A circular coil of raduis R carries a current I. The magnetic field at the center is B. At what distance from the center, on the axis of the coil, magnetic field will be B/8?

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Solution

B(axis) = uoIr2/2(r2+x2)3/2
B(center) = uoI/2r

B(axis) = 1/8 (B(center) so
r2/(r2+x2)3/2 = 1/8r
8r3 = (r2+x2)3/2
(2r)2 = r2 + x2
x = r(root3)

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A circular coil of raduis R carries a current I. The magnetic field at the center is B. At what distance from the center, on the axis of the coil, magnetic field will be B/8?

B(axis) = uoIr2/2(r2+x2)3/2 B(center) = uoI/2r B(axis) = 1/8 (B(center) so r2/(r2+x2)3/2 = 1/8r 8r3 = (r2+x2)3/2 (2r)2 = r2 + x2 x = r(root3)

B(axis) = uoIr2/2(r2+x2)3/2
B(center) = uoI/2r

B(axis) = 1/8 (B(center) so
r2/(r2+x2)3/2 = 1/8r
8r3 = (r2+x2)3/2
(2r)2 = r2 + x2
x = r(root3)