v 0 = 10 5 m/s, entering a region of space where the electric field and the mag- netic ﬂux density are constant and described by = E 0 ê 3 , where E 0 = –10 4 V/m; and = B 0 ê 1 , where B 0 = 10 –2 Wb/m 2 .  #### You've reached the end of your free preview.

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