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

**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}

^{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)_{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.

_{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*Ω*

_{e}T) = 6.28 rad = 2p rad

*Ω*

_{e}T/ 4) =1.57 rad = p / 2

**= 0**

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

From the
information provided we see that:

**v**

_{d}**=**(1 +

**r**

_{¼}^{2 }

**v**⊥

^{2 })

**E**

_{o }

**X B/ B**

^{2}

^{ }So the other information needed to obtain

**E**

_{o}would be the drift velocity,

**v**

_{d}**.**

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