(a) Explanation Given Info: The central conductor carries current ( I 1 ) and tube carries currents ( I 2 ) in same direction, radius of solid conductor is ( a ) , inner radius of conducting tube is ( b ) and outer radius of conducting tube is ( c ) . Formula to determine the net current enclosed in the conductor by a closed curve, ∮ B → ⋅ d → l = μ 0 I encl (I) μ 0 is the magnetic constant, I encl is the current enclosed in the closed curve, ∮ d l → is the differential length of the conductor. The direction of I 1 and I 2 is positive and directed into the page, so the integration path will be in the clockwise direction. Between the outer and inner conductor consider a circular line integral with radius r is equal to the circumference of the circle. Due to symmetry the magnetic field will be constant around the path. Then, the field is due to only the current I 1 . in the inner conductor. The net current enclosed is. I encl = I 1 Substitute I 1 for I encl in equation (I) ∮ B → ⋅ d → l = μ 0 I 1 (II) Take magnitude for equation (II) (b) To determine The expression for magnitude of magnetic field at points outside the tube.

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Author: YOUNG

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Chapter 28, Problem 28.44E

(a)

To determine

The expression for magnitude of magnetic field at points outside the central solid conductor but inside the tube.

(b)

To determine

The expression for magnitude of magnetic field at points outside the tube.

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Chapter 28, Problem 28.43E

Chapter 28, Problem 28.45E

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The expression for magnitude of magnetic field at points outside the central solid conductor but inside the tube.

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The expression for magnitude of magnetic field at points outside the central solid conductor but inside the tube.

The expression for magnitude of magnetic field at points outside the tube.

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