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# CH30 - CHAPTER 30 SOURCES OF THE MAGNETIC FIELD...

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CHAPTER 30—SOURCES OF THE MAGNETIC FIELD ActivPhysics can help with these problems: Activities 13.1–13.3, 13.5 Section 30 - 1:—The Biot-Savart Law Problem 1. A wire carries 15 A. You form the wire into a single-turn circular loop with magnetic field 80 T μ at the loop center. What is the loop radius? Solution Equation 30-3, with x = 0, gives the magnetic field at the center of a circular loop, B I a = μ 0 2 = (with direction along the axis of the loop, consistent with the sense of circulation of the current and the right-hand rule). Thus, the radius is a I B = = × = - μ π μ 0 7 2 2 2 10 15 80 118 = = ( / )( ) ( ) . . N A A T cm Problem 2. A single-turn wire loop is 2.0 cm in diameter and carries a 650-mA current. Find the magnetic field strength (a) at the loop center and (b) on the loop axis, 20 cm from the center. Solution Equation 30-3 gives: (a) at the center, x B I a = = = × × = - 0 2 4 10 650 2 1 40 8 0 7 2 , ( / )( ) ( ) . μ π μ = = N A mA cm T; (b) on the axis, x B Ia x a I = = + = × + = - - 20 1 20 1 5 09 1 2 0 2 2 2 3 2 1 2 0 2 2 2 3 2 cm, cm cm cm nT μ μ ( ) ( ) [( ) ( ) ] . . = = Problem 3. A 2.2-m-long wire carrying 3.5 A is wound into a tight, loop-shaped coil 5.0 cm in diameter. What is the magnetic field at its center? Solution The magnetic field at the center of each of the tightly-wound turns is the same (Equation 30-3 with x = 0 ), so the net field is B N I a = ( ), μ 0 2 = where N is the number of turns. If the wire has length L and the coil has diameter D , then N L D = = π and B L D I D IL D = = = × = - ( )( ) ( )( . )( . ) ( ) . = = = = π μ μ π 0 0 2 7 2 2 4 10 3 5 2 2 5 123 N/A A m cm mT. Problem 4. What is the current in a long wire if the magnetic field strength 1.2 cm from the wire’s axis is 67 T? μ Solution Equation 30-5 (or Equation 30-8) gives the magnetic field strength near a long straight wire, B I r = μ π 0 2 = , from which we find I rB = = × = - 2 12 67 2 10 4 02 0 7 2 π μ μ = = ( . )( ) ( / ) . cm T N A A. Problem 5. Suppose Earth’s magnetic field arose from a single loop of current at the outer edge of the planet’s liquid core (core radius 3000 km), concentric with Earth’s center. What current would be necessary to give the observed field strength of 62 T μ at the north pole? (The currents responsible for Earth’s field are more complicated than this problem suggests.) Solution The north pole is on the axis of the hypothetical circular current loop, with radius a = 3 0 . Mm, at a distance x R E = = 6 37 . Mm from the earth’s center. Equation 30-3 gives a current of I B x a a = + = 2 2 2 3 2 0 2 ( ) = = μ 2 62 6 37 3 0 4 10 3 0 3 83 2 2 3 2 7 2 ( )[( . ) ( . ) ] ( / ( . ) . T Mm Mm T m A) Mm GA. μ π + × = - = =

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192 CHAPTER 30 Problem 6. Earth’s magnetic dipole moment is 8 0 10 22 2 . . × A m What is the magnetic field strength on Earth’s surface at either pole? Solution The magnetic field strength at points on the axis of a magnetic dipole is given by Equation 30-4. The magnetic poles are two such points on the Earth’s surface; see Fig. 30-6. Thus, B R E = = × F H I K × × = × = - - μ π μ 0 3 7 2 22 2 6 3 5 2 2 10 8 10 6 37 10 6 20 10 0 62 N A A m m T G ( ) ( . ) . . .
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