Get Solution of These Packages & Learn by Video Tutorials on EXERCISE-1

Similar documents
200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes

PHYSICS ASSIGNMENT-9

- 5 - TEST 2. This test is on the final sections of this session's syllabus and. should be attempted by all students.

DIRECT CURRENT CIRCUITS

Exam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B

Problem Solving 7: Faraday s Law Solution

Version 001 HW#6 - Electromagnetism arts (00224) 1

Sample Exam 5 - Skip Problems 1-3

Physics 2135 Exam 3 April 21, 2015

Version 001 HW#6 - Electromagnetic Induction arts (00224) 1 3 T

13.4 Work done by Constant Forces

Unique Solutions R. All about Electromagnetism. C h a p t e r. G l a n c e

Physics 24 Exam 1 February 18, 2014

EMF Notes 9; Electromagnetic Induction ELECTROMAGNETIC INDUCTION

Physics 2135 Exam 1 February 14, 2017

Physics Honors. Final Exam Review Free Response Problems

Time : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A

PhysicsAndMathsTutor.com

Physics 1402: Lecture 7 Today s Agenda

KINEMATICS OF RIGID BODIES

SECTION B Circular Motion

Version 001 Exam 1 shih (57480) 1

west (mrw3223) HW 24 lyle (16001) 1

ME 141. Lecture 10: Kinetics of particles: Newton s 2 nd Law

ragsdale (zdr82) HW2 ditmire (58335) 1

#6A&B Magnetic Field Mapping

The momentum of a body of constant mass m moving with velocity u is, by definition, equal to the product of mass and velocity, that is

Prof. Anchordoqui. Problems set # 4 Physics 169 March 3, 2015

Summary of equations chapters 7. To make current flow you have to push on the charges. For most materials:

Mathematics Extension 1

Instructor(s): Acosta/Woodard PHYSICS DEPARTMENT PHY 2049, Fall 2015 Midterm 1 September 29, 2015

IMPORTANT. Read these directions carefully:

Physics Graduate Prelim exam

A wire. 100 kg. Fig. 1.1

Physics 121 Sample Common Exam 1 NOTE: ANSWERS ARE ON PAGE 8. Instructions:

15 - TRIGONOMETRY Page 1 ( Answers at the end of all questions )

AP Physics C: Electricity & Magnetism 1999 Free-Response Questions

CAPACITORS AND DIELECTRICS

Homework Assignment 9 Solution Set

Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016

Electricity and Magnetism

SOLUTIONS TO CONCEPTS CHAPTER 10

Math 8 Winter 2015 Applications of Integration

5.2 Volumes: Disks and Washers

Name Class Date. Match each phrase with the correct term or terms. Terms may be used more than once.

16 Newton s Laws #3: Components, Friction, Ramps, Pulleys, and Strings

F is on a moving charged particle. F = 0, if B v. (sin " = 0)

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 16 CHAPTER 16

Physics 212. Faraday s Law

Mathematics Extension Two

A formula sheet and table of physical constants is attached to this paper. Linear graph paper is available.

Reference. Vector Analysis Chapter 2

Mathematics Extension 2

Applications of Bernoulli s theorem. Lecture - 7

1 Which of the following summarises the change in wave characteristics on going from infra-red to ultraviolet in the electromagnetic spectrum?

CONIC SECTIONS. Chapter 11

THREE-DIMENSIONAL KINEMATICS OF RIGID BODIES

Candidates must show on each answer book the type of calculator used.

HIGHER SCHOOL CERTIFICATE EXAMINATION MATHEMATICS 4 UNIT (ADDITIONAL) Time allowed Three hours (Plus 5 minutes reading time)

Mathematics Extension 2

Electromagnetism Answers to Problem Set 10 Spring 2006

C D o F. 30 o F. Wall String. 53 o. F y A B C D E. m 2. m 1. m a. v Merry-go round. Phy 231 Sp 03 Homework #8 Page 1 of 4

NOT TO SCALE. We can make use of the small angle approximations: if θ á 1 (and is expressed in RADIANS), then

Version 001 HW#6 - Circular & Rotational Motion arts (00223) 1

, MATHS H.O.D.: SUHAG R.KARIYA, BHOPAL, CONIC SECTION PART 8 OF

R(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of

Study Guide Final Exam. Part A: Kinetic Theory, First Law of Thermodynamics, Heat Engines

In-Class Problems 2 and 3: Projectile Motion Solutions. In-Class Problem 2: Throwing a Stone Down a Hill

AP Physics 1. Slide 1 / 71. Slide 2 / 71. Slide 3 / 71. Circular Motion. Topics of Uniform Circular Motion (UCM)

First Semester Review Calculus BC

Method of Localisation and Controlled Ejection of Swarms of Likely Charged Particles

4-6 ROTATIONAL MOTION

Reading from Young & Freedman: For this topic, read the introduction to chapter 24 and sections 24.1 to 24.5.

Lecture 13 - Linking E, ϕ, and ρ

Physics 9 Fall 2011 Homework 2 - Solutions Friday September 2, 2011

7.6 The Use of Definite Integrals in Physics and Engineering

2. VECTORS AND MATRICES IN 3 DIMENSIONS

Potential Formulation Lunch with UCR Engr 12:20 1:00

Year 12 Mathematics Extension 2 HSC Trial Examination 2014

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

SOLUTIONS TO CONCEPTS CHAPTER

ELE B7 Power Systems Engineering. Power System Components Modeling

Problem Set 4: Mostly Magnetic

PHYSICS 211 MIDTERM I 21 April 2004

Hints for Exercise 1 on: Current and Resistance

Mathematics Extension 2

P 1 (x 1, y 1 ) is given by,.

l 2 p2 n 4n 2, the total surface area of the

Math 0230 Calculus 2 Lectures

Physics Jonathan Dowling. Lecture 9 FIRST MIDTERM REVIEW

Mathematics. Area under Curve.

DO NOT OPEN THIS EXAM BOOKLET UNTIL INSTRUCTED TO DO SO.

JURONG JUNIOR COLLEGE

2/20/ :21 AM. Chapter 11. Kinematics of Particles. Mohammad Suliman Abuhaiba,Ph.D., P.E.

Dynamics: Newton s Laws of Motion

Physics 202, Lecture 10. Basic Circuit Components

pivot F 2 F 3 F 1 AP Physics 1 Practice Exam #3 (2/11/16)

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

Due to both the rings the magnetic field will be in same direction B = 0 i

Transcription:

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com EXECISE- SECTION : FUX AND FAADAY S AWS OF EECTOMA OMAGNETIC INDUCTION A. A. A 3*. A 4. A 5. A 6. A 7. Consider the sitution shown in fig. The resistnceless wire A is slid on the fied rils with constnt velocity. If the wire A is replced by resistnceless semicirculr wire, the mgnitude of the induced current will increse () remin the sme (C) decrese (D) increse or decrese depending on whether the semicircle bulges towrds the resistnce or wy from it. A conducting squre loop of side l nd resistnce moves in its plne with uniform velocity v perpendiculr to one of its sides. A uniform nd constnt mgnetic field eists long the perpendiculr to the plne of the loop in fig. The current induced in the loop is lυ/ clockwise (D) zero () lυ/ nticlockwise (C) lυ/ nticlockwise A conducting loop is plced in uniform mgnetic filed with its plne perpendiculr to the field. An emf is induced in the loop if it is trnslted () it is rotted bout its is (C) it is rotted bout dimeter (D) it is deformed Some mgnetic flu is chnged from coil of resistnce 0 ohm. As result n induced current is developed in it, which vries with time s shown in figure. The mgnitude of chnge in flu through the coil in Webers is () 4 (C) 6 (D) 8 Consider the sitution shown in fig. If the switch is closed nd fter some time it is opened gin, the closed loop will show n nticlockwise current-pulse () clockwise current-pulse (C) n nticlockwise current-pulse nd then clockwise current-pulse (D) clockwise current-pulse nd then n nticlockwise current-pulse Solve the previous question if the closed loop is completely enclosed in the circuit contining the switch. n nticlockwise current-pulse () clockwise current-pulse (C) n nticlockwise current-pulse nd then clockwise current-pulse (D) clockwise current-pulse nd then n nticlockwise current-pulse Consider the following sttements: An emf cn be induced by moving conductor in mgnetic field () An emf cn be induced by chnging the mgnetic field. oth A nd re true () A is true but is flse (C) is true but A is flse (D) oth A nd re flse i (mp) A 8. A smll, conducting circulr loop is plced inside long solenoid crrying current. The plne of the loop contins the is of the solenoid. If the current in the solenoid is vried, the current induced in the loop is clockwise () nticlockwise (C) zero (D) clockwise or nticlockwise depending on whether the resistnce in incresed or decresed. A 9. A semicirculr conducting wire is plced in yz plne in uniform mgnetic field directed long positive z- direction. An induced emf will be developed between the ends of the wire if it is moved long: positive direction () positive y direction (C) positive z direction (D) none of these Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't. 4 0. t(s) Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 0

z FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com A 0. A. A. A 3. A 4. A 5. A 6. A 7. A 8. A 9. A 0. If flu in coil chnges by φ, nd the resistnce of the coil is, prove φ tht the chrge flown in the coil during the flu chnge is. (Note : It is independent of the time tken for the chnge.) The flu of mgnetic field through closed conducting loop of resistnce 0.4 Ω chnges with time ccording to the eqution Φ = 0.0t + 0.40t + 0.60 where t is time in seconds. Find (i) the induced emf t t = s. (ii) the verge induced emf in t = 0 to t = 5 s. (iii) chrge pssed through the loop in t = 0 to t = 5s (iv) verge current in time intervl t = 0 to t = 5 s (v) het produced in t = 0 to t = 5s. A wire - loop confined in plne is rotted in its own plne with some ngulr velocity. A uniform mgnetic field eist in the region. Find the emf induced in the loop A closed circulr loop of 00 turns of men dimeter 50 cm & hving totl resistnce of 0 Ω is plced with its plne t right ngles to mgnetic field strength 0 Tesl. Clculte the quntity of electric chrge pssing through it when the coil is turned through 80º bout n is in its plne. A solenoid hs cross sectionl re of 6.00 4 m, consists of 400 turns per meter, nd crries current of 0.40 A. A 0 turn coil is wrpped tightly round the circumference of the solenoid. The ends of the coil re connected to.5 Ω resistor. Suddenly, switch is opened, nd the current in the solenoid dies to zero in time 0.050 s. Find the verge current in the coil. A hert pcing device consists of coil of 50 turns & rdius mm just inside the body with coil of 000 turns & rdius cm plced concentriclly just outside the body. Clculte the induced EMF in the internl coil, if current of A in the eternl coil collpses in 0 milliseconds. Figure illustrtes plne figures mde of thin conductors which re locted in uniform mgnetic field directed wy from reder beyond the plne of the drwing. The mgnetic induction strts diminishing. Find how the currents induced in these loops re directed. () (b) (c) (d) An infinitely long stright conductor lies in the plne of squre loop with n ohmic resistnce & side length t distnce r 0 prllel to one of the loop's sides. The current flowing in the conductor vries ccording to the lw i = αt 3 where α is constnt. Find the current in the loop t time t. Prove tht the emf induced in wire of ny shpe in uniform mgnetic field hs mgnitude V. ( ) where V is the velocity of the wire nd is the position vector of one end of the wire reltive to the other end. The mgnetic field in the cylindricl region shown in figure increses t constnt rte of 0.0 m T/s. Ech side of the squre loop cid nd def hs length of.00 cm nd resistnce of 4.00 Ω Find the current (mgnitude nd sense ) in the wire d if () the switch S is closed but S is open (b) S is open but S is closed (c) both S nd S re open nd (d) both S nd S re closed. A circulr loop of rdius r is fied to rottion is long the z direction, so tht the plne of the loop lwys perpendiculr to the y plne. The loop is rotting in nticlockwise sense with ngulr velocity ω, t t = 0 loop lies in yz plne. Given uniform & constnt eternlly pplied mgnetic field. i r 0 y Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com A. = ( cosα î + sinα kˆ ) in the loop.. Evlute the mgnetic flu φ(t) through the loop & EMF ε(t) induced A uniform mgnetic field eist in cylindricl region or rdius 0cm s shown in figure. A uniform wire of length 80 cm nd resistnce 4.0 Ω is bent into squre frme nd is plced with one side long dimeters of the cylindricl region. If the mgnetic field increses t constnt rte of 0.00 T/s find the current induced in the frme. A 3. A constnt current Ι is flowing in circulr ring of rdius. A second ring whose rdius r is much smller thn tht of the first, is moving with constnt velocity V long the is in such mnner tht the plne of the ring remins prllel to the plne of the ring during the course of the motion. Find the mimum EMF induced in the ring. A 4. A plne spirl with gret number N of turns wound tightly to one nother is locted in uniform mgnetic field perpendiculr to the spirl s plne. The outside rdius of the spirl s turns is equl to. The mgnetic induction vries with time s = 0 sin ωt, where 0 nd ω re constnts. Find the mplitude of emf induced in the spirl. A 5. In the figure, long thin wire crrying vrying current i = i 0 sin ωt lies t distnce y bove one edge of rectngulr wire loop of length nd width W lying in the - z plne. Wht emf is induced in the loop. SECTION () : EMF IN A MOVING OD.. 3. 4 5 A conducting rod is moved with constnt velocity υ in mgnetic field. A potentil difference ppers cross the two ends if υ l () if υ (C) if l (D) none of these A conducting rod A of length l = m is moving t velocity v = 4 m/s mking n ngle 30º with its length. A uniform mgnetic field = T eists in direction perpendiculr to the plne of motion. Then V A V = 8V () V A V = 4V (C) V V A = 8V (D) V V A = 4V v 30º A In the given rrngement, the loop is moved with constnt velocity v in uniform mgnetic field in restricted region of width. The time for which the emf is induced in the circuit is: b ν ( + b) (C) ν () ν ( b) (D) ν A solid conducting sphere of rdius is moved with velocity V in uniform mgnetic field of strength such tht is perpendiculr to V. The mimum e.m.f. induced between two points of the sphere is : V () V (C) V (D) V A verticl rod of length l is moved with constnt velocity v towrds Est. The verticl component of the Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com erth's mgnetic field is nd the ngle of dip is θ. The induced e.m.f. in the rod is: ( A ) l v cot θ () l v sin θ (C) l v tn θ (D) l v cos θ 6 A uniform mgnetic field eists in region given by = 3 î + 4 ĵ + 5kˆ. A rod of length 5 m is plced long y is is moved long is with constnt speed m/sec. Then induced e.m.f. in the rod will be: zero () 5 v (C) 0 v (D) 5 v 7. 8. 9. 0.. Two stright conducting rils form right ngle where their ends re joined. A conducting br in contct with the rils strts from verte t the time t = 0 & moves with constnt velocity of v m/s to the right s shown in figure. A mgnetic field = 0 (Tesl) points out of the pge. Clculte: () The flu through the tringle formed by the rils & br t t = 3.0 s. (b) The EMF round the tringle t tht time. (c) In wht mnner does the EMF round the tringle vry with time. A uniform mgnetic field of induction eists in circulr region of rdius. A loop of rdius encloses the mgnetic field t t = 0 nd then pulled t uniform speed v in the plne of the pper. Find the direction of induced current nd induced EMF in the loop s function of time. A stright wire with resistnce of r per unit length is bent to form n ngle α. A rod of the sme wire perpendiculr to the ngle bisector (of α) forms closed tringulr loop. This loop is plced in uniform mgnetic field of induction. Clculte the current in the loop when the rod moves t constnt speed V. A wire bent s prbol y = k is locted in uniform mgnetic field of induction, the vector being perpendiculr to the plne, y. At the moment t = 0 connector strts sliding trnsltion wise from the prbol pe with constnt ccelertion (figure). Find the emf of electromgnetic induction in the loop thus formed s function of y. A Π-shped conductor is locted in uniform mgnetic field perpendiculr to 0 d the plne of the conductor nd vrying with time t the rte = 0.0 T/s. A conducting connector dt strts moving with n ccelertion w = 0 cm/s long the prllel brs of the conductor. The length of the connector is equl to l = 0 cm. Find the emf induced in the loop t =.0 s fter the beginning of the motion, if t the moment t = 0 the loop re nd the mgnetic induction re equl to zero. The self inductnce of the loop is to be neglected. SECTION (C) : ENZ S AW C. Fig. shown horizontl solenoid connected to bttery nd switch. A copper ring is plce on frictionless trck, the is of the ring being C. C 3. long the is of the solenoid. As the switch is closed, the ring will remin sttioner () move towrds the solenoid (C) move wy from the solenoid (D) move towrds the solenoid or wy from it depending on Two circulr coils A nd re fcing ech other s shown in figure. The current i through A cn be ltered there will be repulsion between A nd if i is incresed () there will be ttrction between A nd if i is incresed (C) there will be neither ttrction nor repulsion when i is chnged (D) ttrction or repulsion between A nd depends on the direction of current. It does not depend whether the current is incresed or decresed. Two identicl conductors P nd Q re plced on two frictionless fied conducting A ~ y 90 + 0 v m/s i Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 3 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com C 4. C 5. C 6. C 7. C 8. C 9. rils nd S in uniform mgnetic field directed into the plne. If P is moved in the direction shown in figure with constnt speed, then rod Q will be ttrcted towrds P () will be repelled wy from P (C) will remin sttionry (D) my be repelled or ttrcted towrds P Two identicl coil circulr loops crry current i ech circulting in the sme direction. If the loops pproch ech other the current in ech loop will decrese () the current in ech loop will increse (C) the current in ech loop will remin the sme (D) the current in one loop will increse nd in the other loop will decrese A squre coil ACDE with its plne verticl is relesed from rest in horizontl uniform mgnetic field of length. The ccelertion of the coil is less thn g for ll the time till the loop crosses the mgnetic field completely () less thn g when it enters the field nd greter thn g when it comes out of the field (C) g ll the time (D) less thn g when it enters nd comes out of the field but equl to g when it is within the field In the figure shown, the mgnet is pushed towrds the fied ring long the is of the ring nd it psses through the ring. when mgnet goes towrds the ring the fce becomes south pole nd the fce A becomes north pole () when mgnet goes wy from the ring the fce becomes north pole nd the fce A becomes south pole (C) when mgnet goes wy from the ring the fce A becomes north pole nd the fce becomes south pole (D) the fce A will lwys be north pole. A metllic ring (non mgnetic) with smll cut is held horizontlly nd mgnet is llowed to fll verticlly through the ring, then the ccelertion of the mgnet is : lwys equl to g () initilly less thn g but greter thn g once it psses through the ring (C) initilly greter thn g but less thn g once it psses through the ring (D) lwys less thn g A nd re two metllic rings plced t opposite sides of n infinitely long stright conducting wire s shown. If current in the wire is slowly decresed, the direction of induced current will be : clockwise in A nd nticlockwise in () nticlockwise in A nd clockwise in (C) clockwise in both A nd (D) nticlockwise in both A & A br mgnet is relesed from rest long the is of very long, verticl copper tube. After some time the mgnet will stop in the tube () will move with lmost constnt speed (C) will move with n ccelertion g (D) will oscillte SECTION (D) : CICUIT POEMS & MECHANICS D D A constnt force F is being pplied on rod of length ' l ' kept t rest on two prllel conducting rils connected t ends by resistnce in uniform mgnetic field s shown. the power delivered by force will be constnt with time () the power delivered by force will be incresing first nd then will decrese (C) the rte of power delivered by the eternl force will be incresing continuously (D) the rte of power delivered by eternl force will be decresing continuously. v ACD is fied conducting smooth ril plced in verticl plne. PQ is conducting rod which is free to slide on the rils. A horizontl uniform mgnetic field eists in spce s shown. If the rod PQ in relesed from rest then, A S C P A N Q D E & A re right & left fces respectively S Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 4 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com D-3. D-4. D 5. D 6. D 7. D 8. D 9. D 0. The rod PQ will move downwrd with constnt ccelertion () The rod PQ will move upwrd with constnt ccelertion (C) The rod will move downwrd with decresing ccelertion nd finlly cqcure constnt velocity (D) either A or. A rectngulr frme of wire bcd hs dimensions 3 cm 8.0 cm nd totl resistnce of.0 Ω. It is pulled out of mgnetic field = 0.0 T by pplying force of 3. 0 5 N (figure). It is found tht the frme moves with constnt speed. Find () this constnt speed, (b) the emf induced in the loop, (c) the potentil difference between the points nd b nd (d) the potentil difference between the points c nd d. A rectngulr loop with sliding connector of length l is locted in uniform mgnetic field perpendiculr to the loop plne (figure). The mgnetic induction is equl to. The connector hs n electric resistnce, the sides A nd CD hve resistnces nd respectively. Neglecting the self-inductnce of the loop, find the current flowing in the connector during its motion with constnt velocity v. Consider the sitution shown in figure. The wires P Q nd P Q re mde to slide on the rils with the sme speed 5 cm/s. Find the electric current in the 9 Ω resistor if () both the wires move towrds right nd (b) if P Q moves towrds left but P Q moves towrds right. Suppose the 9Ω resistor of the previous problem is disconnected. Find the current through P Q in the two situtions () nd (b) of tht problem. Consider the sitution shown in figure. The wire PQ hs negligible resistnce nd is mde to slide on the three rils with constnt speed of 5 cm/s. Find the current in the 0 Ω resistor when the switch S is thrown to () the middle ril (b) bottom ril. The current genertor Ι g, shown in figure, sends constnt current i through the circuit. The wire cd is fied nd b is mde to slide on the smooth, thick rils with constnt velocity v towrds right. Ech of these wires hs resistnce r. Find the current through the wire cd. A wire of mss m nd length l cn slide freely on pir of smooth, verticl rils (figure). A mgnetic field eists in the region in the direction perpendiculr to the plne of the rils. The rils re connected t the top end by n initilly unchrged cpcitor of cpcitnce C. Find the ccelertion of the wire neglecting ny electric resistnce. Figure shows smooth pir of thick metllic rils connected cross bttery of emf ε hving negligible internl resistnce. A wire b of length l nd resistnce r cn slide smoothly on the rils. The entire system lies in horizontl plne nd is immersed in uniform verticl mgnetic field. At n instnt t, the wire is given smll velocity ν towrds right. () Find the current in the wire t this instnt. (b) Wht is the force cting on the wire t this instnt. Show tht fter some time the wire b will slide with constnt velocity. Find this velocity. A 4 cm cm cm P Q ε Ω P P Ω V Q Q =.0 T i Ι g d c 0 Ω D C 9 Ω S =.0T b v C b Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 5 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com D. D. D 3. D 4. D 5. D 6. D 7. D 8. D 9. Figure shows wire sliding on two prllel, conducting rils plced t seprtion l. A mgnetic field eists in direction perpendiculr to the plne of the rils. Wht force is necessry to keep the wire moving t constnt velocity v? Figure shows long U-shped wire of width l plced in perpendiculr mgnetic field. A wire of length l is slid on the U-shped wire with constnt velocity v towrds right. The resistnce of ll the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shped wire. Drw n equivlent circuit digrm t time t, showing the induced emf s bttery. Clculte the current in the circuit. Consider the sitution of the previous problem. () Clculte the force needed to keep the sliding wire moving with constnt velocity v. (b) If the force needed just fter t = 0 is F 0, find the time t which the force needed will be F 0 /. A wire b of length l, mss m nd resistnce slides on smooth, thick pir of metllic rils joined t the bottom s shown in figure. The plne of the rils mkes n ngle θ with the horizontl. A verticl mgnetic field eists in the region. If the wire slides on the rils t constnt speed v, show tht = mg sinθ vl cos θ In the figure, CDEF is fied conducting smooth frme in verticl plne. A conducting uniform rod GH of mss m cn move verticlly nd smoothly without losing contct with the frme. GH lwys remins horizontl. It is given velocity u upwrds nd relesed. Tking the ccelertion due to grvity s g nd ssuming tht no resistnce is present other thn. Find out time tken by rod to rech the highest point. Two prllel verticl metllic rils A nd CD re seprted by m. They re connected t the two ends by resistnce nd s shown in the figure. A horizontl metllic br of mss 0. kg slides without friction, verticlly down the rils under the ction of grvity. There is uniform horizontl mgnetic field of 0.6T perpendiculr to the plne of the rils. It is observed tht when the terminl velocity is ttined, the power dissipted in nd re 0.76 W nd. W respectively. Find the terminl velocity of br nd vlue nd. A br of mss m is pulled horizontlly (in z plne) cross set of prllel rils by mssless string tht psses over n idel pulley nd is ttched to freely suspended mss M. At t = 0 br is t rest. Find the horizontl speed of the br s function of time t. is constnt nd is long +y is. Two prllel long smooth conducting rils seprted by distnce l re connected by movble conducting connector of mss 'm'. Terminls of the rils re connected by the resistor & the cpcitor C s shown. A uniform mgnetic field perpendiculr to the plne of the rils is switched on. The connector is drgged by constnt force F. Find the speed of the connector l F C s function of time if the force F is pplied t t = 0. Also find the terminl velocity of the connector. A long stright wire crries current Ι 0. t distnce nd b from it there re two other wires, prllel to the former one, which re interconnected by resistnce (figure). A connector slides without friction long the wires with constnt velocity ν. Assuming the resistnces of the wires, the connector, the sliding contcts, nd the self-inductnce of the frme to be negligible, find; () The mgnitude nd the direction of the current induced in the connector; (b) The force required to mintin the connector s velocity constnt. (c) Point of ppliction of mgnetic force on sliding wire due to the long wire. l l θ v v b Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 6 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com SECTION (E) : EMF INDUCED IN A OD O OOP IN NONUNIFOM MAGNETIC FIED E E-. For the sitution shown in the figure, flu through the squre loop is : µ 0i µ 0ib l n () ln π b π b µ 0ib µ 0i (C) l n (D) ln π b π b A circulr copper-ring of rdius r trnsltes in its plne with constnt velocity v. A uniform mgnetic field eists in the spce in direction perpendiculr to the plne of the ring. Consider different pirs of dimetriclly opposite points on the ring. () etween which pir of points is the emf mimum? (b) etween which pir of points is the emf minimum? Wht is the vlue of this minimum emf? E 3. The mgnetic field in region is given by = kˆ 0 y where is fied length. A conducting rod of length lies long the Y-is between the origin nd the point (0,, 0). If the rod moves with velocity = v i, find E 4. E 5. E 6. E 7. E 8. the emf induced between the ends of the rod. A conducting rod slides on pir of thick metllic rils lid prllel to n infinitely long fied wire crrying constnt current i. The center of the rod is t distnce from the wire. The ends of the rils re connected by resistor of resistnce. () Wht force is needed to keep the rod sliding t constnt speed v, s shown in figure? (b) In this sitution wht is the current in the resistnce? (c) Find the rte of het developed in the resistor. (d) Find the power delivered by the eternl gent eerting the force on the rod. Figure shows squre frme of wire hving totl resistnce r plced complnrly with long, stright wire. The wire crries current i given by i = i 0 sin ωt. Find () the flu of the mgnetic field through the squre frme, (b) the emf induced in the frme nd (c) the 0 π het developed in the frme in the time intervl 0 to. ω A rectngulr metllic loop of length l nd width b is plced complnry with long wire crrying current i (figure). The loop is moved perpendiculr to the wire with speed v in the plne contining the wire nd the loop. Clculte the emf induced in the loop when the rer end of the loop is t distnce from the wire. Solve by using Frdy s lw for the flu through the loop nd lso by replcing different segments with equivlent btteries. v 0 A current of 0 A is flowing in long stright wire situted ner rectngulr circuit whose two side of length 0. m re prllel to the wire. One of them is t distnce of 0.05 m nd the other t distnce of 0. m from the wire. The wire is in plne of the rectngle. Find the mgnetic flu through the rectngulr circuit. If the current decys uniformly to zero in 0.0 sec, find the EMF induced in the circuit nd indicte the direction in which the induced current flow. Two infinite long stright prllel wires A nd re seprted by 0. m distnce nd crry equl current in opposite directions. A squre loop of wire C of side 0. m lies in the plne A nd. The loop of wire C is kept prllel to both A nd t distnce of 0. m from the nerest wire. Clculte the EMF induced in loop C while the current in A nd is incresing t the sme rte of 0 3 As. Also indicte the direction of current in loop C SECTION (F) : INDUCED EMF IN A OD, ING, DISC OTATING TING IN A UNIFOM MAGNETIC FIED F. A rod length l rottes with smll but uniform ngulr velocity ω bout its perpendiculr bisector. A uniform mgnetic field eists prllel to the is of rottion. The potentil difference between the centre of the rod nd n end is zero () 8 ωl (C) ωl (D) ωl i i b i b i Infinitely long wire b l b V Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 7 v Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

F. A rod of length l rottes with uniform ngulr velocity ω bout its perpendiculr bisector. A uniform mgnetic field eists prllel to the is of rottion. The potentil difference between the two ends of the rod is FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com F 3. F 4. F 5. F 6. F 7. F 8. F 9. F 0. F. F. zero () ωl (C) ωl (D) ωl A rod of length 0 cm mde up of conducting nd non-conducting mteril (shded prt is non-conducting). The rod is rotted with constnt ngulr velocity 0 rd/sec bout point O, in constnt mgnetic field of tesl s shown in the figure. The induced emf between the point A nd of rod will be 0.09 v () 0. v (C) 0.05 v (D) 0.064 v A semicirculr wire of rdius is rotted with constnt ngulr velocity ω bout n is pssing through one end nd perpendiculr to the plne of the wire. There is uniform mgnetic field of strength. The induced e.m.f. between the ends is: ω / () ω (C) is vrible (D) none of these Two identicl cycle wheels (geometriclly) hve different number of spokes connected from centre to rim. One is hving 0 spokes nd other hving only 0 (the rim nd the spokes re resistnceless). One resistnce of vlue is connected between centre nd rim. The current in will be: double in first wheel thn in the second wheel () four times in first wheel thn in the second wheel (C) will be double in second wheel thn tht of the first wheel (D) will be equl in both these wheels. In the figure there re two identicl conducting rods ech of length rotting with ngulr speed ω in the directions shown. One end of ech rod touches conducting ring. Mgnetic field eists perpendiculr to the plne of the rings. The rods, the conducting rings nd the led wires re resistnceless. Find the mgnitude nd direction of current in the resistnce. 3cm A ω cm O A circulr coil of one turns of rdius 5.0 cm is rotted bout dimeter with constnt ngulr speed of 80 revolutions per minute. A uniform mgnetic field = 0.00 T eists in direction perpendiculr to the is of rottion. Find () the mimum emf induced. (b) the verge emf induced in the coil over long period nd (c) the verge of the squres of emf induced over long period. A bicycle is resting on its stnd in the est-west direction nd the rer wheel is rotted t n ngulr speed of 00 revolutions per minute. If the length of ech spoke is 30.0 cm nd the horizontl component of the erth s mgnetic field is.0 0 5 T, find the emf induced between the is nd the outer end of spoke. Neglect centrepetl force cting on the free electrons of the spoke. Figure shows conducting disc rotting bout its is in perpendiculr mgnetic ω field. A resistor of resistnce is connected between the centre nd the rim. Clculte the current in the resistor. Does it enter the disc or leve it t the centre? The rdius of the disc is 5.0 cm, ngulr speed ω = 0 rd/s, = 0.40 T nd = 0 Ω. A metl disc of rdius r = 0.m is plced perpendiculr to uniform mgnetic D field of induction = 0.50 T. It is cpble of rottion bout n is XY C prllel to the induction, the is is pssing through its centre. Using sliding contcts C & D the disc is connected to resistnce =.5 Ω. Determine the mechnicl power required in rotting the disc if current of 0.0 A flows through. Also find the ngulr velocity of rottion of the disc. ////////////////////// Friction cn be neglected. A thin wire & smll sphericl bob constitute simple pendulum of effective length l. If this pendulum is mde to swing through semi-verticl ngle θ, under grvity in plne norml to uniform mgnetic field of induction, find the mimum potentil difference between the ends of the wire. A conducting disc of rdius is rolling without sliding on horizontl surfce with constnt velocity ' v '. A uniform mgnetic field of strength is pplied norml to the plne of the disc. Find the EMF induced between () P & Q (b) P & C. ( C is centre, P&Q re opposite points on verticl dimeter of the disc) C D Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 8 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com F 3. F 4. F 5. A circulr coil of one turns of rdius 5.0 cm is rotted bout dimeter with constnt ngulr speed of 80 revolutions per minute A uniform mgnetic field = 0.00 T eists in direction perpendiculr to the is of rottion. Find () the mimum emf induced. (b) the verge emf induced in the coil over long period nd (c) the verge of the squres of emf induced over long period. Suppose the ends of the coil in the previous problem re connected to resistnce of 00 Ω. Neglecting the resistnce of the coil find the het produced in the circuit in one minute. A wire shped s semi-circle of rdius rottes bout n is OO' with n ngulr velocity ω in uniform mgnetic field of induction (figure). The rottion is is perpendiculr to the field direction. The totl resistnce of the circuit is equl to. Neglecting the mgnetic field of the induced current, find the men mount of therml power being generted in the loop during rottion period. SECTION (G) : FIXED OOP IN A TIME VAV AYING MAGNETIC FIED & INDUCED EECTIC FIED G. A cylindricl spce of rdius is filled with uniform mgnetic induction prllel to the is of the cylinder. If chnges t constnt rte, the grph showing the vrition of induced electric field with distnce r from the is of cylinder is E r () E r G. In cylindricl region uniform mgnetic field which is perpendiculr to the plne of the figure is incresing with time nd conducting rod PQ is plced in the region. Then P will be t higher potentil thn Q. () Q will be t higher potentil thn P. (C) oth P nd Q will be equipotentil. (D) no emf will be developed cross rod s it is not crossing / cutting ny line of force. G 3 A uniform mgnetic field of induction is confined to cylindricl region of rdius. G 4 G 5. G 6. G 7. (C) d The mgnetic field is incresing t constnt rte of (tesl/second). An electron dt of chrge q, plced t the point P on the periphery of the field eperiences n ccelertion : (C) e d m dt e d m dt towrd left towrd left E r (D) E e d () towrd right m dt (D) zero A nd CD re fied conducting smooth rils plced in verticl plne nd joined by constnt current source t its upper end. PQ is conducting rod which is free to slide on the rils. A horizontl uniform mgnetic field eists in spce s shown. If the rod PQ in relesed from rest then, The rod PQ will move downwrd with constnt ccelertion () The rod PQ will move upwrd with constnt ccelertion (C) The rod will move downwrd with decresing ccelertion nd finlly cqcure constnt velocity (D) either A or. A circulr loop of rdius m is plced in vrying mgnetic field given s = 6t Tesl. Find the emf induced in the coil if the plne of the coil is perpendiculr to the mgnetic field. In the bove question find the verge electric field in the tngentil direction, induced due to the chnging mgnetic field. In the bove question find the current in the loop if its resistnce isω/m. r 0 + ω 0' Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 9 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com G 8. G 9. The current in n idel, long solenoid is vried t uniform rte of 0.0 A/s. The solenoid hs 000 turns/m nd its rdius is 6.0 cm. () Consider circle of rdius.0 cm inside the solenoid with its is coinciding with the is of the solenoid. Write the chnge in the mgnetic flu through this circle in.0 seconds. (b) Find the electric field induced t point on the circumference of the circle. (c) Find the electric field induced t point outside the solenoid t distnce 8.0 cm from its is. A squre wire loop with m sides in perpendiculr to uniform mgnetic field, with hlf the re of the loop in the field. The loop contins 0 V bttery with negligible internl resistnce. If the mgnitude of the field vries with time ccording to = 4 t, with in Tesl & t in sec. () Wht is the totl EMF in the circuit? (b) Wht is the direction of the current? SECTION (H) : SEF INDUCTION,, SEF INDUCTANCE SEF INDUCED EMF & MAGNETIC ENEGY DENSITY H.* Two different coils hve self-inductnce = 8 mh, = mh. The current in one coil is incresed t constnt rte. The current in the second coil is lso incresed t the sme rte. At certin instnt of time, the power given to the two coils is the sme. At tht time the current, the induced voltge nd the energy stored in the first coil re i, V nd W respectively. Corresponding vlues for the second coil t the sme instnt re i, V nd W respectively. Then H. H 3*. H 4. H 5. H 6. H 7. H 8. H 9. H 0. H. i i = () W = 4 i 4 i (C) V = 4 W (D) V Two inductors nd re connected in prllel nd time vrying current i flows s shown. The rtio of currents i /i t ny time t is / () / (C) ( + ) (D) ( + ) A constnt current i is mintined in solenoid. Which of the following quntities will increse if n iron rod is inserted in the solenoid long is? mgnetic field t the centre () mgnetic flu linked with the solenoid (C) self-inductnce of the solenoid (D) rte of Joule heting Figure shows squre loop of side 0.5 m nd resistnce 0 Ω. The mgnetic field hs mgnitude =.0T. The work done in pulling the loop out of the field uniformly in.0 s is 3.5 0 3 J () 6.5 0 4 J (C).5 0 J (D) 5.0 0 4 J A coil of inductnce H nd negligible resistnce is connected to source of supply whose voltge is given by V = 4t volt. If the voltge is pplied when t = 0, then find the energy stored in the coil in 4 second. 5 J () 56 J (C) 04 J (D) 44 J The dimensions of the quntity /(CV) is... Find the self inductnce of solenoid which hs 0 turns per cm. Its length is m nd rdius cm. The figure shows n inductor of H through which current which is incresing t the rte of 5A/sec, is flowing. Find the potentil difference V X -V Y. Figure shows prt of circuit. Find the rte of chnge of the current, shown. 0V 5V A Ω 0H 3V An verge emf of 0 V is induced in n inductor when the current in it is chnged from.5 A in one direction to the sme vlue in the opposite direction in 0. s. Find the self-inductnce of the inductor. A mgnetic flu of 8 0 4 weber is linked with ech turn of 00 turn coil when there is n electric current of 4A in it. Clculte the self-inductnce of the coil. 0V i = 4 i i i =0 i, incresing with the rte 5A/sec = H y Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 30 v Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com H. The current in solenoid of 40 turns, hving length of cm nd rdius of cm, chnges t rte of 0.8 A/s. Find the self emf induced in it. H 3. H 4. H 5. H 6. H 7. Current in n inductor of self inductnce 6H chnges from A to A in sec. Find the increse in the stored energy in the inductor. Find the rte of increse in the stored energy t t= sec in n inductor 5H if the current pssing through it is given s i = t 3 + 5t. In the circuit shown find () the power drwn from the cell, (b) the power consumed by the resistor which is converted into het nd (c) the power given to the inductor. A current of.0 A is estblished in tightly wound solenoid of rdius cm hving 000 turns/metre. Find the mgnetic energy stored in ech metre of the solenoid. Consider smll cube of volume mm 3 t the centre of circulr loop of rdius 0 cm crrying current of 4A. Find the mgnetic energy stored inside the cube. H 8. A long wire crries current of 4.00 A. Find the energy stored in the mgnetic field inside volume of.00 mm 3 t distnce of 0.0 cm from the wire. H 9. H 0. H. A long wire crries current of uniform density. et i be the totl current crried by the wire. Show tht the mgnetic energy per unit length stored within the wire equls µ 0 i. (Note tht it does not depend 6 π on the wire dimeter). Wht is the mgnetic energy density (in terms of stndrd constnt & r) t the centre of circulting electron in the hydrogen tom in first orbit. (dius of the orbit is r) Suppose the EMF of the bttery, the circuit shown vries with time t so the current is given by i (t) = 3 + 5 t, where i is in mperes & t is in seconds. Tke = 4 Ω, = 6 H & find n epression for the bttery EMF s function of time. SECTION (I) : CICUIT CUIT CONTAINING INDUCTANCE, ESISTANCE & ATTE TTEY,, GOWTH AND DECAY OF CUENT IN A CICUIT CUIT CONTAINING INDUCTO Ι. Ι. Ι 3., C nd represent the physicl quntities inductnce, cpcitnce nd resistnce combintions hve dimensions of frequency? () (C) (D) C/ C C An circuit with bttery is connected t t =0. Which of the following quntities is not zero just fter the connection? current in the circuit () mgnetic field energy in the inductor (C) power delivered by the bttery (D) emf induced in the inductor In n circuit current t t = 0 is 0 A. After s it reduces to 8 A. The time constnt of the circuit is (in second): 0 ln 9 () (C) 0 ln 9 Ι 4. In the given circuit find the rtio of i to i. Where i is the initil (t t = 0) current, nd i is stedy stte (t t = ) current through the bttery :.0 () 0.8 (C). (D).5 0 (D) ln 9 0 V 5V 6Ω A H 4Ω mh 3Ω Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 3 4Ω Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com Ι 5. Ι 6. Ι 7*. Ι 8. Ι 9. In the circuit shown in figure, switch S is closed t t = 0. Then: fter long time intervl potentil difference cross cpcitor nd inductor will be equl. () fter long time intervl chrge on cpcitor will be E C. (C) fter long time intervl current in the inductor will be E /. (D) fter long time intervl current through bttery will be sme s the current through it initilly. In series growth circuit, if mimum current nd mimum induced emf in n inductor of inductnce 3mH re A nd 6V respectively, then the time constnt of the circuit is : ms. () /3 ms. (C) /6 ms (D) / ms A circuit consisting of constnt e.m.f. 'E', self induction '' nd resistnce '' is closed t t = 0. The reltion between the current I in the circuit nd time t is s shown by curve '' in the figure. When one or more of prmeters E, & re chnged, the curve 'b' is obtined.the stedy stte current is sme in both the cses. Then it is possible tht: E & re kept constnt nd is incresed. () E & re kept constnt nd is decresed (C) E & re both hlved nd is kept constnt (D) E & re kept constnt nd is decresed A solenoid of resistnce 50 Ω nd inductnce 80 Henry is connected to 00 V bttery. How long will the current tke to rech 50 % of its finl equilibrium vlue? Clculte the mimum energy stored. Find the vlue of t/τ for which the current in n circuit builds up to () 90%, (b) 99% nd (c) 99.9% of the stedy-stte vlue (given ln 0 =.3) () 90%, (b) 99% (c) 99.9% ln 0 =.3) Ι 0. An inductor-coil crries stedy-stte current of.0 A when connected cross n idel bttery of emf 4.0 V. Ι. If its inductnce is.0 H, find the time constnt of the circuit. A coil of resistnce 40 Ω is connected cross 4.0 V bttery, 0.0 s fter the bttery is connected, the current in the coil is 63 ma. Find the inductnce of the coil. [e 0.37] Ι. (i) An circuit hs =.0 H nd = 0 Ω. It is connected cross n emf of.0 V t t = 0. Find di/dt t () t = 0, (b) t = 50 ms nd (c) t. Ι 3. Ι 4. Ι 5. Ι 6. I 7. (ii) Wht re the vlues of the self-induced emf in the circuit of the previous problem t the times indicted therein? The current in dischrging circuit without the bttery drops from.0 A to.0 A in 0.0 s. () Find the time constnt of the circuit. (b) If the inductnce of the circuit is 4.0 H, wht is its resistnce? Consider the circuit shown in figure. () Find the current through the bttery long time fter the switch S is closed. (b) Suppose the switch is opened t t = 0. Wht is the time constnt of the decy circuit? (c) Find the current through the inductor fter one time constnt. A superconducting loop of rdius hs self inductnce. A uniform & constnt mgnetic field is pplied perpendiculr to the plne of the loop. Initilly current in this loop is zero. The loop is rotted bout its dimeter by 80º. Find the current in the loop fter rottion. Show tht if two inductors with equl inductnce re connected in prllel then the equivlent inductnce of the combintion is /. The inductors re seprted by lrge distnce. A closed circuit consists of source of constnt emf E nd choke coil of inductnce connected in series. The ctive resistnce of the whole circuit is equl to. It is in stedy stte t the moment t = 0 the choke coil inductnce ws decresed bruptly η times. Find the current in the circuit s function of time t. ε S Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 3 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com Ι 8. Ι 9. In figure, ξ = 00 V, = 0 Ω, = 0 Ω, 3 = 30 Ω nd = H. Find the vlue of i & i. () immeditely fter switch S w is closed (b) long time fter (c) immeditely fter S w is opened gin (d) long time lter. A conducting frme ACD is kept in verticl plne. A conducting rod EF of mss m cn slide smoothly on it remining horizontl lwys. The resistnce of the loop is negligible nd inductnce is constnt hving vlue. The rod is left from rest nd llowed to fll under grvity nd inductor hs no initil current. A uniform mgnetic field of mgnitude is present throughout the loop pointing inwrds. Determine. () position of the rod s function of tim e ssuming initil position of the rod to be = 0 nd verticlly downwrd s the positive X-is. (b) mimum current in the circuit (c) mimum velocity of the rod. SECTION (J) : MUTUA INDUCTION & MUTUA INDUCTANCE J. Two coils re t fied loctions. When coil hs no current nd the current in coil increses t the rte 5.0 A/s the e.m.f. in coil in 5.0 mv, when coil hs no current nd coil hs current of 3.6 A, flu linkge in coil is 6 mwb () 0 mwb (C) 4.00 mwb (D) 6.00 mwb J. Two coils A nd hve coefficient of mutul inductnce M = H. The mgnetic flu pssing through coil A chnges by 4 Weber in 0 seconds due to the chnge in current in. Then chnge in current in in this time intervl is 0.5 A () the chnge in current in in this time intervl is A (C) the chnge in current in in this time intervl is 8 A (D) chnge in current of A in coil A will produce chnge in flu pssing through by 4 Weber J.3 A rectngulr loop of sides nd b is plced in y plne. A very long wire is lso plced in y plne such tht side of length of the loop is prllel to the wire. The distnce between the wire nd the nerest edge of the loop is d. The mutul inductnce of this system is proportionl to: () b (C) /d (D) current in wire J.4 Two coils of self inductnce 00 mh nd 400 mh re plced very close to ech other. Find the mimum mutul inductnce between the two when 4 A current psses through them 00 mh () 300 mh (C) 00 mh (D) none of these J.5 A long stright wire is plced long the is of circulr ring of rdius. The mutul inductnce of this system is () µ 0 π (D) 0 µ 0 (C) µ 0 J 6. The verge emf induced in the secondry coil is 0. V when the current in the primry coil chnges from to A in 0. s. Wht is the mutul inductnce of the coils. J 7. J 8. J 9. The mutul inductnce between two coils is.5 H. It the current in one coil is chnged t the rte of A/s, wht will be the emf induced in the other coil? Find the mutul inductnces between the stright wire nd the squre loop of figure. A solenoid of length 0 cm, re of cross-section 4.0 cm nd hving 4000 turns is plced inside nother solenoid of 000 turns hving cross-sectionl re 8.0 cm nd length 0 cm. Find the mutul inductnce between the solenoids. i b Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 33 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com SECTION (K) : C OSCIATIONS K. K. The frequency of oscilltion of current in the inductor is: 3 C () 6 π C (C) C (D) π C In the given C circuit if initilly cpcitor C hs chrge Q on it nd C hs chrge Q.The polrities re s shown in the figure. Then fter closing switch S t t = 0 energy will get eqully distributed in both the cpcitor just fter closing the switch. () initil rte of growth of current in inductor will be Q/3 C (C) mimum energy in the inductor will be 3 Q / C (D) none of these MISCENOUS * ONE O MOE THEN ONE MAY E COECT :. An inductor coil stores energy U when current i is pssed through it nd dissiptes energy t the rte of P. The time constnt of the circuit when this coil is connected cross bttery of zero internl resistnce is 4U P () P U U (C) P U (D) U. A rectngulr loop with sliding connector of length l =.0 m is situted in uniform mgnetic field = T perpendiculr to the plne of loop. esistnce of connector is r = Ω. Two resistnces of 6Ω nd 3Ω re connected s shown in figure. The eternl force required to keep the connector moving with constnt velocity v = m/s is 6N () 4N (C) N (D) N 3. A metl rod of resistnce 0 Ω is fied long dimeter of conducting ring of rdius 0. m nd lies in -y plne. There is mgnetic field = (50T) kˆ. The ring rottes with n ngulr velocity ω = 0 rd/s bout its is. An eternl resistnce of 0 Ω is connected cross the centre of the ring nd rim. The current through eternl resistnce is 4 A () A (C) 3 A (D) zero 4. Two concentric nd coplnr circulr coils hve rdii nd b (>> ) s shown in figure. esistnce of the inner coil is. Current in the outer coil is incresed from 0 to i, then the totl chrge circulting the inner coil is : µ i π 0 () b µ 0 ib µ 0 i π (C) (D) µ 0 ib π 5. Two inductor coils of self inductnce 3H nd 6H respectively re connected with resistnce 0Ω nd bttery 0 V s shown in figure. The rtio of totl energy stored t stedy stte in the inductors to tht of het developed in resistnce in 0 seconds t the stedy stte is(neglect mutul inductnce between nd ): 0 () 00 (C) 000 6*. A metl disc of rdius rottes with constnt ngulr velocity ω bout its is. The potentil difference between the centre nd the rim of the disc in stedy stte is (m = mss of electron, e = chrge on electron): mω e () mω e (C) eω m (D) 7. The rdius of the circulr conducting loop shown in figure is. Mgnetic field is decresing t constnt rte α. esistnce per unit length of the loop is ρ. Then current in wire A is (A is one of the dimeters) α α from A to () from to A (C) ρ ρ α ρ from A to 6Ω eω (D) m (D) Zero A S t=0 C + - + - C v b 3Ω Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 34 Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

FEE Downlod Study Pckge from website: www.tekoclsses.com & www.mthsysuhg.com 8. A non conducting ring of rdius nd mss m hving chrge q uniformly distributed over its circumference is plced on rough horizontl surfce. A verticl time vrying uniform mgnetic field = 4t is switched on t time t=0. The coefficient of friction between the ring nd the tble, if the ring strts rotting t t = sec, is : 4qm g () qm g 8q (C) mg 9. A conducting wire frme is plced in mgnetic field which is directed into the pper. The mgnetic field is incresing t constnt rte. The directions of induced currents in wires A nd CD re : to A nd D to C () A to nd C to D (C) A to nd D to C (D) to A nd C to D 0. When the current in the portion of the circuit shown in the figure is A nd incresing t the rte of A/s, the mesured potentil difference V b = 8V. However when the current is A nd decresing t the rte of A/s, the mesured potentil difference V b = 4V. The vlues of nd re : 3 ohm nd henry respectively () ohm nd 3 henry respectively (C) 0 ohm nd 6 henry respectively (D) 6 ohm nd henry respectively. The bttery shown in the figure is idel. The vlues re ε = 0 V, = 5Ω, = H. Initilly the current in the inductor is zero. The current through the bttery t t = s is A () 7 A (C) 3 A (D) none of these *. In the circuit digrm shown time constnt is / (C) stedy stte current in inductor is ε/ (D) q mg () time constnt is / (D) stedy stte current in inductor is ε/ 3. In the figure shown squre loop PQS of side '' nd resistnce 'r' is plced in ner n infinitely long wire crrying constnt current Ι. The sides PQ nd S re prllel to the wire. The wire nd the loop re in the sme plne. The loop is rotted by 80º bout n is prllel to the long wire nd pssing through the mid points of the side Q nd PS. The totl mount of chrge which psses through ny point of the loop during rottion is : µ 0Ι µ 0Ι µ 0Ι ln () ln (C) πr πr πr (D) cnnot be found becuse time of rottion not give. 4*. A conducting loop rottes with constnt ngulr velocity bout its fied dimeter in uniform mgnetic field in direction perpendiculr to tht fied dimeter. The emf will be mimum t the moment when flu is zero. () The emf will be '0' t the moment when flu is mimum. (C) The emf will be mimum t the moment when plne of the loop is prllel to the mgnetic field (D) The phse difference between the flu nd the emf is π/ A b i > < i C Teko Clsses, Mths : Suhg. Kriy (S.. K. Sir), hopl Phone : 0 903 903 7779, 0 98930 5888. pge 35 D Successful People eplce the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.