Question
Asked Dec 23, 2019
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A conducting bar of length l moves to the right on two
frictionless rails, as shown in Figure P20.30. A uniform magnetic
field directed into the page has a magnitude of 0.30 T.
Assume ,l = 35 cm and R = 9.0 Ω. (a) At what constant speed
should the bar move to produce an 8.5-mA current in the
resistor? What is the direction of this induced current? (b) At

what rate is energy delivered to the resistor? (c) Explain the
origin of the energy being delivered to the resistor.

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Figure P20.30
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app Figure P20.30

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Expert Answer

Step 1

part A:

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Given Info: The magnetic field is 0.30 T, length of the bar is 35 cm, the resistance is 9.0 N, and the current produce in the bar is 8.5 mA. The motion emf induced in the bar is, e = IR (1) e is the emf, I is the current in the bar, Ris the resistance, The relation between emf and velocity of the bar is, Blv (2) • Bis the magnetic field directed into the page, • lis the length of the bar, • vis the speed of the moving bar,

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Step 2
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Compare equation (1) and (2) to rewrite in terms of v. IR = Blv IR BI Substitute 0.30 T for B, 35 cm for 1, 9.0 N for R, and 8.5 mA for I. ImA(9.02) (0.30 T)(35 cm)( (8.5 mA) ( 1 mA -2 m = 0.73 m/s Therefore, the speed of the moving bar is 0.73 m/s.

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Step 3
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Conclusion: The flux through the closed loop formed by the rails, the bar, and the resistor is directed into the page and is increasing in magnitude. To oppose the increasing inward flux, the induced current will generate a magnetic field directed out of the page through the area enclosed by the loop so that the current passing through counterclockwise direction. Therefore, the direction of the induced current is counterclockwise.

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Science

Physics

Current Electricity