# 18 - Kapoor(mk9499 oldhomework 18 Turner(60230 This...

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

Kapoor (mk9499) – oldhomework 18 – Turner – (60230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points What is the maximum torque on a 450-turn circular coil oF radius 0 . 85 cm that carries a current oF 1 mA and resides in a uniForm magnetic feld oF 0 . 2 T? Correct answer: 2 . 04282 × 10 5 N · m. Explanation: Let : N = 450 , B = 0 . 2 T , r = 0 . 85 cm = 0 . 0085 m , and I = 1 mA = 0 . 001 A . μ = N I A = N I π r 2 τ max = μ B = N I π r 2 B = (450) (0 . 001 A) π × (0 . 0085 m) 2 (0 . 2 T) = 2 . 04282 × 10 5 N · m . 002 (part 1 oF 3) 10.0 points A particle with charge q and mass m is un- dergoing circular motion with speed v . At t = 0, the particle is moving along the nega- tive x axis in the plane perpendicular to the magnetic feld V B , which points in the positive z direction (up From the plane oF the fgure). x y z Vv V B ±ind the magnitude oF the centripetal ac- celeration ( b Va b ≡ a ). 1. a = q v B m correct 2. a = q B m v 3. a = q m v B 4. a = v B q m 5. a = q v B m 6. a = q v m B Explanation: Basic Concepts: Newton’s 2nd Law: F = m a . Centripetal acceleration: F c = m v 2 r ±orce on charge q in magnetic feld (no electric feld): V F B = qVv × V B . Solution: The trajectory oF the particle will be bent into a circle by the magnetic feld. ±rom this we understand there has to be a centripetal acceleration a to keep the particle in this circle (just like a string is needed to provide the tension to keep a ball whirling in a circle). Since F = m a and F B = q v B , the particle moves perpendicularly to the mag- netic feld, so that m a = q v B , a = q v B m . 003 (part 2 oF 3) 10.0 points ±ind the period T oF oscillation; i.e. , the time it takes For the particle to complete one revo- lution. 1. T = q B m 2. T = 2 π m q B correct 3. T = 2 π m B q

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

View Full Document
Kapoor (mk9499) – oldhomework 18 – Turner – (60230) 2 4. T = 2 π q B m 5. T = m q B 6. T = m B q Explanation: The centripetal force F c is provided by F B , so m v 2 r = q v B , v r = q B m . the period of oscillation is T = 2 π r v , which is intuitive since the particle traverses a distance 2 π r in a revolution and, moving at speed v , takes the time T to do so. We know the ratio v r , so T = 2 π r v = 2 π m q B . 004 (part 3 of 3) 10.0 points Find the direction of the instantaneous accel- eration h a at t = 0 if q is negative. 1. h a = ˆ j correct 2. h a = ˆ k 3. h a = ˆ i 4. h a = ˆ k 5. h a = ˆ i 6. h a = ˆ j Explanation: The particle is moving along the negative x -axis in this instant: Vv = v ( ˆ i ) ; since it is moving in a circle, we need to talk about instantaneous direction.
This is the end of the preview. Sign up to access the rest of the 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.

### Page1 / 8

18 - Kapoor(mk9499 oldhomework 18 Turner(60230 This...

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

View Full Document
Ask a homework question - tutors are online
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}