Hwksolns3

Hwksolns3 - dI dI not I and V and L are both known we can...

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34.12 a. No. Since dI VL dt Δ= and Δ V and L are both known, we can only find dI dt , not I . b. Yes, through the right-hand inductor, since dI V dt L Δ = . c. Yes, it is possible to tell: The current is decreasing. If 0 dI dt < then L 0 V Δ > and the input side is more negative, and the potential increases in the direction of the current.

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34.16. Model: A changing magnetic field creates an electric field. Visualize: Solve: (a) Apply Equation 34.26. For a point on the axis, r = 0 m, so E = 0 V/m. (b) For a point 2.0 cm from the axis, () 0.020 m 4.0 T/s 0.040 V/m 22 rdB E dt ⎛⎞ == = ⎜⎟ ⎝⎠
34.19. Model: Assume that the current changes uniformly. Visualize: We want to increase the current without exceeding a maximum potential difference. Solve: Since we want the minimum time, we will use the maximum potential difference: () 3 max 3.0 A 1.0 A 200 10 H 1.0 ms 400 V dI I I VL L t L dt t V ΔΔ Δ= = = × = Assess: If we change the current in any shorter time the potential difference will exceed the limit.

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34.21. Visualize: The solenoid has inductance and when a current flows there is energy stored in the magnetic field. Solve: The inductance and energy of the solenoid are () ( ) 22 7 2 4 0 sol 24 2 5 11 L 4 10 H/m 200 0.015 m 2.96 10 H 0.12 m (2.96 10 H)(0.80 A) 9.5 10 J NA L l UL I ππ μ −− × == = × × = ×
34.25. Visualize: Please refer to Figure Ex34.25. This is a simple LR circuit if the resistors in parallel are treated as an equivalent resistor in series with the inductor. Solve: We can find the equivalent resistance from the time constant since we know the inductance. We have 3 eq 6 eq 3.6 10 H 360 10 10 s LL R R τ × =⇒= = = Ω × The equivalent resistance is the parallel addition of the unknown resistor R and 600 Ω . We have ( )( ) eq 600 360 11 1 900 600 600 360 R RR ΩΩ =+ = = Ω Ω

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34.26. Visualize: Please refer to Figure Ex34.26. This is a simple LR circuit if the two resistors in series are treated as an equivalent resistor in series with the inductor. Solve: The time constant of the equivalent circuit is 3 4 eq 1 2 50 10 H 1.00 10 s 500 LL RR R τ × == = If the current is initially at the maximum, then it will decay exponentially with time according to / 0 () t It Ie = . The time 1/2 t when the current drops to half the initial value is found as follows: () ( ) / 45 11 00 1 / 2 1 / 2 22 ln / ln 2 1.00 10 s 0.69 6.9 10 s t II e t t ττ −− =⇒ = = = × = × Assess: This is less than , as expected, since the time constant is the amount of time it takes to decay to 0.37 of the initial current and we are only going to 1/2. This result is quite general and is characteristic of all exponential decay phenomena.
34.59. Model: Use Faraday’s law of electric induction. Visualize: Equation 34.23, dB Ed s A dt ⋅= G G ú , describes the relationship between an induced electric field and the flux through a fixed area A. To solve this equation, choose a clockwise direction around a circle of radius r as the closed curve. The electric field vectors, as can be seen from the figure, are everywhere tangent to the curve. The line integral of E G then is () 2 E ds r E π G G ú

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Hwksolns3 - dI dI not I and V and L are both known we can...

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