Model only the two currents enclosed by the closed

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Model: Only the two currents enclosed by the closed path contribute to the line integral. Visualize: Please refer to Figure EX33.22. Solve: Ampere’s law gives the line integral of the magnetic field around the closed path: 5 0 through 1.38 10 T m B d s I μ = = × G G ú ( ) ( ) ( ) 7 0 2 3 3 4 10 T m/A 8.0 A I I I μ π = + = × + ( ) 5 3 7 1.38 10 T m 8.0 A 4 10 T m/A I π × + = × I 3 = 3.0 A, out of the page. Assess: The right-hand rule was used above to assign a positive sign to I 2 . Since I 3 is also positive, it is in the same direction as I 2 .
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33.23. Model: The magnetic field is that of a current flowing into the plane of the paper. The current- carrying wire is very long. Visualize: Please refer to Figure EX33.23. Solve: Divide the line integral into three parts: f i semicircle right line left line B d s B d s B d s B d s = + + G G G G G G G G The magnetic field of the current-carrying wire is tangent to clockwise circles around the wire. B G is everywhere perpendicular to the left line and to the right line, thus the first and third parts of the line integral are zero. Along the semicircle, B G is tangent to the path and has the same magnitude B = μ 0 I /2 π d at every point. Thus 7 f 6 0 0 i (4 10 T m/A)(2.0 A) 0 0 ( ) 1.26 10 T m 2 2 2 I I B ds BL d d μ μ π π π × = + + = = = = × G G where L = π d is the length of the semicircle, which is half the circumference of a circle of radius d .
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33.24. Model: Assume that the solenoid is an ideal solenoid. Solve: We can use Equation 33.16 to find the current that will generate a 3.0 mT field inside the solenoid: 0 solenoid NI B l μ = solenoid 0 B l I N μ = Using l = 0.15 m and 0.15 m 0.0010 m 150, N = = ( ) ( ) ( ) ( ) ( ) 3 7 3.0 10 T 0.15 m 2.4 A 4 10 T m/A 150 I π × = = Assess: This is a reasonable current to pass through a good conducting wire of diameter 1 mm.
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33.25. Model: Assume that the solenoid is an ideal solenoid. Solve: We can use Equation 33.16 to find the current that will generate a magnetic field of 1.5 T inside the solenoid: 0 solenoid NI B l μ = solenoid 0 B l I N μ = Using l = 1.8 m and 1.8 m 0.0020 900, N = = ( )( ) ( ) 7 1.5 T 1.8 m 2.4 kA 4 10 T m/A 900 I π = = Assess: Large currents can be passed through conducting wires. The current density through the superconducting wire is ( ) ( ) 2 8 2 2400 A 0.001 m 7.6 10 A/m π = × . This value is reasonable for a superconducting wire.
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33.26. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure EX33.26. Solve: (a) The force on the charge is ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 19 7 on q 7 19 13 ˆ ˆ ˆ 1.60 10 C 1.0 10 m/s cos45 sin45 0.50 T 1.0 10 m/s ˆ ˆ ˆ ˆ ˆ 1.60 10 C 0.50 T 5.7 10 N 2 F qv B i k i i i k i j = × = × × ° + ° × × = × × + × = + × G G G (b) Because the cross product ˆ ˆ i i × in the equation for the force is zero, on q 0 N. F = G G
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33.27. Model: A magnetic field exerts a magnetic force on a moving charge. Visualize: Please refer to Figure EX33.27.
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