A current is set up in a wire loop consisting of a semicircle of radius 2.22 cm,

Question

A current is set up in a wire loop consisting of a semicircle of radius 2.22 cm, a smaller concentric semicircle, and two radial straight lengths, all in the same plane. Part (a) of the figure shows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is 45.42 ?T. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane (part (b) of the figure). The magnetic field produced at the (same) center of curvature now has magnitude 15.14 ?T, and its direction is reversed. What is the radius of the smaller semircircle?

Explanation / Answer

The B field from a semicircular current (radius R and current I) is B = 0I / 4R .

If a is the outer
radius and b is the inner radius, then the B field for (a) is Ba = 1/4 *0I ( 1/ a -1/ b )
and Bb = 1/4 *0I ( 1/ a -1/ b )

The ratio of (b) to (a) is Bb / Ba = (1/ a +1/ b)) / (1/ a 1/ b) or
Bb / Ba = (a + b) / (b a) . Using Bb / Ba = 45.42/15.14 = -3,

we obtain b = 2a/4 = a/2 = 1.11 cm.

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