# 22 - Kapoor(mk9499 – homework22 – Turner –(60230 1 This print-out should have 11 questions Multiple-choice questions may continue on the next

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Kapoor (mk9499) – homework22 – Turner – (60230) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An infinite sheet lying in yz plane is carrying a current which is uniform along the y direction, as illustrated below. The current is flowing along the negative z direction. The permeability of free space is 1 . 25664 × 10 − 6 N / A 2 . b a x y A C D F E z What is the direction of magnetic field vector B at point A, which is located on the negative x-axis ( i.e. , x < 0)? Caution: See the sketch to confirm the location of the point A. 1. hatwide B = ( ˆ i − ˆ j ) √ 2 2. hatwide B = − ( ˆ i − ˆ j ) √ 2 3. hatwide B = ( ˆ i + ˆ j ) √ 2 4. hatwide B = − ˆ k 5. hatwide B = − ˆ ı 6. hatwide B = − ˆ 7. hatwide B = ˆ correct 8. hatwide B = − ( ˆ i + ˆ j ) √ 2 9. hatwide B = ˆ ı 10. hatwide B = ˆ k Explanation: Basic Concepts: Right-hand rule for de- termining the direction of vector B . Amp´ ere’s Law: contintegraldisplay vector B · vector ds = μ I . Solution: Consider that the current sheet consists of infinite number of current wires connected in parallel. We can easily deter- mine the direction of vector B for one wire with right-hand rule. Then the direction of vector B for two wires set together in parallel can be found (as illustrated below) by superposition prin- ciple. By symmetry, the direction of vector B due to an infinite current sheet is therefore along positive y direction (ˆ ) in the part of space where x < 0. 002 (part 2 of 3) 10.0 points Ampere’s Law states that contintegraldisplay vector B · dvectors = μ I Denote the magnetic field at A by B A . Eval- uate the loop integral along the rectangular loop CDEFC , where EF = a and FC = b . The magnitude of this loop integral is given by 1. contintegraldisplay vector B · dvectors = a + b 2 B A . 2. contintegraldisplay vector B · dvectors = b − a 2 B A . 3. contintegraldisplay vector B · dvectors = radicalbig a 2 + b 2 B A . 4. contintegraldisplay vector B · dvectors = 2 a B A ....
View Full Document

## This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

### Page1 / 5

22 - Kapoor(mk9499 – homework22 – Turner –(60230 1 This print-out should have 11 questions Multiple-choice questions may continue on the next

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online