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Anna romba romba nantri anna i thought what do i learn this big question and the steps involved in this question onnumme puriyala thank you so much anna romba nantri
3:20 - Sir If Negative is High Then Postive is Low The Cathode Ray Repeles To Upward And If Positive Is High And Negative Is Low .The Cathode Ray Is Attracted By Postive And Repelled By Negative So It Goes Upward again How it Delflects On Both Side ?
I need important questions for volume 2 sir .2marks, 3marks and 5 marks .6, 7th lessons ku video upload pana mari..plz sir upload soon .i am waiting and keep rocking😊
1. In 1887, J.J. Thomson measured the specific charge (e/m) of electron. 2. The specific charge is defined as the charge per unit mass of the particle. Principle: 3. In the presence of electric and magnetic fields, the cathode rays are deflected. Arrangement of J.J. Thomson experiment to determine the specific charge of an electron 4. A highly evacuated discharge tube is used and cathode rays (electron beam) produced at the cathode are attracted towards anode disc A. 5. An anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. 6. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. 7. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. 8. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. 9. This is achieved by coating the screen with zinc sulphide. • Determination of velocity of cathode rays: 1. For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O 2. i.e the magnitude of electric force is balanced by the magnitude of force due to the magnetic field. eE = eBv v=E/B 3. Electric force balances the magnetic force - the path of the electron beam is a straight line. • Determination of specific charge: 1. Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. 2. Let V be the potential difference between anode and cathode, then the potential energy is eV. 3. Then from law of conservation of energy. ev=1/2 mv^2 e/m = 1/2V x v^2 4. Substituting the value of velocity from equation (1), we get e/m =1/2V x E^2/B^2 5. Substituting the values of E, B and V, the specific charge can be determined as e/m =1.7x10^11 C/kg
Im from kerala.u must be a college professor. I have a suggestion to include more about the parabolic areas including singly and doubly ionised concepts.
I have a small doubt Anna..... 3:01 in that place...... Since election was negative charge and negative plate is strong it repel each other and electron Mela pogum and positive plate strong ahh irrundha attraction naala mela thaan pogum apram electron keezha varathu ku kaana chance sae illa yae.... But adhu eppadi keezha varuthu Please adha mattum clear pannunga len
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0:06 "Chapter 9"
It's old version book. Our books are new ones
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Anna romba romba nantri anna i thought what do i learn this big question and the steps involved in this question onnumme puriyala thank you so much anna romba nantri
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Thala unga neraya video pathan vera level thala mass katuringa very help ful
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Yes video is too length but u may see it in 2x playback speed. It's what I was doing. Your mark in 12th senior?
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3:20 - Sir If Negative is High Then Postive is Low The Cathode Ray Repeles To Upward And If Positive Is High And Negative Is Low .The Cathode Ray Is Attracted By Postive And Repelled By Negative So It Goes Upward again How it Delflects On Both Side ?
same question
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I can't understand your question?
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Thanks for this topic also cover all topics 12th physics
2:59 sir positive athigama irunthalum upward thana sir move aagum epdi downward varum??
I need important questions for volume 2 sir .2marks, 3marks and 5 marks .6, 7th lessons ku video upload pana mari..plz sir upload soon .i am waiting and keep rocking😊
Study all the questions don't study important question.
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1. In 1887, J.J. Thomson measured the specific charge (e/m) of electron.
2. The specific charge is defined as the charge per unit mass of the particle.
Principle:
3. In the presence of electric and magnetic fields, the cathode rays are deflected.
Arrangement of J.J. Thomson experiment to determine the specific charge of an electron
4. A highly evacuated discharge tube is used and cathode rays (electron beam) produced at the cathode are attracted towards anode disc A.
5. An anode disc is made with pin hole in order to allow only a narrow beam of cathode rays.
6. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage.
7. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other.
8. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed.
9. This is achieved by coating the screen with zinc sulphide.
• Determination of velocity of cathode rays:
1. For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O
2. i.e the magnitude of electric force is balanced by the magnitude of force due to the magnetic field.
eE = eBv
v=E/B
3. Electric force balances the magnetic force - the path of the electron beam is a straight line.
• Determination of specific charge:
1. Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode.
2. Let V be the potential difference between anode and cathode, then the potential energy is eV.
3. Then from law of conservation of energy.
ev=1/2 mv^2
e/m = 1/2V x v^2
4. Substituting the value of velocity from equation (1), we get
e/m =1/2V x E^2/B^2
5. Substituting the values of E, B and V, the specific charge can be determined as
e/m =1.7x10^11 C/kg
*Diagram must
Sir.. I am from class 11th... But this video helps me to understand the derivation of e/m ratio... Thank you so much sir😁
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Sir we use Fleming's left hand rule right to know movement of electrons in magnetic field
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Easy to understand concept tnk u sir
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Sir oru vela nama antha magnet um parallel plates um vaikalana kuda electron straight ah travel aaga chances iruku la sir
Im from kerala.u must be a college professor. I have a suggestion to include more about the parabolic areas including singly and doubly ionised concepts.
School teacher with 3years experience
@@MurugaMP hi sir
24-25 batch........ >like
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Sir ,I have one dought
Neenga board la write Pannadha mattum with diagram
Exam la write Panna mark poduvanga la .. konjam sollunga
Sir enga settukku expected questions make panni podunga sir plz...
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When magnetic field only there what happened
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Sir is the value of V B and E
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Downstream motion concept video podunga
Sir last point (deflection of charge only due to uniform magnetic field )???
Yaa sir....?
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Sir.what is mean by deflection in this topic(iii)?
2:59 positive athigama iruntha atracth thana agum sir?
Same doubt
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Sir why didn't the electron beam gets attracted by anode?!!
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Anna I have doubt(3.09), if it is positive it should be attracted then why it is replusive and travelled towards negative
Same doubt.
Y electric fieldum and magnetic fieldum apply panurom...
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Sir, in the equation of eE=eBv, the velocity (V) is represented what? which one's velocity?
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I have a small doubt Anna..... 3:01 in that place...... Since election was negative charge and negative plate is strong it repel each other and electron Mela pogum and positive plate strong ahh irrundha attraction naala mela thaan pogum apram electron keezha varathu ku kaana chance sae illa yae.... But adhu eppadi keezha varuthu
Please adha mattum clear pannunga len
Enakkum same doubt
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Sir engaluku question la only determine the specific charge e/m ketrukanga so nanga e/m = 1.7× 10^11 Ckg idhu varaikum pota ok va sir
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Ithu mattum elithana public la 5 mark kidaikuma sir