BIOTECHNOLOGY- Study Materials & Lectures

Study material

Biotechnology

JVC-Janta Vedic College

chemiStry

Chemistry zoology botany

Click for more details

IITK-Indian Institute of Technology Kanpur

Transcription Translation and ...

This presentation gives an overview of Transcription, Translation and recombinant DNA Technology. The idea for recombinant DNA was first proposed by Peter Lobban, a graduate student of Prof. Dale Kaiser in the Biochemistry Department at Stanford University Medical School.[20] The first publications describing the successful production and intracellular replication of recombinant DNA appeared in 1972 and 1973. Recombinant DNA is widely used in biotechnology, medicine and research. Today, recombinant proteins and other products that result from the use of rDNA technology are found in essentially every western pharmacy, doctor's or veterinarian's office, medical testing laboratory, and biological research laboratory. This presentation starts with the description of various macro molecules, namely Protein, Nucleic Acids and etc, and explain their differneces. <a rel="nofollow" href="http://www-class.unl.edu/biochem/gp2/m_biology/animation/gene/gene_a2.html">Click this line, there is a good presntation over the Transcription is presented in a very attratcive way. </a>

Click for more details

BEC-Balaji Engineering College

Fundelmentals Of Bio Technolog ...

When you go through this lecture about Fundelmentals Of Bio Technology, there is a fine need for high concentration due to the high importance of this topic and the level of understanding that you require to get the subject precisely. The subject is very important for all branches of engineering since it is quite closely related to a number of other syllabus topics. A document has been attached that is complete in all means for the help of students.

Click for more details

BEC-Balaji Engineering College

Fundelmentals Of BioTechnology

In this lecture, we have explained the Fundelmentals Of BioTechnology and the various aspects of it including the challenges faced and the procedural analysis of it. The topic covers a lot of marks in the syllabus and also is very essential for some other topics as well hence this is important that a student understand it correctly. By downloading the document attached with this lecture you can start learning this topic from the scratch.

Click for more details

SMU-Sikkim Manipal University

Introduction to Biotechnology

This lecture is a technical terminology specific for the Introduction to Biotechnology which is a preliminary topic when you start with the subject. It is very important to get every detail about this topic correctly since this works as the backbone for many of the additional topics in the overall syllabus. To cover the entire topic in just one book and to make it easier to understand a document has been attached for the convenience of students.

Click for more details

SMU-Sikkim Manipal University

Bio Process Principle Ebook

Bio Process Principle is considered as the basic knowledge topic in the entire engineering and to properly understand the complete subject it is very much recommended that you go through the depths of this topic first. One cannot simply skip for the higher order topics of discipline unless they know the working mechanism of such an important topic. The attached document will help you in understanding the lecture and the practical knowledge required for this field.

Click for more details

BEC-Bapatla Engineering College

CLOUD COMPUTING basic

You might have seen network diagrams with a puffy cloud representing the behind-the-scenes components. The cloud conveys the notion that you don’t really need to know the location or number of servers delivering the service. Your experience is that you request a service

Click for more details

BU-Bangalore University

Genetic Engineering Exam Sampl ...

Genetic Engineering is the paper of M.Sc., 3rd Semester in Bangalore university.This paper is of 80 marks and time allotted here is 3 hours. This Paper consists of 3 sections. Section 1 has 7 questions, out of which 5 are to be answered. This section has short notes questions. The Section B has 6 questions. Out of which only 4 is to be answered. The question of this section is of 5 marks each. The final section has long descriptive type questions. There are 5 questions of 15 marks each, out of which candidates require answering any 3.

Click for more details

BIT-Birla Institute of Technology Mesra

1st Semester Biomolecular Mode ...

Biomolecular Modeling (BIO C417) is the 1st Sem. subject in Birla Institute of Technology and Science, Pilani.The paper is of 2 hours where each question is compulsory and has 70 marks for it. The second part has 30 marks and 1 hours is given to solve this section..

Click for more details

AIIMS-All India Institute of Medical Sciences

work and energy notes for All ...

Definitions Work can be defined as transfer of energy. In physics we say that work is done on an object when you transfer energy to that object. If one object transfers (gives) energy to a second object, then the first object does work on the second object. Work is the application of a force over a distance. Lifting a weight from the ground and putting it on a shelf is a good example of work. The force is equal to the weight of the object, and the distance is equal to the height of the shelf (W= Fxd). Work-Energy Principle --The change in the kinetic energy of an object is equal to the net work done on the object. Energy can be defined as the capacity for doing work. The simplest case of mechanical work is when an object is standing still and we force it to move. The energy of a moving object is called kinetic energy. For an object of mass m, moving with velocity of magnitude v, this energy can be calculated from the formula E= 1/2 mv^2. Types of Energy There are two types of energy in many forms: Kinetic Energy = Energy of Motion Potential Energy = Stored Energy Forms of Energy Solar Radiation -- Infrared Heat, Radio Waves, Gamma Rays, Microwaves, Ultraviolet Light Atomic/Nuclear Energy -energy released in nuclear reactions. When a neutron splits an atom's nucleus into smaller pieces it is called fission. When two nuclei are joined together under millions of degrees of heat it is called fusion Electrical Energy --The generation or use of electric power over a period of time expressed in kilowatt-hours (kWh), megawatt-hours (NM) or gigawatt-hours (GWh). Chemical Energy --Chemical energy is a form of potential energy related to the breaking and forming of chemical bonds. It is stored in food, fuels and batteries, and is released as other forms of energy during chemical reactions. Mechanical Energy -- Energy of the moving parts of a machine. Also refers to movements in humans Heat Energy -- a form of energy that is transferred by a difference in temperature What is Power Power is the work done in a unit of time. In other words, power is a measure of how quickly work can be done. The unit of power is the Watt = 1 Joule/ 1 second. One common unit of energy is the kilowatt-hour (kWh). If we are using one kW of power, a kWh of energy will last one hour. Calculating Work, Energy and Power WORK = W=Fd Because energy is the capacity to do work , we measure energy and work in the same units (N*m or joules). POWER (P) is the rate of energy generation (or absorption) over time:P = E/t Power's SI unit of measurement is the Watt, representing the generation or absorption of energy at the rate of 1 Joule/sec. Power's unit of measurement in the English system is the horsepower, which is equivalent to 735.7 Watts. <div>  </div>

Click for more details

AIIMS-All India Institute of Medical Sciences

magnetic force and field notes ...

<h1 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading1">30. THE MAGNETIC FORCE AND FIELD.</a></h1> <h2 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading2">30.1. The magnetic force</a></h2> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Up to now we have only considered the electrostatic forces acting on charges at rest. When the charges are in motion, an extra force acts on them. This extra force is called the magnetic force. The magnetic force between two charges q1 and q2, moving with velocities v1 and v2, is equal to</center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3001.gif" /></center> (30.1) where u0 is called the permeability constant which is equal to 4[pi] x 10-7 Ns2/C2, and r is the distance between the two charges (see Figure 30.1). The ratio R of the magnetic force and the electric force is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3002.gif" /></center> (30.2) <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3003.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.1. Relevant vectors for definition of magnetic force.  </center> Substituting the numerical values of [epsilon]0 and u0 into eq.(30.2), the ratio R can be rewritten as <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3004.gif" /></center> (30.3) where c is the velocity of light in vacuum (c = 3 x 108 m/s). Clearly, the magnetic force is small compared with the electric force unless the speed of the particles is high (a significant fraction of the velocity of light). A magnetic field B can be associated with the magnetic force. The magnetic field at some point in the vicinity of a moving charge can be determined by placing a test charge at that point and moving it with some velocity v. The test charge will experience, besides the electric force, a magnetic force Fmag. Per definition, the magnetic field B is related to the magnetic force Fmag via <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3005.gif" /></center> (30.4) A measurement of the magnetic force acting on the test charge for various directions of v can be used to determine the magnetic field B. The magnetic force is always perpendicular to the velocity vector and the direction of the magnetic field. The unit of magnetic field strength is the Tesla (T). Comparing eq.(30.1) and eq.(30.4) we can determine the magnetic field generated by a point charge q2 moving with a velocity v2: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3006.gif" /></center> (30.5) Similar to electric field lines we can graphically represent the magnetic field by field lines. The density of field lines indicates the strength of the magnetic field. The tangent of the field lines indicates the direction of the magnetic field. The magnetic field lines form closed loops, that is they do not begin or end anywhere in the way that the electric field lines begin and end on positive and negative charges. This immediately implies that the magnetic flux through an arbitrary closed surface is equal to zero: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3007.gif" /></center> (30.6) The principle of superposition is also valid for the magnetic field. <h3 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading3">Example: Problem 30.10</a></h3> At the surface of a pulsar, or neutron star, the magnetic field may be as strong as 108 T. Consider the electron in a hydrogen atom on the surface of the neutron star. The electron is at a distance of 0.53 x 10-10 m from the proton and has a speed of 2.2 x 106 m/s. Compare the electric force that the proton exerts on the electron with the magnetic force that the magnetic field of the neutron star exerts on the electron. Is it reasonable to expect that the hydrogen atom will be strongly deformed by the magnetic field ? The electron in a hydrogen atom is at a distance r equal to 0.53 x 10-10 m from the proton. The electric force acting on the electron is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3008.gif" /></center> (30.7) The maximum magnetic force acting on the electron occurs when the direction of the electron is perpendicular to the direction of the magnetic field. The maximum magnetic force is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3009.gif" /></center> (30.8) Comparing eq.(30.7) and eq.(30.8) we conclude that the magnetic field is significantly stronger hat the electric field, and we expect that the orbits of the electrons are strongly affected by the intense magnetic field. <h2 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading4">30.2. The Biot-Savart Law</a></h2> The definition of the magnetic force showed that two moving charges experience a magnetic force. In other words, a moving charge produces a magnetic field which results in a magnetic force acting on all charges moving in this field. A current flowing through a wire is equivalent to a collection of electrons moving with a certain velocity along the direction of the wire. Each of the moving electrons produces a magnetic field that is given by eq.(30.5). Consider a small segment of the wire with a length dL (see Figure 30.2). At any given time, a charge dq will be located in this segment. The magnetic field, dB, generated by this charge at point P is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3010.gif" /></center> (30.9) where v is the velocity of the charge carriers. The time dt that it takes for all original charge carriers to leave the segment dL is given by <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3011.gif" /></center> (30.10) The current I in the wire can now be obtained easily <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3012.gif" /></center> (30.11) <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3013.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.2. Calculation of magnetic field produced by an electric current.   </center> This equation can be rewritten as <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3014.gif" /></center> (30.12) and substituted into eq.(30.9): <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3015.gif" /></center> (30.13) Equation (30.13) is called the Biot-Savart Law. <h3 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading5">Example: Problem 30.33</a></h3> Helmholtz coils are often used to make reasonably uniform magnetic fields in laboratories. These coils consist of two thin circular rings of wire parallel to each other and on a common axis, the z-axis. The rings have a radius R and they are separated by a distance which is also R. These rings carry equal currents in the same direction. Find the magnetic field at any point on the z-axis. <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3016.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.3. Calculation of magnetic field produced by one ring.  </center> The first step to calculate the field of a pair of Helmholtz coils is to calculate the magnetic field produced by each ring. Suppose the ring is located in the x-y plane and we are interested in the field at point P, a distance z above the x-y plane (see Figure 30.3). The net magnetic field of the ring at point P will be directed along the z-axis. The magnitude dB of the magnetic field produced by a small segment of the ring with length dL is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3017.gif" /></center> (30.14) To obtain eq.(30.14) we have used the fact that for any point on the ring, the position vector r is perpendicular to the direction of dL. The z-component of the magnetic field dB is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3018.gif" /></center> (30.15) The magnitude of the position vector r is related to R and z: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3019.gif" /></center> (30.16) The angle a is also related to R and z: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3020.gif" /></center> (30.17) Combining eqs.(30.15), (30.16) and (30.17) we obtain <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3021.gif" /></center> (30.18) Integrating eq.(30.18) over the whole ring we obtain for the total field generated by the ring <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3022.gif" /></center> (30.19) Figure 30.4 shows the magnetic field generated by one coil with a radius of 1 m, located at z = 0 m. To find the field generated by a pair of Helmholtz coils, we assume that the coils are centered at z = 0 and at z = R. The magnetic field generated by the coil located at z = 0 is given by eq.(30.19). The magnetic field generated by the coil located at z = R is given by <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3023.gif" /></center> (30.20) The total field on the axis of a pair of Helmholtz coils is equal to the sum of the field generated by coil 1 and the field generated by coil 2: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3024.gif" /></center> (30.21) The total magnetic field generated by a pair of Helmholtz coils is shown in Figure 30.5 where also the contributions of the two coils are shown individually. We observe that the field is very homogeneous between the coils (0 < z < R).   <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3025.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.4. Magnetic field generated by a coil with R = 1 m.   </center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3026.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.5. Magnetic field generated by a pair of Helmholtz coils.   </center> <h3 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading6">Example: Problem 30.22</a></h3> A very long wire is bent at a right angle near its midpoint. One branch of it lies along the positive x-axis and the other along the positive y-axis (see Figure 30.6). The wire carries a current I. What is the magnetic field at a point in the first quadrant of the x-y plane ? <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3027.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.6. Problem 30.22   </center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3028.gif" /></center> <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> Figure 30.7. Field generated by wire.  </center> The first step to solve this problem is to look at the magnetic field produced by this single wire (see Figure 30.7). The direction of the magnetic field generated by a small segment of the wire is pointing out of the paper. The magnitude of the field, dB, is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3029.gif" /></center> (30.22) The position x of the segment under consideration is determined by the angle a: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3030.gif" /></center> (30.23) or <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3031.gif" /></center> (30.24) From eq.(30.24) we can obtain a relation between dx and da: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3032.gif" /></center> (30.25) Furthermore, <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3033.gif" /></center> (30.26) Substituting eq.(30.25) and eq.(30.26) into eq(30.22) we obtain <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3034.gif" /></center> (30.27) The total field can be obtained by integrating eq.(30.27) over the wire. The integration limits are <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3035.gif" /></center> (30.28) and <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3036.gif" /></center> (30.29) The result of the integration is <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3037.gif" /></center> (30.30) The field of the vertical wire can be obtained in a similar fashion: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3038.gif" /></center> (30.31) The magnitude of the total field is thus equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3039.gif" /></center> (30.32) <h2 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading7">30.3. The magnetic dipole</a></h2> The magnetic field on the axis of a current loop was discussed in Problem 30.33. At large distances from the current loop (z >> R) the field is approximately equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3040.gif" /></center> (30.33) which shows that the magnetic field strength decreases as 1/z3. This dependence of the magnetic field strength on distance is similar to the dependence observed for the electric field strength of an electric dipole: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3041.gif" /></center> (30.34) Equation (30.33) is often rewritten as <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3042.gif" /></center> (30.35) where <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3043.gif" /></center> (30.36) is called the magnetic dipole moment of the loop. In general, the dipole moment of a current loop is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3044.gif" /></center> (30.37) Magnetic dipole moments exist for objects as small as electrons and as large as the earth. <h3 style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; "> <a name="Heading8">Example: Problem 30.44</a></h3> An amount of charge Q is uniformly distributed over a disk of paper of radius R. The disk spins about its axis with angular velocity [omega]. Find the magnetic dipole moment of the disk. The first step to solve this problem is to determine the dipole moment of a ring of the disk, with radius r and with a width dr. The amount of charge dq on this ring is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3045.gif" /></center> (30.38) The angular velocity of the disk is [omega] and its period T is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3046.gif" /></center> (30.39) During one period the charge dq will pass any given point on the ring. The current dI is thus equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3047.gif" /></center> (30.40) The magnetic dipole moment du of the ring is equal to <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3048.gif" /></center> (30.41) The total dipole moment of the disk can be found by integrating eq.(30.41) between r = 0 and r = R: <center style="color: rgb(0, 0, 0); font-family: 'Times New Roman'; font-size: medium; "> <img border="1" src="http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter30/Chapter3049.gif" /></center> (30.42)

Click for more details

Latest Study Materials

By Course

By Specialization

By Specialization :

A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
Others

By Course :

A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
T
U
V
W
X
Y
Z
Others

POST BLOG

What is there for you?

Study material
Colleges are sharing lecture notes, study material, file, assignment etc.
Clarify Doubts
Ask any study related doubt and get answered by college students and faculty.
Educational videos
College student sharing a great video to help peers study.
Previous Year Papers
Access previous year papers for various courses.
Info. update by students of the respective college only.