Physics Solution Manual for 1100 and 2101

Chapter 22 answers to focus on concepts questions 13 c

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Unformatted text preview: EASONING Since the two wires are next to each other, the net magnetic field is everywhere parallel to ∆l in Figure 21.37. Moreover, the net magnetic field B has the same magnitude B at each point along the circular path, because each point is at the same distance from the wires. Thus, in Ampère's law (Equation 21.8), B|| = B , I = I1 + I 2 , and we have ΣB|| ∆l = B ( Σ∆l ) = µ0 ( I1 + I 2 ) But Σ∆l is just the circumference (2πr) of the circle, so Ampère's law becomes B ( 2π r ) = µ0 ( I1 + I 2 ) This expression can be solved for B. SOLUTION a. When the currents are in the same direction, we find that ( ) –7 µ0 ( I1 + I 2 ) 4π × 10 T ⋅ m/A ( 28 A + 12 A ) B= = = 1.1 × 10 –5 T 2π r 2π ( 0.72 m ) Chapter 21 Problems 1179 b. When the currents have opposite directions, a similar calculation shows that ( ) –7 µ0 ( I1 – I 2 ) 4π × 10 T ⋅ m/A ( 28 A–12 A ) B= = = 4.4 × 10 –6 T 2π r 2π ( 0.72 m ) 69. SSM REASONING AND SOLUTION Ampère's law in the form of Equation 21.8 indicates that ΣB||∆l = µ0 I . Since the magnetic field is everywhere perpendicular to the plane of the paper, it is everywhere perpendicular to the circular path and has no component B|| that is parallel to the circular path. Therefore, Ampère's law reduces to ΣB||∆l = 0 = µ0 I , so that the net current passing through the circular surface is zero . 70. REASONING Both parts of this problem can be solved using Ampère’s law with the circular closed paths suggested in the Hint given with the problem statement. The circular closed paths are used because of the symmetry in the way the current is distributed on the copper cylinder. SOLUTION a. An end-on view of the copper cylinder is a circle, as the drawing at the right shows. The dots around the circle represent the current coming out of the paper toward you. The larger dashed circle of radius r is the closed path used in Ampère's law and is centered on the axis of the cylinder. Equation 21.8 gives Ampère's law as ΣB||∆l = µ0 I . Because of the symmetry of the arrangement in the drawing, we have B|| = B for all ∆l on the circular path, so that Ampère's law becomes Circular path, radius = r Copper cylinder ΣB||∆l = B ( Σ∆l ) = µ0 I In this result, I is the net current through the circular surface bounded by the dashed path. In other words, it is the current I in the copper tube. Furthermore, Σ∆l is the circumference of the circle, so we find that µI B ( Σ∆l ) = B ( 2π r ) = µ0 I or B= 0 2π r 1180 MAGNETIC FORCES AND MAGNETIC FIELDS Circular path, radius = r b. The setup here is similar to that in part a, except that the smaller dashed circle of radius r is now the closed path used in Ampère's law (see the drawing at the right). With this change, the derivation then proceeds exactly as in part a. Now, however, there is no current through the circular surface bounded by the dashed path, because all of the current is outside the path. Therefore, I = 0 A, and B= Copper cylinder µ0I = 0T 2π r 71. REASONING AND SOLUTION The drawing at the right shows an end-on view of the solid cylinder. The dots represent the current in the cylinder coming out of the paper toward you. The dashed circle of radius r is the closed path used in Ampère's law and is centered on the axis of the cylinder. Equation 21.8 gives Ampère's law as ΣB||∆l = µ0 I . Because of the symmetry of the arrangement in the drawing, we have B|| = B for all ∆l on the circular path, so that Ampère's law becomes Solid cylinder Circular path, radius = r ΣB||∆l = B ( Σ∆l ) = µ0 I In this result, Σ∆l = 2π r, the circumference of the circle. The current I is the part of the total current that comes through the area π r 2 bounded by the dashed path. We can calculate this current by using the current per unit cross-sectional area of the solid cylinder. This current per unit area is called the current density. The current I is the current density times the area π r 2 : I r2 I I = 02 π r2 = 0 2 R π 4 R4 123 ( ) current density Thus, Ampère's law becomes B ( Σ∆l ) = µ0 I or I r2 B ( 2π r ) = µ0 0 2 R or B= µ0 I 0 r 2π R 2 Chapter 21 Problems 72. REASONING The magnitude B of the magnetic field is given by B = 1181 F q0 v sin θ (Equation 21.1), and we will apply this expression directly to obtain B. SOLUTION The charge q0 = −8.3 ×10−6 C travels with a speed v = 7.4 × 106 m/s at an angle of θ = 52° with respect to a magnetic field of magnitude B and experiences a force of magnitude F = 5.4 × 10−3 N. According to Equation 21.1, the field magnitude is B= F 5.4 ×10−3 N = = 1.1× 10−4 T q0 v sin θ −8.3 × 10−6 C 7.4 × 106 m/s sin 52° ( ) Note in particular that it is only the magnitude q0 of the charge that appears in this calculation. The algebraic sign of the charge does not affect the result. 73. REASONING AND SOLUTION The magnitude B of the magnetic field at a distance r from a long straight wire carrying a current I is B = µ0I/(2π r). Thus, the distance is c b hg h 4...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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