# Model the resistor and capacitor form an rc circuit

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Model: The resistor and capacitor form an RC circuit. Assume that the length of the wire next to the dot is much larger than 1.0 cm. Visualize: Please refer to Figure P33.44 Solve: As the capacitor discharges through the resistor, the current through the circuit will decrease as ( )( ) ( ) ( ) ( ) ( ) ( ) 5 2 F 10 s 7 10 s 10 s 4 10 s 0 0 50 V 10 A 5 2.0 10 T m/A 10 A 10 A 2.0 10 T 2 2 0.010 m t t RC t t t t I e e e R I B e e e r r μ μ μ μ μ ε μ μ π π Ω = = = Ω × = = = = × A graph of this equation is shown above.

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33.45. Model: Assume that the superconducting niobium wire is very long. Solve: The magnetic field of a long wire carrying current I is 0 wire 2 I B d μ π = We’re interested in the magnetic field of the current right at the surface of the wire, where d = 1.5 mm. The maximum field is 0.10 T, so the maximum current is ( ) ( ) ( ) 3 wire 7 0 2 1.5 10 m 0.10 T 2 750 A 4 10 T m/A d B I π π μ π × = = = × Assess: The current density in this superconducting wire is of the order of 1 × 10 8 A/m 2 . This is a typical value for conventional superconducting materials.
33.46. Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P33.46. Solve: (a) The Biot-Savart law (Equation 33.6) for the magnetic field of a current segment s Δ G is 0 2 ˆ 4 I s r B r μ π Δ × = G G where the unit vector ˆ r points from current segment Δ s to the point, a distance r away, at which we want to evaluate the field. For the two linear segments of the wire, s Δ G is in the same direction as ˆ, r so ˆ 0. s r Δ × = G For the curved segment, s Δ G and ˆ r are always perpendicular, so ˆ . s r s Δ × = Δ G Thus 0 2 4 I s B r μ π Δ = Now we are ready to sum the magnetic field of all the segments at point P. For all segments on the arc, the distance to point P is r = R . The superposition of the fields is 0 0 0 2 2 arc 4 4 4 I IL I B ds R R R μ μ μ θ π π π = = = where L = R θ is the length of the arc. (b) Substituting θ = 2 π in the above expression, 0 0 loop center 2 4 2 I I B R R μ π μ π = = This is Equation 33.7, which is the magnetic field at the center of a 1-turn coil.

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33.47. Model: Use the Biot-Savart law for a current carrying segment. Visualize: Please refer to Figure P33.47. The distance from P to the inner arc is r 1 and the distance from P to the outer arc is r 2 . Solve: As given in Equation 33.6, the Biot-Savart law for a current carrying small segment s Δ G is 0 2 ˆ 4 I s r B r μ π Δ × = G G For the linear segments of the loop, B Δ s = 0 T because ˆ 0. s r Δ × = G Consider a segment s Δ G on length on the inner arc. Because s Δ G is perpendicular to the ˆ r vector, we have 2 0 0 1 0 0 0 0 arc 1 2 2 1 1 1 1 1 1 2 4 4 4 4 4 4 I s Ir I Id I I B B r r r r r r π π μ μ θ μ θ μ θ μ μ π π π π π π Δ Δ Δ = = = = = = A similar expression applies for arc 2 B . The right-hand rule indicates an out-of-page direction for B arc 2 and an into- page direction for B arc 1 . Thus, 0 0 0 1 2 1 2 1 1 , into page , out of page , into page 4 4 4 I I I B r r r r μ μ μ = + = G
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• Winter '10
• E.Salik
• Current, Magnetic Field, ΔS

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