CBSE Physics

Question
CBSEENPH12039297

Two long, straight, parallel conductors carry steady currents, I1 and I2 , separated by a distance . If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other? Obtain the expression for this force. Hence, define one ampere.

Solution

                                       
Magnetic field induction at some point P on wire 2 due to current I1 passing through wire 1 is given by, 
                                       straight B subscript 1 space equals space fraction numerator straight mu subscript straight o space 2 space straight I subscript 1 over denominator 4 space straight pi space straight d end fraction 
Magnetic field is produced by wire 1 and current carrying wire 2 lies in magnetic field B1.
 The unit length of wire 2 will experience a force, given by 
                              F= B1 I2 x 1 = fraction numerator straight mu subscript straight o over denominator 4 straight pi end fraction space fraction numerator 2 space I subscript 1 space I subscript 2 over denominator d end fraction

According to Fleming's left-hand rule, the force on wire 2 acts in the plane of paper perpendicular to wire 2, directed towards wire 1. Similarly, wire 1 also experiences the same force towards wire 2. Thus, both the conducting wires attract each other with the same force F.
One ampere can be defined as the amount of current flowing through two parallel conductors, which are in the same direction or opposite directions, placed at a distance of one metre in free space, and both the wires attract or repel each other with a force of 2 x 10-7 per metre of their lengths. 

Question
CBSEENPH12039298

How are em waves produced by oscillating charges?
 
Draw a sketch of linearly polarised em waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.

                                                      OR

Write Maxwell's generalisation of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is I = straight epsilon subscript straight o space dϕ subscript straight E over dt, where straight ϕ subscript straight E is the electric flux produced during charging of the capacitor plates. 

Solution

A moving charge produces both electric and magnetic fields, and an oscillating charge produces oscillating magnetic and electric fields. These oscillating electric and magnetic fields with respect to space and time produce electromagnetic waves. 
The propagation of electromagnetic waves can be shown as:


                                                  OR

Maxwell's generalization of ampere's circuital law given by,

contour integral space B with rightwards harpoon with barb upwards on top. space space stack d l with rightwards harpoon with barb upwards on top space equals space mu subscript o space left parenthesis I space plus space I subscript D right parenthesis thin space equals space straight mu subscript straight o space left parenthesis space straight I space plus space straight epsilon subscript straight o space dφ over dt right parenthesis
Consider that a parallel capacitor C is charging in a circuit. 
The magnitude of electric field between the two plates will be, 
straight E space equals space fraction numerator straight q over denominator straight epsilon subscript straight o straight A end fraction , is perpendicular to the surface of the plate. 
Electric flux through the surface will be, 
straight ϕ subscript straight E space equals space straight E with rightwards harpoon with barb upwards on top. space straight A with rightwards harpoon with barb upwards on top space

space space space space space equals space EA space cos space straight theta

space space space space space equals space fraction numerator straight q over denominator straight epsilon subscript straight o space straight A end fraction space straight x space straight A space

space space space space space equals space straight q over straight epsilon subscript straight o

rightwards double arrow space dϕ subscript straight E over dt space equals space fraction numerator straight d space open parentheses begin display style straight q over straight epsilon subscript straight o end style close parentheses over denominator dt end fraction

rightwards double arrow space dq over dt space equals space straight epsilon subscript straight o space dϕ subscript straight E over dt
That space is comma space dq over dt space is space the space rate space of space change space with space time. space

rightwards double arrow space straight I thin space equals space straight epsilon subscript straight o space dϕ subscript straight E over dt

Question
CBSEENPH12039305

(i) An a.c. source of voltage V = Vo sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain an expression for impedance of a circuit and the phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?

ii) In a series LR circuit, XL = R and the power factor of the circuit is P1. When capacitor with capacitance C, such that XL = XC is put in series, the power factor becomes P2. Calculate P1 / P2.

Solution

Voltage of the source is given by, 
V = Vo sin ωt 
                                           
Let current of the source be I = Io sin ωt
Maximum voltage across R is VR = Vo R, represented along OX
Maximum voltage across L = VL  = IO XL, represented along OY and is 90o ahead of Io.
Maximum voltage across C = VC = Io XC, represented along OC and is lagging behind Io by 90o
Hence, reactive voltage is VL - VC, represented by OB'

the vector sum of VR, VL and VC is resultant of OA and OB', represented along OK.
OK = Vosquare root of OA squared plus OB squared end root
i.e., Vosquare root of straight V subscript straight R squared plus left parenthesis straight V subscript straight L minus straight V subscript straight C right parenthesis squared end root space equals space square root of left parenthesis I subscript O R right parenthesis squared plus left parenthesis I subscript o X space minus space V subscript C right parenthesis squared end root
rightwards double arrow space straight V subscript straight o space equals space straight I subscript straight o space square root of straight R squared space plus space left parenthesis straight X subscript straight L minus straight X subscript straight c right parenthesis squared end root
Impedance, Z = straight V subscript straight o over straight I subscript straight o space equals space R square root of straight R squared space plus space left parenthesis straight X subscript straight L minus straight X subscript straight c right parenthesis squared end root
When, X= XC ,  the voltage and current are in the same phase. 
In such a situation, the circuit is known as non-inductive circuit. 
ii) 
Given,
Power factor, P1 = R/Z
rightwards double arrow space straight P subscript 1 space equals space fraction numerator straight R over denominator square root of straight R squared space plus space straight X squared end root end fraction space equals space fraction numerator straight R over denominator square root of 2 straight R squared end root end fraction space equals space fraction numerator 1 over denominator square root of 2 end fraction

space space space space space straight P subscript 2 space equals space straight R over straight Z

rightwards double arrow space straight P subscript 2 space equals space fraction numerator straight R over denominator square root of straight R squared plus left parenthesis straight X subscript straight L minus straight X subscript straight C right parenthesis squared end root end fraction space equals space 1 space
Thus, 
straight P subscript 1 over straight P subscript 2 space equals space fraction numerator 1 over denominator square root of 2 end fraction


Question
CBSEENPH12039306

(i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.

(ii) The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is also 100. The input voltage and power are 220 V and 1100 W, respectively. Calculate the:

a) number of turns in secondary

b) current in primary

c) voltage across secondary

d) current in secondary

e) power in secondary

Solution

i) A transformer is a device that changes a low alternating voltage into a high alternating voltage or vice versa. 
Transformer works on the principle of mutual induction. 
A changing alternate current in primary coil produces a changing magnetic field, which induces a changing alternating current in secondary coil. 
 
Energy losses in transformer: 
Flux leakage due to poor structure of the core and air gaps in the core.
Loss of energy due to heat produced by the resistance of the windings.
Eddy currents due to alternating magnetic flux in the iron core, which leads to loss due to heating. 
Hysteresis, frequent and periodic magnetisation and demagnetization of the core, leading to loss of energy due to heat.
ii)
a)  Number of turns in secondary coil is given by, 

space space space space space space straight N subscript straight s over straight N subscript straight P space equals space n
rightwards double arrow space space N subscript S over 100 space equals space 100 space
rightwards double arrow space space space space space N subscript S space equals space 10 comma 000 
b) Current in primary is given by, 
straight I subscript straight P space straight V subscript straight P space equals space straight P
rightwards double arrow space straight I subscript straight P space equals space 1100 over 220 space equals space 5 space straight A

c) Voltage across secondary  is given by, 
space space space space space space space fraction numerator straight V subscript straight S over denominator space straight V subscript straight P end fraction space equals space N subscript S over N subscript P space equals space n

rightwards double arrow space space space straight V subscript straight S space equals space 100 space straight x space 220 space equals space 22 comma 000 space straight V
d) Current in secondary is given by, 
space space space space space straight V subscript straight S space straight I subscript straight S space equals space straight P

rightwards double arrow space straight I subscript straight S space equals space straight P over straight V subscript straight S space equals space 1100 over 22000 space equals space 0.05 space straight A
e) In an ideal transformer,
Power in secondary = Power in primary = 1,100 W

Sponsor Area

Question
CBSEENPH12039309

(i) Define the term drift velocity.

(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend?

(iii) Why alloys like constantan and manganin are used for making standard resistors?

Solution

i) 
Drift velocity is the average velocity of the free electrons in the conductor with which they get drifted towards the positive end of the conductor under the influence of an external electric field.
ii) 
Free electrons are in continuous random motion. They undergo changethey undergo change in direction at each collision and the thermal velocities are randomly distributef in all directions. 
straight u space equals space fraction numerator straight u subscript 1 space space plus space straight u subscript 2 space plus space.... plus space straight u subscript straight n over denominator straight n space end fraction space equals space 0 space space space space space space space space space space... left parenthesis 1 right parenthesis

Electric field, E = -eE 
Acceleration of each electron, straight a with rightwards harpoon with barb upwards on top space equals space fraction numerator negative e E over denominator m end fraction         ... (2) 
Here, 
m = mass of an electron
e = charge on an electron
Drift velocity is given by, straight v subscript straight d space equals space fraction numerator straight v subscript 1 space plus space straight v subscript 2 space plus space.... plus space straight v subscript straight n over denominator straight n end fraction
straight v subscript straight d space equals space fraction numerator left parenthesis straight u subscript 1 space space plus aτ subscript 1 right parenthesis plus space left parenthesis straight u subscript 2 plus aτ subscript 2 right parenthesis plus....... plus thin space left parenthesis straight u subscript straight n space plus space aτ subscript straight n right parenthesis over denominator straight n end fraction space

straight v subscript straight d space equals space fraction numerator left parenthesis straight u subscript 1 space plus space straight u subscript 2 plus.... plus space straight u subscript straight n right parenthesis over denominator straight n end fraction plus fraction numerator straight a left parenthesis straight tau subscript 1 space plus space straight tau subscript 2 plus.... plus straight tau subscript straight n right parenthesis over denominator straight n end fraction

Since comma space straight v subscript straight d space equals space aτ space space space space space space space space space space space space space space space space space space... space left parenthesis 3 right parenthesis space

Here comma space

straight tau space equals space fraction numerator straight tau subscript 1 space plus space straight tau subscript 2 plus..... plus space straight tau subscript straight n over denominator straight n end fraction comma space is space the space average space time space elapsed
Substituting space for space straight a space from space equation space left parenthesis 2 right parenthesis comma space

straight v subscript straight d space equals space fraction numerator negative eE over denominator straight m end fraction space straight tau space space space space space space space space space space space space space space space space space space space space space space space space... space left parenthesis 4 right parenthesis 
Electrons are accelerated because of the external electric field. 
They move from one place to another and current is produced.
For small interval dt, we have
I dt = -q ; where q is the total charge flowing
Let, n be the free electrons per unit area. Then, total charge crossing area A in time dt is given by,
Idt = neAvd dt
Substituting the value of vd, we get 
I.dt = neA open parentheses eE over straight m close parentheses d t 
Current density, J  = straight n space straight e squared over straight m open vertical bar straight E close vertical bar straight tau
From ohm's law, we have
J = straight sigma space straight E
Here, straight sigma is the conductivity of the material through whic the current is flowing,
Thus, 
straight sigma space equals space straight n space straight e squared over straight m straight tau
straight sigma space equals space 1 over straight rho space or space straight rho space equals space 1 over straight sigma

Substituting space the space value space of space conductivity comma space we space have

straight rho space equals space fraction numerator straight m over denominator ne squared straight tau end fraction space semicolon space straight tau space is space the space relaxation space time
iii) 
Alloys like constantan and manganin are used for making standard resistors because:
a) they have high value of resistivity
b) temperature coefficient of resistance is less. 

Sponsor Area