### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# GeneralPhysicsII PHYS104

Drexel

GPA 3.54

### View Full Document

## 30

## 0

## Popular in Course

## Popular in Physics 2

This 22 page Class Notes was uploaded by Vernice Schuster on Wednesday September 23, 2015. The Class Notes belongs to PHYS104 at Drexel University taught by GuoliangYang in Fall. Since its upload, it has received 30 views. For similar materials see /class/212517/phys104-drexel-university in Physics 2 at Drexel University.

## Popular in Physics 2

## Reviews for GeneralPhysicsII

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/23/15

Chapter 23 Mirrors and Lenses Our eyes can see objects of certain sizes at certain distances and at certain locations To overcome these limitations optical instruments have been developed Many of these instruments are based on mirrors and lenses When we look at an object via a mirror or a lens we look at the IMAGE of the object Images can be larger than smaller than or same as the object Images can have the same orientation or the inverse orientation as the object Images can be real or virtual 1 Flat Mirrors When an object is placed in front of a at mirror an image of the object is formed39 The image is as far behind as the object is in 39ont image dirtance q object distance 9 The image is unmagni ed The image is virtual light does not pass through the image point light appears to diverge from that point The image is upright having the same orientation as the object There is an apparent Ie right reversal How to find the location of the image using light rays Steps to determine the image Draw a special ray from the object to the mirror Draw the re ected ray according to the law of re ection Draw a second ray from the object to the mirror Draw the re ected ray Extend both re ected rays in the reverse direction The intersection is the location ofthe image it 1s always necessary to follow at least two rays of light as they re ect from the mirror Example In order to see the full height of yourself how large a at l mirror will you need and how will you hang the mirror Question How does the rearview mirror work for the Day and Night settings gas L mirror 2 Images formed by Spherical Mirrors Muni A spherical mirrorhas the shape of a quot segment of a sphere mm 4 meuu A concave spherical mirrorhas the silvered surface on the inner concave side ofthe curve A convex spherical mirror has the silvered surface on the outer convex side ofthe curve A concave mirror Image formed by a concave mirror Hmm Steps to determine the image location Draw two special rays from the object to the mirror Draw the re ected rays according to the law of re ection The intersection is the location ofthe image Mirror Equation x The magni cation is i r 7 The relationship between q and is The mm39or equation When the object is very far away the incident rays become parallel This point is called the focal point F and the distance qR2 is called the focal length f m w 3 Convex Mirrors and Sign Conventions A convex mirror is a diverging mirror um The focal point Fand the center of H quotquott the back side ofthe mirror The image is a Virtual image v liq 7 The Mirror Equation works for both concave and convex mirrors by following the sign convention rimquot 0 31 TABLE 231 ml mit rlu 5 ltimm mnrlinugzitiw 4 lmulim liglu lump w 1 i i m i mm lt tum hum w v NcllcuLrl liglil MWquot M u 7 Inlniw m mm m um Hll c How to determine the direction ofthe magnetic force F qVBSlne Using the right hand rule 1 r Li B Right hand rule 1 u 1 Point ngers in the direction of v For the negative 2 Curl ngers in the direction of B charge the gtl direction is the 3 Thumb is pointing to F opposite Magnetic force on a moving charge Special cases Magnitude gt F qusine e 0 gt F 0 Direction gt Rihthandml 6900 gt Fmaximum g e 91800 gt Fo Example A proton moves at 80106 ms along the Xaxis The magnetic field is 25 T and directed at 60 with the x direction What is the force and acceleration ofthe proton 4 Magnetic Force on a CurrentCarrying V re A magnetic field exerts a force on a currentcarrying wire due to the moving charges inside the wire rm 9 Faramtxirpamcles l v aivdenAo 939 l 7 r I We know that d 7 quotIA This is the maximum force with the wire L B Direction ofF can be found using the right hand rule Therefore 1 Point ngers in the direction of 2 Curl ngers in the direction of B If B is nntperpendicular to I the force becomes u 3 Thumb is pointing to F FMS BIKsme Question which way will the Wire de ect if i N V a current is passed throu Example A wire carrying 22 A of current runs from east to west The Earth s magnetic field is horizontal from south to north with a magnitude of 05x10394 T a What is the force on a 36m length of the wire 3 How does the force change if the current is from west to east North in Wesl East xxxxxx W xxxxxx yv yyyyyvvvvvyxvyv Scum oul 5 Torque on a Current Loop a What is a torque e To move an object gt requires a force 0 To rotate an object gt requires a torque Fsin 4 F F777 Torque distanceJ component of the force r i r u E FL r F rsin an d Flcosd b What happens when a current loop is in a magnetic field Top View Side View F F l 5 a gsine x Y39 Pal 1 F2 quot 39 1 F1a2F2a2 BIba2BIba2 t F am h all B I A sin 9 BIab BIA For a loop of arbitrary shape in a magnetic field I 1 BIA sin 9 B When there are N turns in the loop n 1 N BIA sin 9 9 1 is not constant rm B I A liminl 0 Electric motors are based on this principle 6 Motion of a Charged Particle in a Magnetic Field a What kind of path does a charged particle follow in a magnetic field When a charged particle enters a magnetic eld x 3915 x x x x x x x The force on the char e is F B x x X X X X 9 V x x x R x39 x 1 an x F Is always perpendIcularto V x x x x x x X x x 4 x x What type ofpath willthe particlefollow f i i Q j 1 f f A Circular path Centripetal force F qu Centripetal acceleration a vZIr q m vzlr Thus the radius of the orbit is X Question How will the kinetic energy ofthe particle change b What happens if v is not perpendicular to By Helical I pain I The path is a helix Example The mass spectrometer to separate or detect charged particles by their mass Two particles each with charge e m1 167X1027 kg m 334X1027 kg are both accelerated to V 10X1U3 ms B 010 T How far are they separated when reaching detector P 5 Generators To convert mechanical energy to electrical energy a What is the working principle of an electric generator A Wire 100p rotated by some external means falling water heat by burning coal to produce steam etc in a magnetic eld AG ltDBAcosmt and 5 ENE ZNBAW WW b ac generators and dc generators dc generator ac generator gt Slip lino iquot 1 a 39r 39 J ll r Brushes t quotI I r l 7 l W To produce a steady current many loops and commutators are used 6 SelfInductance How does the current change in a circuit affect itself Will the current reach s R E 3 immediame assisclnsed No x if g I I j a N 1 5 312 The circuit s opposition to sudden current change in itself is called self induction and the induced emfis called the selfinduced em The selfinduced emf is L gt inductance of the loop The unit ofL is the henry H 1 H 1 V sA The inductance L can be calculated using Faraday s Law The magnetic eld inside B uonl u0NlI The magnetic ux CD BA u0NDAI So 2 LNNONIIAI 01 A L does not depend on I only on geometric parameters 8 Energy Stored in a Magnetic Field For a device inductor with an inductance L and a current I the energy stored is Chapter 22 Reflection and Refraction of Light Light is a electromagnetic wave but it also behaves like a beam of particles photons under certain circumstances gt WaveParticle Duality of Light 1 The Speed oflight a How fast does light travel In vacuum approximately in air the speed of light is c30X108ms which is equivalent to 300000 krns or 671224362 mph b How does light travel Light travels along a straight line until it encounters a boundary 2 Reflection and Refraction a Wnat happens when a light beam encounters a boundary A light beam is re ected at a boundary a a A a 45 quot m Specular Re ection Diffuse Re ection b In which direction does the reflected beam go Navmm Normal a line perpendicular to the surface 9 angle of incidence 9391 angle of reflection Law of reflection The incident ray and the re ected ray are in the same plan e39 91 Question In which direction will the re ected light beam go gt c What happens when a light beam goes through a boundary Incident Normal Re ected my my Light ray changes A direction at the n quotquot boundary 7 l Part of the beam is 1 reflected and the other i 5 part is refracted 206 Re 92 angle of refraction Light beam refracts because it has different speeds in different media Min391 The angle of refraction is related to the quoth quotu speeds via I quot 5 11162 v2 33 397 7 sm 61 v1 n v1 speed of light in medium 1 air v2 speed of light in medium 2 glass 1 b Light path through a boundary is reversible Normal Normal ii gt12 a lt 02 H 112 Air Air Glass Grass i 3 J in i 3 The Law of Refraction a What is the index of refraction We define the index of refraction n Speed of light in vacuum 0 Speed of light in the medium v n is an optical property of the material n gt1 for all materials For air n a 1 Chapter 18 Direct Current Circuits Assignments for the week Reading 181 184 Recitation problems Ch18 CO 2 5 10 13 2122 P2 6 81112151620 Homework problems Due Monday 021 12008 Ch18 P14 26 Dlrectanent gtdirection and magnitude of the current do not change Current gt both direction and magnitude of the current change with time 1 Sources of emf To keep a current flow a source is required to make the charges circulating in the I a circuit Such a device is called a source of em 1 I 4 HI I39 39 39 force 39 39r l Examples battery generator solar cell a How much voltage can we get from a battery A real battery has internal resistance Thus the terminal voltage is not equal to the em but AVah 6 II E gt engfofthe battery I gt internal resistance Terminal voltage a gt emf r gt internal resistance Rngt load resistance Active gurelS 1 In an open circuit 1 0 8 AVab opencircuit voltage b How much current can we get out of a battery From circuit diagram we know 8 eAvAvR1r1R gt I R r The current output of a battery depends on both the load resistance R and the internal resistance r Example A battery with 9 90 V delivers i 17 mA when connected to load resistor of 72 Q What is the internal resistance c How much power can we get out of a battery The total power output of the source 15 is dissipated by both the load resistor 12R and the internal resistor Pr Is 12R Zr For a good battery r is very small and negligible However as the battery is getting old r becomes signi cant and the power output is reduced Combination of Resistors Varinlls types armismrs 2 Resistors in Series 4 Resistors connected in series 1 1 gt 16quot Current T1 IZ I Voltage AV AV1 AVZ W R Currem 1 11 12 From Ohm s IaW39 Voltage AV1 AV2 AV gt 1 1112 Reg R1 R2 For two resistors connected in parallel For any number of resistors connected in parallel Chapter 18 Direct Current Clrcults Cont Question ifthe iightbuibs are identical which one is the brightest What if quotAquot burns out What if C burns out Question How does a 3way iightbuib work How are Christmas tree lights connected WW 120v zdmm mnw Munm WW unv I anSW yummvsw quotl l 15quot 3011 5311 90in 77 b Series connection Req gt R i 1 2 3 Parallel connection Req lt R i 1 2 3 4 Kirchhoff Rules ln a complex circuit where a simple equivalent 39 how can quot the parameters on the resistors such as l P DV Gustav Kirchhoff 1824 1887 Kirchhoffs Rules total current leavmg that Jun closet1100p must be zero 1 At any junction the39 total current entering the junction must equal the 39 ction 2 The sum of the potential differences across all the elements around any Junction rule At anyjuncfion the total current enteringthejunction must 4 V I1 IZ 13 The direction of a current is assumed at the beginning The results will tell if the assumption is correct 4M 24111 rule The sum ofthe potential differences across all the elements around any closed loop must be zero Ifthe loop direction is determined according to these diagrams lExample Find the current passing through each resistor First we assign symbols to the currents to be found and assume a direction for each current Second use thejunction rule l 2 ll 391 Third use the loop rule LoopA HOV6E2 l 29 13 l2 Lam LoopB 4 212 lav69 l l4V0 3 Fourth solve the equations simultaneously Example Find the current passing through each resistor using Kirchhoff rules f 5 m 39

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I made $350 in just two days after posting my first study guide."

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.