{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Is Curl of A perpendicular to A

Is Curl of A perpendicular to A - 1 Is it correct to say...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1. Is it correct to say that ~ ∇ × ~ A is a vector field that is perpendicular to ~ A because it is a cross product of ~ and ~ A ? Answer: Answer is NO! For example, one can show ~ ∇ × ~ A is not necessarily perpendicular to the vector ~ A , as one would expect from a cross product. We will show this by an explicit construction. Choose ~ A = - ay ˆ i + bx ˆ j + c ˆ k ; where a , b and c are non-zero positive constants. ~ ∇ × ~ A = ( a + b ) ˆ k . The scalar product of ~ ∇ × ~ A and ~ A is ( a + b ) c 6 = 0. The field lines of the vector field ~ A look like a cork screw and its curl points towards the advancing direction of the screw given by the right hand rule. On the other hand the field also has a non-zero component in the direction in which the screw advances. It is not surprising that they are not perpendicular. What then makes ~ ∇ × ~ A a vector field? ~ ∇ × ~ A is a vector field because its components transform like one. For a similar reason ~ ∇ · ~ A is not a simple dot product. For example, ~ ∇ · ~ A 6 = ~ A · ~ . It is a scalar field because it does not change under a general rotation in the three dimensional space. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2. A charge Q is distributed uniformly over a sphere of radius R. Express the resulting volume charge density involving Dirac- δ function. Answer: The charge density is independent of the angles θ and φ and a function of r alone, i.e. ρ ( ~ r ) ρ ( r ). Charge density is zero for all values of r except at r = R . When charge density is integrated over all space (integration over the entire domains of θ and φ gives 4 π ,) one must get the total charge Q . Thus the integration of ρ ( r ) over r is finite even though the function is nonzero only ar r = R . It is possible only if charge density has an integrable singularity at r = R . This implies ρ ( r, θ ) = (
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}