**Problem Statement:**A segment of wire 7.7 cm long is moving at a velocity of 5.80 m/s toward a parallel (much longer) wire which carries a current of 3.51 A. At a particular instant, the wire segment is 6.4 cm away from the long, current-carrying wire. What is the induced emf (voltage) between the two ends of the wire segment?**Options:**- ○ 4.90E-06 V- ○ 1.47E-06 V- ○ 3.43E-06 V- ○ 6.86E-06 V**Explanation:**This problem involves electromagnetic induction, where the movement of a wire segment in a magnetic field generated by a current-carrying wire induces an electromotive force (emf). The key elements to consider are the velocity of the wire segment, the length of the wire, the current in the stationary wire, and the perpendicular distance between the two wires.

concept and formula here we use the formula of induced emf and magnetic field due to long straight…

moving at a velocity of 5.80 m/s toward a parallel (much longer) wire which carries a current of 3.51 A. At a particular instant the wire segment is 6.4 cm away from the long, current-carrying wire. What is the induced emf (voltage) between the two ends of the wire segment? O 4.90E-06 V

moving at a velocity of 5.80 m/s toward a parallel (much longer) wire which carries a current of 3.51 A. At a particular instant the wire segment is 6.4 cm away from the long, current-carrying wire. What is the induced emf (voltage) between the two ends of the wire segment? O 4.90E-06 V

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Answer :(a) Magnetic Field.T: Explanation and equation: ...... (b) CurrentAmp; Explanation and equation:.