midterm 03-solutions

midterm 03-solutions - Version 085 – midterm 03 – chiu...

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: Version 085 – midterm 03 – chiu – (56565) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A wire loop of radius R carrying a counter- clockwise (when viewed from the right) cur- rent I is moving to the right along the x-axis at a speed v . It passes around a second, sta- tionary wire loop of radius a where a ≪ R . Choose the plot that correctly displays the qualitative behavior of the induced current I ( t ) in the stationary loop. On the plots, the + I axis represents clockwise current, while the − I axis represents counter-clockwise cur- rent (when viewed from the right on the + x- axis). 1. VIII : 2. I : correct 3. V : 4. II : 5. IV : 6. VI : 7. III : Version 085 – midterm 03 – chiu – (56565) 2 8. VII : Explanation: I is the correct choice. Before the current loop passes the wire ring, the flux through the ring points to the right and is increas- ing; to resist this change in flux, a clockwise (positive) current is induced in the ring. Af- ter the loop passes the ring, the flux still points to the right but is decreasing , induc- ing a counter-clockwise (negative) current in the ring. To get a sense of the functional form of the change in flux, note that we are entitled to use B loop ∝ 1 ( x 2 + R 2 ) 3 / 2 since a ≪ R . (Since we are only inter- ested in the qualitative behavior, we can leave off the constant factors and concern our- selves only with proportionality.) Since the area of the ring is not changing, d Φ B /dt = A ring dB loop /dt . Again ignoring the constant factor A loop , we calculate dB loop /dt , dB loop dt ∝ d dt parenleftbigg 1 ( x 2 + R 2 ) 3 / 2 parenrightbigg dB loop dt ∝ − 3 2 2 x ( x 2 + R 2 ) 5 / 2 dx dt dB loop dt ∝ − 3 xv ( x 2 + R 2 ) 5 / 2 Examining this result, we see that the pres- ence of x in the numerator means our plot must pass through 0, while to either side of t = 0 we should expect some polynomial- like behavior. Combined with what we know about the sign of the current for times t < and t > 0, I is the only correct option. 002 10.0 points A very long thick wire of radius R carries a current I , as in the following figure. I 1 R Find the magnitude of the magnetic field inside the wire, a distance r from the center of the wire, where r < R . Assume current density is the same at every point inside the wire. 1. μ 2 π IR 2 r 3 2. μ 2 π IR r 2 3. μ 4 π Ir 2 R 3 4. μ 2 π Ir 2 R 5. μ 2 π Ir R 2 correct 6. μ 2 π Ir 2 R 3 7. μ 4 π IR r 2 8. μ 4 π Ir 2 R 9. μ 4 π Ir R 2 10. μ 4 π IR 2 r 3 Explanation: The current density (current/area) is the same throughout the wire. Sketch an Ampe- rian loop within the wire with radius r . Then, the current through this loop i divided by area of this loop is equal to the total current per unit area....
View Full Document

This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

Page1 / 13

midterm 03-solutions - Version 085 – midterm 03 – chiu...

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