Electrodynamics Solutions (Pt.. 3)

1)Take the electric field E to be in the x-direction and write out an expression for curl E.

Solution: For the case of E_x alone we have:

 

curl E  =   [ e _x         e _y              e _z]

                    [0             0          ¶/¶ z]

                    [   E _x         0                0  ]

=   e _y [¶E_x  /¶ z ]

2)For E in three dimensions (x, y, z) show that:

div curl E = 0;

Solution: First, obtain curl E:

curl E  =   [ e _x         e _y                e _z]

                    [¶/¶ x       ¶/¶ y        ¶/¶ z]

                    [   E _x         E _y                E _z]

= e _x (¶E _z  /¶ y - ¶E _y  /¶ z ) +  e _y (¶E _x  /¶ z - ¶E _z  /¶ x )

+  e _z (¶E _y  /¶ x - ¶E _x  /¶ y )

Taking the divergence:

¶/¶ x (¶E _z  /¶ y - ¶E _y  /¶ z ) + ¶/¶ y (¶E _x  /¶ z - ¶E _z  /¶ x )

+ ¶/¶ z (¶E _y  /¶ x - ¶E _x  /¶ y )  =  

¶ ²E _z /¶ x ¶ y  -   ¶ ²E _y /¶ x ¶ z  + ¶ ²E _x /¶ y ¶ z  - ¶ ²E _z /¶ x ¶ y 

+ ¶ ²E _y /¶ z ¶ x -    ¶ ²E _x /¶ z ¶ y   =    0

(Since like positive and negative terms cancel) 

3)For a particular solar active region the magnetic diffusivity is h  »     327.6 m² /s

If the length scale is L »   10 ⁷ m  and the Alfven speed is V_A  = 10³   m /s, then find the magnetic Reynolds number for the region. From this assess whether the magnetic field is frozen in or not.

Soln.

The magnetic Reynolds number is defined from the text:   Â_m  =  L V_A / h

Where L is a typical length scale for a given solar environment, V_A   is the Alfven velocity and h  is the magnetic diffusivity.  Hence, solving using the parameters given:

Â_m  =    [(10 ⁷ m ) (10³   m /s)]/  (327.6 m² /s)  »   3 x 10 ⁸ 

 Since Â_m   >> 1 then diffusion can be ignored and the frozen-in condition is satisfied.

4) Write out the full mathematical form for curl A in rectangular coordinates.

curl A  =   [e _x         e _y                 e _z]

                    [¶/¶x        ¶/¶y         ¶/¶ z]

                    [   A _x         A _y             A _z  ]

= e _x (¶A _z  /¶ y - ¶A _y  /¶ z ) +  e _y (¶A _x  /¶ z - ¶A _z  /¶ x )

+  e _z (¶A _y  /¶ x - ¶A _x  /¶ y )

 

5) (a) A solar loop has an estimated diameter of 1.1 x 10 ⁹   cm. If the longitudinal magnetic field (estimated by vector magnetograph) is B_z  » 0.1 T, estimate the total current.

(b) A steady current I flows through a hollow cylinder of radius a  and is uniformly distributed around the tube. Let r be the distance from the axis of symmetry of the tube to a given point.Find the magnitude of the magnetic field B at a point inside the tube. 

Solutions:

a)     By def. the total current is:  I_T  =  2 pr B _z  / 0.012

So:    I_T   =   2 p (1.1 x 10 ⁷  m)  (0.1 T)/ 0.012  =  5.7 x 10 ⁸   A

b) By Ampere’s relation:    ∮ B d ℓ =   ò ò_S     J  dS  =  m ₀  I   

Where I denotes the enclosed current. Then:

     ∮ B d ℓ = 0  and      ÷ B÷   = 0 

6) For the problem 5(a), using the same quantities, estimate the force free parameter, a.  Typical solar values of  a associated with coronal loops are of magnitude »  10 ^-10 m ^–1.   How does the value you obtained compare?

Soln.:

The solar version of Ampere’s law is such that:

 curl B  =  aB

 And since: curl B =  m ₀ J   

 Then:   a = m ₀  J  / B

Find magnitude of J from:  J = I/ A = I/ (pr ²)

÷ J÷  =  (5.7 x 10 ⁸   A)/ p (1.1 x 10 ⁷   m) ² = 

(5.7 x 10 ⁸   A)/ p (1.2 x 10 ¹⁴   m ²) = 1.5 x 10 ^-6   Am ^-2

 

Then:   a = m ₀  ÷ J÷  / B _z  =

[(4p  x 10^-7  H/m) (1.5 x 10 ^-6   Am ^-2] / 0.1T  »  1.9 x 10 ^-11   m ^-1

Which is almost an order of magnitude less than the typical solar loop value of:

a = 10 ^-10   m ^-1