A segment of wire 5.7 cm long is moving at a velocity of 2.89 m/s toward a parallel (much longer) wire which carries a current of 5.45 A. At a particular instant the wire segment is 5.7 cm away from the long, current-carrying wire. What is the induced emf (voltage) between the two ends of the wire segment? O 4.10E-06 V O 3.15E-06 V O 9.45E-07 V O 2.21E-06 V

The figure below shows a simplified model of a cardiac defibrillator, a device used to resuscitate patients in ventricular fibrillation. S R Rorso + When the switch S is toggled to the left, the capacitor C charges through the resistor R. When the switch is toggled to the right, the capacitor discharges current through the patient's torso, modeled as the resistor Rorso allowing the heart's normal rhythm to be reestablished. HINT The voltage across a charging capacitor is AV = E(1 - e-t/(RC)) and the voltage across a discharging capacitor is AV = AV maxe-t(RC) Click the hint button again to remove this hint. (a) If the capacitor is initially uncharged with C = 8.25 µF and Ɛ = 1270 V, find the value of R (in ohms) required to charge the capacitor to a voltage of 795 V in 1.20 s. (b) If the capacitor is then discharged across the patient's torso with Rorso = 1240 0, calculate the voltage (in V) across the capacitor after 4.50 ms. V

Two long, parallel wires, each with a mass per unit length of 40 g/m, are supported in a horizontal plane by 6.0-cm-long strings, as shown in Figure P19.72. Each wire carries the same current I, causing the wires to repel each other so that the angle 0 between the supporting strings is 16°. (a) Are the currents in the same or opposite directions? (b) Determine the magnitude of each current.

& = 24.0 V b 0.10 2 R, 5.0 Ω. E- 48.0 V R 20 Ω R a 0.50 2 40 Q 20.20 2 R 78 Ω E3 6.0 V Er 36.0 V i 0.05 Q h Write the equation which results from applying the loop rule to loop akledcba. 0 = If the current going through the center of the loop is 45 A, what is the current going through the top, in amps? I =