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
Ex alone we have:
curl
E = [ e x e
y e z]
[0 0
¶/¶ z]
[ E x 0 0 ]
= e y [¶Ex
/¶ 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 VA = 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: Âm = L VA / h
Where L is a typical length scale for a given solar
environment, VA is the Alfven velocity and h is the
magnetic diffusivity. Hence, solving
using the parameters given:
Âm = [(10 7
m ) (103 m /s)]/ (327.6 m2 /s) » 3 x 10 8
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 Bz » 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:
IT = 2 pr B z / 0.012
So: IT = 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 òS 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:
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
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