A cross section of earth\'s magnetic field can be represented as a vector field

Question

                    A cross section of earth's magnetic field can be represented as a vector field in which the center of earth is located at the origin and the positive                     y-axis points in the direction of the magnetic north pole. The equation for this field is F(x,y)= (m/(x^2+y^2)^(5/2)) *(3xy i + (2y^2 - x^2)j)                 

                    where m is the magnetic moment of earth.                 

                    Find the work in moving a particle from (200, 400) to (300, 500) in terms of m.                 

                    (Hint: It is a conservative vector field.)                 

Explanation / Answer

dW= F.ds

= (m/(x^2+y^2)^(5/2)) *(3xy i + (2y^2 - x^2)j) *(1//(x^2+y^2)^(1/2)) *(x i + yj)

= (m/(x^2+y^2)^(3))*(3x^2y + 2y^3 - x^2y)

= (m/(x^2+y^2)^(3))*(2x^2y + 2y^3)

= (2my/(x^2+y^2)^(2))

since its a conservative vector filed

W = 2*m * 400/ (200^2 + 400^2)^2 - 2*m*500/(300^2 + 500^2)^2

=2*10^-8m- 8.65*10^-9m

= 11.35*10^-9 m

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