Model the coils are identical parallel and carry

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Model: The coils are identical, parallel, and carry equal currents in the same direction. The magnetic field is that of the currents in these coils. Solve: (a) The on-axis magnetic field of an N -turn current loop at a distance z from the loop center is ( ) 2 0 loop 3/ 2 2 2 = 2 NIR B z R μ + If the spacing between the loops is R , then the midpoint between them is z = R /2. Since the currents are in the same direction, the field of each loop is in the same direction and the net field is simply twice the field due to a single loop. Thus ( ) 2 3/ 2 0 0 3/ 2 2 2 = (1.25) / 4 NIR NI B R R R μ μ = + (b) The magnetic field is ( ) ( ) ( ) 7 4 4 10 T m/A 10 1.0 A 0.716 1.80 10 T 0.050 m B π = = ×
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33.53. Model: The magnetic field is that of a current in the wire. Visualize: Please refer to Figure P33.53. Solve: As given in Equation 33.6 for a current carrying small segment , s Δ G the Biot-Savart law is 0 2 ˆ 4 I s r B r μ π Δ × = G G For the straight sections, ˆ 0 s r Δ × = G because both s Δ G and ˆ r point along the same line. That is not the case with the curved section over which s Δ G and r G are perpendicular. Thus, 0 0 0 2 2 4 4 4 I s IR d I d B r R R μ μ θ μ θ π π π Δ = = = where we used s R R d θ θ Δ = Δ for the small arc length Δ s . Integrating to obtain the total magnetic field at the center of the semicircle, /2 0 0 0 /2 4 4 4 I d I I B R R R π π μ θ μ μ π π π = = =
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33.54. Model: The toroid may be viewed as a solenoid that has been bent into a circle. Visualize: Please refer to Figure P33.54. Solve: (a) A long solenoid has a uniform magnetic field inside and it is roughly parallel to the axis. If we bend the solenoid to make it circular, we will have circular magnetic field lines around the inside of the toroid. However, as explained in part (c) the field is not uniform. (b) A top view of the toroid is shown. Current is into the page for the inside windings and out of the page for the outside windings. The closed Ampere’s path of integration thus contains a current NI where I is the current flowing through the wire and N is the number of turns. Applied to the closed line path, the Ampere’s law is 0 through 0 B d s I NI μ μ = = G G ú Because B G and d s G are along the same direction and B is the same along the line integral, the above simplifies to 0 0 2 2 NI B ds B r NI B r μ π μ π = = = ú (c) The magnetic field for a toroid depends inversely on r which is the distance from the center of the toroid. As r increases from the inside of the toroid to the outside, B toroid decreases. Thus the field is not uniform.
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33.55. Model: The magnetic field is that of the current which is distributed uniformly in the hollow wire. Visualize: Ampere’s integration paths are shown in the figure for the regions 0 m < r < R 1 , R 1 < r < R 2 , and R 2 < r . Solve: For the region 0 m < r < R 1 , 0 through B d s I μ = G G ú . Because the current inside the integration path is zero, B = 0 T. To find I through in the region R 1 < r < R 2 , we multiply the current density by the area inside the integration path that carries the current. Thus, ( ) ( ) 2 2 through 1 2 2 2 1 I I r R R R π π = where the current density is the first term. Because the magnetic field has the same magnitude at every point on
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