A circular loop with radius R is in a magnetic field that is perpendicular to it....
A circular loop with radius R is in a magnetic field that is perpendicular to it. Assume that the normal to the loop is in the same direction as the field. The strength of the field, as a function of radial distance from the center r, is given by B=kr (where k is a constant). What is the flux over the loop?
My plan for this problem was to take the integral of the function of the magnetic field B=kr. So I did dB = B(r)*dA = 2*pi*k*R^2 dr
I took the integral of dB from 0 to R which gave me B = 2*pi*k*(R^3/3)
Since magnetic flux is = B.A when B is constant, I multiplied the B from above by A = pi*R^2 which gave me 2*pi*k*(R^3/3)*pi*R^2
But this is not the right answer
Any help would be great
Thank you!
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