1) Find the ion and electron gyrofrequencies for an ion and electron in solar plasma with a magnetic induction (field strength) of

**B**= 0.0001 T.
Solution:

**The ion gyrofrequency**will be:*Ω*

_{i}

**= qB/ m**

_{i }= [(1.6 x 10

^{-19}C) (0.0001 T) ]/ 1.7 x 10

^{-27}kg

*Ω*

_{i}= 9.4 x 10

^{3}/s

And the

**electron gyrofrequency**is:*Ω*

_{e}

**= qB/ m**

_{e }= [(1.6 x 10

^{-19}C) (0.0001 T) ]/ 9.1 x 10

^{-31}kg

*Ω*

_{e}

**= 1.7 x 10**

^{7}/s

2) If the perpendicular velocity component (

**v**⊥) is 10^{5 }m/s*for the electron*, find its Larmor radius and its gyro-period.**v**⊥ / B] =

**v**⊥/ (qB/ m

_{e}) =

**v**⊥/

*Ω*

_{e}

r = (10

^{5}m/s) / 1.7 x 10^{7}/s = 0.0056 m or: 0.56 cm
Gyro-period: T = 2 p /

*Ω*_{e}*=*2 p / 1.7 x 10^{7}/s = 3.5 x 10^{-7}s
3) Thence or otherwise obtain the gyration energy in eV.

(1.6 x 10

^{-19}J = 1 eV)
Gyration energy E = m

(9.1 x 10

_{e}(**v**⊥)^{2}/ 2 =(9.1 x 10

^{-31}kg) (10^{5}m/s)^{ 2}/ 2
E = 4.5 x 10

^{-21}J
Or: in

**:***Electron volts*
(4.5 x 10

^{-21}J )/ (1.6 x 10^{-19}J/eV) = 0.028 eV
4) Find the guiding center positions for the electron referenced above (previous problems) if t = T/4.

Guiding center positions:

x – x

_{o}= r sin (*Ω*_{e}t) = (0.0056 m) sin (*Ω*_{e }T/4) == (0.0056 m) sin [(1.7 x 10

^{7}/s) (3.5 x 10

^{-7}s/ 4)] = 0.0056 m

y – y

_{o}= r cos (*Ω*_{e}t) =
(0.0056 m) sin [(1.7 x 10

^{7}/s) (3.5 x 10^{-7}s/ 4)] = 0
To see how these values can be, note that: (

*Ω*_{e}T) = 6.28 rad = 2p rad
So: (

*Ω*_{e}T/ 4) =1.57 rad = p / 2
But: sin (p /2) = 1 and (cos p /2)

**= 0**
So the value for x – x

_{o}is simply dictated by the value for r
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