Compendium ClassXII Physcis

Compendium/Physics/Class XII 1 © 2010 Vidyamandir Classes

Dear Students,Vidyamandir Classes academic team knows that you are rigourously studying to cover the entire prescribedsyllabus. As the Final Exams approach, this is the time when you need to revisit the concepts you havelearned. At this time, you have to be very focused and directed in your approach.To make your learning process precise, effective and enjoyable, we at Vidyamandir Classes conceptualizedthe compendium series, strategically designed to help you in scoring high grades in examination. TheCompendium is primarily intended to present the concepts of chapter in a concise manner. All key definitions,diagrams and formulae have been integrated for a quick revision of the chapter.To help you to easily master complicated concepts, definitions, diagrams and formulae, we have addedinteresting tips, mnemonics, maps and matrices. Let us take a look at the elements of the Compendium andhow to use them.Knowing these features will make it easier for you to assimilate complex information.

Icon Description How it can help you

Concept map To directly recapitulate main concepts of the

chapter.

Drawing Tips

Drawing Tips

To help you draw and remember diagrams, we have thoughtfully developed some mnemonics to help you to memorise information

Compare Contrast Matrix To help you in comparing different concepts

Memory Tips To make your learning process effective, easy

tips have been provided.

In this compendium, we have also incorporated:• CBSE Blue Print: Type of questions asked and the weightage of different forms of questions.• Analysis of Previous Years CBSE questions: The topic wise analysis of previous years question along

with the marks allocated.• We are confident that this Compendium will prove very helpful in achieving excellent result in your

exams.

All the very best for your exams!Vidyamandir Classes Academic Team

ABOUT THE COMPENDIUM


The unit Magnetic Effect of Current and Magnetism is one of the most important units ofPhysics for Class XII. The weightage of this unit in the board examinations is 8 marks and amix of 1, 2, 3 and 5 marks questions can be asked.

Magnetic Effect of Current and MagnetismAbout Magnetic Effect of Current and Magnetism

CBSE BLUE PRINT: MAGNETIC EFFECT OF CURRENT AND MAGNETISMThe numerical problems based on electric potential are relatively easy. Numerical problemsbased on capacitors, especially capacitor combinations, need cautious approach. Derivationsare simple and learning the same can help you perform better in the examinations.

MOVING CHARGES AND MAGNETISMThe analysis of questions from this chapter for the last five years shows that derivationsand description of devices are given much importance. The following table shows thesame.

Number of Questions Chapters

in this unit

Options Very short

(1 mark)

Short Answer I (2 marks)

Short Answer II (3 marks)

Long Answer

(5marks)

Total Marks

I

1

1

-

1

8

1. Moving Charges and Magnetism

2. Magnetism and

Matter II

-

-

1

1

8

Years Derivations/ Devices (Marks)

Understanding of concepts/ Numericals (Marks)

2004 3 2 2005 3 2 2006 5 - 2007 5 1 2008 4 2


APPROACH TO MOVING CHARGES AND MAGNETISMThe main derivations in this chapter are to arrive at the expressions for magnetic fieldcreated by current carrying conductors. These are based on two laws, viz. Biot-savart’s lawand Ampere’s circuital law. The devices described are cyclotron and moving coil galvanom-eter. Thorough study of these topics will fetch you a lion’s share of marks alloted to thischapter.


(i) Faraday’s laws and Lenz’s law

Force acting on a charged particlemoving in a magnetic field:

Helps you to conclude that:

(i) the induced emf is directly proportional tothe rate of change of magnetic flux.

(ii) effect of is to oppose the change in F= q v×B

(ii) Magnetic moment of a revolving electron

(min)

e

he4 m

B. DERIVATIONS AND DEVICES

Leads you to understand that a circularcurrent loop acts as a magnetic dipole.Helps you think of the variation of magneticmoment with(i) number of turns (ii) area of the loop

Think of revolving electron as a single turncurrent loop.Get expressions for I and A.Use the equation: m = n I A.Include angular momentum term and applyBohr’s quantization condition.

(i ) Biot-Savart’s lawMagnetic field strength at a point due to a small current element of length dl carryingcurrent I,

02

I d sindB4 r

0 - the absolute permeability of free space= –7 –14 10 T mA

A. CONCEPTS AND NUMERICALS


This diagram is vital to thederivation. So practise it well.

Magnetic field due to a circular current loop

Get expressions for dB at P due to current elements dl atC and D.Resolve both into sine and cos components.Cos components cancel out and sine components getadded.Now integrate dB to get the total field due to the entireloop

dl over the circumference will always give 2 r for acircle.Substitute for sin and simplify.Apply x = 0 to find the field at the centre

Remember:Use right hand thumb rule to find the direction of thefield due to a current loop

2 2PC PD d r x

0 0

2 2 2

I d IddB sin 90

4 d 4 r x

0

2 2

I sinB dB sin d

4 r x

20

32 2 2

IrB2 r x

00

IB2r


I I . Ampere’s circuital lawThe line integral of magnetic field around any closed path (or circuit) in vacuum is equal to

0 (absolute permeability of space) times the total current (I) threading the closed path.

0B.d I

Magnetic field due to an infinitely long straight conductor carrying currentDraw a circle of radius R aroundthe wire.

.B dl Bdl , as B is tangential to

the circle at any point.

By Ampere’s law,

0B.d I

MAGANETIC FIELD DUE TO A SOLENOID AND TOROID

. 2 B dl Bdl B dl B R

02B R I

0

2IBR

Take n as the number of turns per unitlength.Draw the diagram highlighting therectangle (MNOP) of length L.Solve B.d

over the closed path MNOP.

For sides, MP and NO, . 0B dl

as

B dl . For PO, B = 0. Only

N

M

B.d

exists. Here, .B dl Bdl

By Ampere’s law,Substitute for I. Equate the RHS of boththe equations.

N O P M

M N O P

B.d B.d B.d B.d B.d

B.d B L

0B.d total current (in the rectangle MNOP)

0 n L I

0B nI

(a) For a point well inside the solenoid

0B.d I


Remember: Toroid is a solenoid in the form of a ring.

Draw the figure.

Find B.d

Apply Ampere’s circuital law by clearly stating the totalcurrent.

Equate the RHS of both the expressions.

Force between two straight, parallel live conductors

Draw the figure clearly indicating the direction of force oneach conductor due to the other.Write the expression for the field (B1)created by the first conductor on the second.Find the force exerted per unit length (F2) on the secondconductor by B1 .Substitute for B1.Also write the force experienced by the first conductor perunit length. Magnitudes of forces are the same.This expression helps you define the SI unit of the currentstrength - Ampere.

(b) At any end point of a solenoid,

0n IB2

(c) For a toroid

0 11

IB2 d

F2 = B1I2 ×1 = B1I2

0 1 22

I IF2 d

0 1 22

I IF2 d

0 1 2I IF2 d

B.d B 2 a

0B.d total current passin g through the circle of radius ' a '

= 0 2 a n I

B = 0nI

Remember:(i) each conductor is placed in the magnetic field created by the other.(ii) F BI , force on a current carrying conductor of length perpendicular to a magnetic field of

strength B and carrying a current I.


Torque on a rectangular current loop in a uniform magnetic field

Draw a neat figure clearly indicating theforces acting on each arm.Find the expressions for F1, F2, F3 and F4.F2 and F4 are equal and opposite and actalong the same line. They cancel out.F1 and F3 are equal and opposite, but theyare acting on two different points, forminga couple.Find the moment of the couple.Substitute for F and put b = A, the area ofthe loop.Generalise for loop of n turns

2 4F F I b Bsin

1 2F F 0

1 3F F F I B

F b

Fbsin IABsin

For n turns,

nIABsin

III. Cyclotron

Use:To accelerate positively charged particles like proton,deuteron, etc. such that they acquire sufficient energyfor nuclear reactions.

Principle:A positively charged particle can acquire sufficiently highenergy with a comparatively small alternating potentialdifference by making them cross the same electric fieldfrequently with the use of a strong magnetic field.


Working:The particle is produced at P.D1 and D2 are given +ve and –ve potentials respectively.Particle gets accelerated to D2 and enters a field free re-gion.It moves with a constant speed v and gets deflected alonga semicircular path of radius r such that mvr

qB

Period of revolution, 2 r 2 mTv qB

Frequency,

c1 qBT 2 m

(called cyclotron frequency)Polarity of the dees is reversed frequently to repeat theprocess.The particle gets accelerated every time with increasingradius of the path and acquiring more and more energy.The accelerated particle is removed with the help of anelectric field. Kinetic energy of the particle at exit aftertaking a path of radius R, 2 2 2

21 q B RK mv2 2m

Parts:D-shaped hollow evacuatedmetal chambers D1 and D2 ,called dees, placedhorizontally with a small gapbetween them.A high frequency oscillatorconnected to the dees.A strong magnet representedby pole pieces N and SApply concepts of centripetalforce and Lorent’s force.

Mention how the particle isaccelerated again and againWrite how the particle isremoved.

IV. Moving coil galvanometer.

Principle: Torque acts on a rectangular coil placed inside a uniform magnetic field.

Draw a neat and simple diagram.

Parts:Rectangular coil of N turns and area A.Phosphor bronze fibre to suspend the coilCylindrical core that houses the coil andcylindrical pole pieces.Lamp and scale arrangement.


Device Nature of device Modification to galvanometer Required diagram

Ammeter Low resistance

A shunt resistance S is connected in parallel to the galvanometer

g

g

I GS

I – I

Voltmeter High resistance

A high resistance R is connected in series with the galvanometer

g

VR – GI

Functions of each part:Phosphor bronze fibre: Enables the coil to execute torsional oscillations and also acts asone current lead to the coil.Cylindrical core and cylindrical pole pieces: To make the field radial so that the plane of thecoil is parallel to the field.Lamp and scale arrangement: to measure the deflection of the coil.

NIAB θC θ

NAB CI

I k

C u rren t s e ns itiv ity :

V o lta ge sens itiv ity :

Find the deflecting couple of the coil andthe restoring couple of the suspensionsystem.Apply the equilibrium condition

The deflection per unit current

The deflection per unit potential differ-ence across the coil

T h eo ry :

where NAB

Ck , the galvanometer constant.

C.S = NAB 1 dI dI C k

G – galvanometer resistance, Ig – current for full deflection of the galvanometerI – maximum value of current to be measured, V – maximum value of voltage to be measured.

Conversion of galvanometer into ammeter and voltmeter:

Voltage sensitivity :

Current sensitivity :

C.SV.S V IR R

d NAB 1C.S I dI C k


ANALYSIS OF CBSE QUESTIONS (2004-2008)

Note :The weightage of this chapter is 4 - 6 marks only. If the total marks for questions from thischapter for a particular year exceeds 6 in the above table, it is because of the fact thatquestions from three or more papers have been analysed.

Concepts: Years

Lorentz’ force 2008 (2 marks), 2007 (2 marks), 2005 (1 mark), 2004 (2 mark),

Cyclotron 2008 (5 marks), 2006 (2 marks), 2005 (5 marks), 2004 (3 marks)

Biot-Savart’s law and its applications 2008 (5 marks), 2007 (5 marks), 2006 (5 marks), 2005 (3 marks), 2004 (3 marks)

Ampere’s circuital law and its applications

2007 (3 marks), 2006 (3 marks), 2005 (3 marks), 2004 (2 marks), 2004 (3 marks)

Magnetic moment of a circular loop 2008 (2 marks), 2004 (3 marks)

Moving coil galvanometer 2007 (5 marks), 2006 (5 marks), 2005 (3 marks), 2004 (3 marks)

Documents

Compendium ClassXII Physcis