we get OD sV×Bda=0/s(J+) Ot′ da Using Stokes Theorem, this is written as OD CB·dl=p0/s(J+x)·da, where C is the curve bounding the surface Let c to be a circle of radius r. Note that Jf+ ot is homogeneous. So OD 2丌B= 丌T OD B=OUf+ a ) ot