We now finish up problem (2) from the previous instalment. We left off noting that , if N denotes the total electrons released, then you will have to find

f(L) = N - [1 - B(0)/ B(L)]^1/2

bearing in mind, B(z=L) = B(z= -L) = 10 (B(0)

This then yields:

f(L) = N - [1 - 1/ 10]^1/2

= N – [9/10]^1/2 = N – (0.9)^1/2 = N – 0.948

Now let N = 1 (Since N >= 1) be the total (normalized) released electrons, then the total lost is:

f(L) = 1 – 0.948 = 0.052

e.g. 5.2% of the released electrons are lost.

Plasma problem (3):

Given a 1Gev electron traveling in space with a magnetic field B = 0.00005 T, find:

a)the frequency generated

b)the radius of the electron’s motion in the B-field

c)the angle theta of the cone (centered on the direction of the instantaneous velocity)

d)the length of time for a pulse received from the electron by an outside observer

A diagram is included to help shed light. This diagram shows the orbital plane of the (relativistic) electron's motion, as well as the orientations of the magnetic induction (B) - out of the plane (toward the reader), and the radiated E-field which is polarized parallel to the orbital plane. (E with arrow direction in the diagram).

The electron velocity, v, is also shown (tangent to the orbit at the position) and the angle subtended by the radiation cone.

Now, note that a 1 Gev (10^9 ev) electron will be marginally “relativistic” and its velocity can be worked out – using:

E = mc_o^2/ SQRT {1 – (v/c)^2}

Where v is the velocity, c is the speed of light and m_o the electron rest mass (in kg).

The reader should be able to show from this that v = 6.7 x 10^6 m/s. Then the frequency generated (f) can then be obtained from:

f = (1/2pi) eB/ m_o SQRT[1 – (v/c)^2]

where e = 1.6 x 10^-19 C, the unit of electron charges, o f will be approximately, 1.4 MHz

b)the radius of the electron’s orbital plane (see diagram included) will be obtained from:

R = c/ 2pi(f) = 34.1 m

c)the angle subtended by the radiation cone, in which bulk of radiation is concentrated (see diagram) will be:

theta = (1.2 x 10^19) [m_oc^2/ E(ev)]

where E(ev)= 10^9, the energy in electron volts.

You will obtain theta = 9.8 x 10^-4 rad (radians), or about 0.001 rad.

To get degrees, multiply this by 57.3 (deg/rad) to get 0.057 deg

Which – multiplied by 60’ (per degree) - yields 3.4 minutes of arc

d)the pulse duration (t) will be obtained using:

t = R(theta)/ e [1 – (v/c)^2]= 2.2 x 10^-7 sec

or about 22 microseconds each

*the fraction lost*from:f(L) = N - [1 - B(0)/ B(L)]^1/2

bearing in mind, B(z=L) = B(z= -L) = 10 (B(0)

This then yields:

f(L) = N - [1 - 1/ 10]^1/2

= N – [9/10]^1/2 = N – (0.9)^1/2 = N – 0.948

Now let N = 1 (Since N >= 1) be the total (normalized) released electrons, then the total lost is:

f(L) = 1 – 0.948 = 0.052

e.g. 5.2% of the released electrons are lost.

Plasma problem (3):

Given a 1Gev electron traveling in space with a magnetic field B = 0.00005 T, find:

a)the frequency generated

b)the radius of the electron’s motion in the B-field

c)the angle theta of the cone (centered on the direction of the instantaneous velocity)

d)the length of time for a pulse received from the electron by an outside observer

A diagram is included to help shed light. This diagram shows the orbital plane of the (relativistic) electron's motion, as well as the orientations of the magnetic induction (B) - out of the plane (toward the reader), and the radiated E-field which is polarized parallel to the orbital plane. (E with arrow direction in the diagram).

The electron velocity, v, is also shown (tangent to the orbit at the position) and the angle subtended by the radiation cone.

Now, note that a 1 Gev (10^9 ev) electron will be marginally “relativistic” and its velocity can be worked out – using:

E = mc_o^2/ SQRT {1 – (v/c)^2}

Where v is the velocity, c is the speed of light and m_o the electron rest mass (in kg).

The reader should be able to show from this that v = 6.7 x 10^6 m/s. Then the frequency generated (f) can then be obtained from:

f = (1/2pi) eB/ m_o SQRT[1 – (v/c)^2]

where e = 1.6 x 10^-19 C, the unit of electron charges, o f will be approximately, 1.4 MHz

b)the radius of the electron’s orbital plane (see diagram included) will be obtained from:

R = c/ 2pi(f) = 34.1 m

c)the angle subtended by the radiation cone, in which bulk of radiation is concentrated (see diagram) will be:

theta = (1.2 x 10^19) [m_oc^2/ E(ev)]

where E(ev)= 10^9, the energy in electron volts.

You will obtain theta = 9.8 x 10^-4 rad (radians), or about 0.001 rad.

To get degrees, multiply this by 57.3 (deg/rad) to get 0.057 deg

Which – multiplied by 60’ (per degree) - yields 3.4 minutes of arc

d)the pulse duration (t) will be obtained using:

t = R(theta)/ e [1 – (v/c)^2]= 2.2 x 10^-7 sec

or about 22 microseconds each

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