Question

Let F be the vector field represented in the figure: y X 1Q0Y, 1X P(-1, 1) X Q3.1 3 Points 2d-Curl F(0,0) > 0 O 2d-Curl F(0,0) = 0 2d-Curl F(0,0) < 0 Q3.2 3 Points OV: F(0,0) > 0 OV: F(0,0) = 0 OV: F(0,0) < 0

Answer 1

SOLUTION:-

3.1: Answer is, 2d-curl F (0, 0) is greater than zero. because around the origin if we consider a circle its direction along the vector field looks like a counter clock wise rotation. A counter clockwise rotation gives a positive 2d- curl and clockwise rotation gives a negative 2d-curl. If there is no rotation, then curl is zero.

Here the 2d- curl F at (0, 0) is positive.

3.2: For the case of Divergence of F at (0, 0) we see that around the circle there is outward flow for the given vector field F. So div F (0, 0) is positive.

$i.e, \bigtriangledown .F (0,0) > 0$