Ch23_h1-solutions - pan (zp695) Ch23 h1 chiu (56565) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: pan (zp695) Ch23 h1 chiu (56565) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A magnetic field near the floor points up and is increasing. Looking down at the floor, does the non-Coulomb electric field curl clockwise or counterclockwise? 1. Clockwise correct 2. Not enough information 3. Counterclockwise Explanation: From the Maxwells equation, we see that change in the magnetic field d B dt in this case is down so the electric field E NC curls clockwise. 002 10.0 points A magnetic field near the ceiling points down and is decreasing. Looking up at the ceil- ing, does the non-Coulomb electric field curl clockwise or counterclockwise? 1. Counterclockwise correct 2. Clockwise 3. Not enough information Explanation: From the Maxwells equation, we see that change in the magnetic field d B dt in this case is toward the floor so the electric field E NC curls counterclockwise. 003 (part 1 of 2) 10.0 points On a circular path of radius 11 cm in air around a solenoid with increasing magnetic field, the emf is 35 V. What is the magnitude of the non-Coulomb electric field on this path? Correct answer: 50 . 6402 V / m. Explanation: Here we will use the definition of emf writ- ten in terms of the electric field. Since the value of potential is given, we can find out the magnitude of the electric field. We begin by writing down the equation for the emf: emf = C E NC d = E NC C d = 2 r E NC = 35 V . So, rearranging this expression, we obtain E NC = 35 V 2 r = 35 V 2 (11 cm) = 50 . 6402 V / m . 004 (part 2 of 2) 10.0 points A wire with resistance 5 is placed along the path. What is the current in the wire? Correct answer: 7 A. Explanation: Now we use Ohms Law to figure out the current flow: I = V R = emf R = 35 V 5 = 7 A . 005 10.0 points A uniform magnetic field of 3 T points 29 degrees away from the perpendicular to the plane of a rectangular loop of wire 0 . 2 m by . 4 m as in the figure below. n 29 pan (zp695) Ch23 h1 chiu (56565) 2 What is the magnetic flux on this loop? Correct answer: 0 . 209909 T m 2 . Explanation: We can write mag = S B ndA = S B cos 29 dA = B cos 29 S dA = B (cos 29 ) A = (3 T)(cos29 )(0 . 2 m)(0 . 4 m) = . 209909 T m 2 . 006 (part 1 of 6) 10.0 points A conventional current I runs through a coil in the direction shown in the figure below. A single loop of copper wire is near the coil. The loop and the coil are stationary. The point P is located on the- x axis inside the copper loop. Power Supply +- y x z Loop P Coil I In this initial state (constant current in coil), what is the direction of the magnetic field at the center of the copper loop, due to the current in the coil?...
View Full Document

Page1 / 8

Ch23_h1-solutions - pan (zp695) Ch23 h1 chiu (56565) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online