Answer:
(a) 
(b) 
Explanation:
(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

The radius of the wire = a

The surface current density 
(b) The current density is inversely proportional
......(1)
k is the constant of proportionality

........(2)
substituting (1) into (2)





substitute 

Answer:
20cm
Explanation:
A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).
According to the lens equation
where;
f is the focal length of the lens
u is the object distance
v is the image distance
If the magnification is - 0.6
mag = v/u = -0.5
v = -0.5u
since v = 10cm
10 = -0.5u
u = -10/0.5
u =-20 cm
Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

Hence the focal length of the convex lens is 20cm
Yes I'm pretty sure you can
Answer:
t.f. are you sure that's english? it looks like not a real thing
Explanation: