MP_Assignment#1_soln

MP_Assignment#1_soln - Assignments Student View 2/15/11...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignments Student View 2/15/11 10:26 AM Summary View Diagnostics View Print View withEdit Assignment Answers Settings per Student MP_Assignment#1 [ Print ] Due: 8:00am on Monday, February 7, 2011 Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy The Cyclotron Description: Several qualitative and quantitative questions: radius and frequency of revolution, energy of the particles. Focuses on classical model but last follow - up statement makes reference to relativistic model. Learning Goal: To learn the basic physics and applications of cyclotrons. Particle accelerators are used to create well - controlled beams of high - energy particles. Such beams have many uses, both in research and industry. One common type of accelerator is the cyclotron , as shown in the figure. In a cyclotron, a magnetic field confines charged particles to circular paths while an oscillating electric field accelerates them. It is useful to understand the details of this process. Consider a cyclotron in which a beam of particles of positive charge and mass is moving along a circular path restricted by the magnetic field (which is perpendicular to the velocity of the particles). Part A Before entering the cyclotron, the particles are accelerated by a potential difference . Find the speed with which the particles enter the cyclotron. Express your answer in terms of , , and . ANSWER: = Part B http://session.masteringphysics.com/myct/assignments Page 1 of 32 Assignments 2/15/11 10:26 AM Find the radius Hint B.1 of the circular path followed by the particles. The magnitude of the magnetic field is Find the force Find the magnitude of the force acting on the particles. Express your answer in terms of ANSWER: Hint B.2 . , , , and . You may or may not use all these variables. = Find the acceleration Find the magnitude of the acceleration Express your answer in terms of of the particles. and the radius of revolution . ANSWER: = Hint B.3 Solving for Use Newton's second law to construct an equation to be solved for Express your answer in terms of , , , and . . ANSWER: = Part C Find the period of revolution Hint C.1 for the particles. Relationship between and The period is the amount of time it takes a particle to make one complete orbit. Since the speed of the particle is constant, the period will be equal to the distance the particle travels in one orbit divided by the particle's speed: . Express your answer in terms of , , and . ANSWER: http://session.masteringphysics.com/myct/assignments Page 2 of 32 Assignments 2/15/11 10:26 AM = Note that the period does not depend on the particle's speed (nor, therefore, on its kinetic energy). Part D Find the angular frequency Hint D.1 of the particles. Relationship between and Recall that the frequency of revolution is equal to ; it represents the number of revolutions a particle makes per second. The angular frequency is equal to ; it is the number of radians the particle traverses per second. Express your answer in terms of , , and . ANSWER: = Part E Your goal is to accelerate the particles to kinetic energy . What minimum radius of the cyclotron is required? Hint E.1 Find in terms of Find an expression for the speed of a particle in terms of its kinetic energy Express your answer in terms of and . . ANSWER: = You can now substitute this value for Express your answer in terms of , into your expression for , , and from Part B. . ANSWER: = http://session.masteringphysics.com/myct/assignments Page 3 of 32 Assignments 2/15/11 10:26 AM Part F If you can build a cyclotron with twice the radius, by what factor would the allowed maximum particle energy increase ? Assume that the magnetic field remains the same. Hint F.1 Find in terms of Using your result from Part E, solve for Express your answer in terms of , in terms of , , and . . ANSWER: = In other words, the kinetic energy ANSWER: is proportional to . 2 4 8 16 Part G If you can build a cyclotron with twice the radius and with the magnetic field twice as strong, by what factor would the allowed maximum particle energy increase ? Hint G.1 Find in terms of and Using your result from Part E, solve for Express your answer in terms of , in terms of , , and and . . ANSWER: = It's now clear that the kinetic energy is proportional to . ANSWER: http://session.masteringphysics.com/myct/assignments Page 4 of 32 Assignments 2/15/11 10:26 AM ANSWER: 2 4 8 16 One limitation of the cyclotron has to do with the failure of the laws of classical mechanics to accurately predict the behavior of high - energy particles. When their speeds become comparable to the speed of light , the angular frequency is no longer what you determined in Part D. Using special relativity, one can show that the angular frequency is actually given by the formula . As you can see, the frequency drops as the energy (and speed) increases; the particles' motion falls out of phase with the pulsating voltage, restricting the cyclotron's ability to accelerate the particles further Magnetic Field from Two Wires Description: Calculate, at several different locations, the net magnetic field due to two straight infinite wires carrying anti- parallel currents. Students are asked to look for a pattern in the results as the points of interest become more and more remote from the wires. Learning Goal: To understand how to use the principle of superposition in conjunction with the Biot - Savart (or Ampere's) law. From the Biot - Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by , where ( ) is the permeability constant, is the current in the wire, and is the distance from the wire to the location at which the magnitude of the magnetic field is being calculated. The same result can be obtained from Ampere's law as well. The direction of vector can be found using the right - hand rule. (Take care in applying the right - hand rule. Many students mistakenly use their left hand while applying the right - hand rule since those who use their right hand for writing sometimes automatically use their "pencil - free hand" to determine the direction of .) In this problem, you will be asked to calculate the magnetic field due to a set of two wires with antiparallel currents as shown in the diagram . Each of the wires carries a current of magnitude . The current in wire 1 is directed out of the page and that in wire 2 is directed into the page. The distance between the http://session.masteringphysics.com/myct/assignments Page 5 of 32 Assignments wires is 2/15/11 10:26 AM . The x axis is perpendicular to the line connecting the wires and is equidistant from the wires. As you answer the questions posed here, try to look for a pattern in your answers. Part A Which of the vectors best represents the direction of the magnetic field created at point K (see the diagram in the problem introduction) by wire 1 alone? Enter the number of the vector with the appropriate direction. ANSWER: 3 Part B Which of the vectors best represents the direction of the magnetic field created at point K by wire 2 alone? Enter the number of the vector with the appropriate direction. http://session.masteringphysics.com/myct/assignments Page 6 of 32 Assignments ANSWER: 2/15/11 10:26 AM 3 Part C Which of these vectors best represents the direction of the net magnetic field created at point K by both wires ? Enter the number of the vector with the appropriate direction. ANSWER: 3 Part D Find the magnitude of the magnetic field Express your answer in terms of , created at point K by wire 1. , and appropriate constants. ANSWER: = Of course, because point K is equidistant from the wires. Part E http://session.masteringphysics.com/myct/assignments Page 7 of 32 Assignments 2/15/11 10:26 AM Find the magnitude of the net magnetic field Express your answer in terms of , created at point K by both wires. , and appropriate constants. ANSWER: = This result is fairly obvious because of the symmetry of the problem: At point K, the two wires each contribute equally to the magnetic field. At points L and M you should also consider the symmetry of the problem. However, be careful! The vectors will add up in a more complex way. Part F Point L is located a distance magnetic field Hint F.1 from the midpoint between the two wires. Find the magnitude of the created at point L by wire 1. How to approach the problem Use the distances provided and the Pythagorean Theorem to find the distance between wire 1 and point L. Express your answer in terms of , , and appropriate constants. ANSWER: = Part G Point L is located a distance magnetic field Hint G.1 from the midpoint between the two wires. Find the magnitude of the net created at point L by both wires. How to approach the problem Sketch a detailed diagram with all angles marked; draw vectors and ; then add them using the parallelogram rule. Hint G.2 Find the direction of the magnetic field due to wire 1 Which of the vectors best represents the direction of the magnetic field created at point L (see the diagram in the problem introduction) by wire 1 alone? Enter the number of the vector with the appropriate direction. http://session.masteringphysics.com/myct/assignments Page 8 of 32 Assignments 2/15/11 10:26 AM ANSWER: Hint G.3 2 Find the direction of the magnetic field due to wire 2 Which of the vectors best represents the direction of the magnetic field created at point L by wire 2 alone? Enter the number of the vector with the appropriate direction. ANSWER: Hint G.4 4 Find the direction of the net magnetic field Which of the vectors best represents the direction of the net magnetic field created at point L by both wires ? Enter the number of the vector with the appropriate direction. http://session.masteringphysics.com/myct/assignments Page 9 of 32 Assignments 2/15/11 10:26 AM ANSWER: 3 Note that the directions of the magnetic fields created by individual wires at point L are different from each other and from those at point K; however, the direction of the net magnetic field at points K and L is the same. Hint G.5 Angle between magnetic field due to wire 1 and the x axis Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point L and the x axis. Hint G.6 Find the angle between magnetic field due to wire 1 and the x axis Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point L and the x axis. Express your answer numerically, in degrees. ANSWER: Hint G.7 = Net magnetic field Consider the symmetry of the magnetic field at point L due to wire 1 and the magnetic field due to wire 2. You should note that the y components of these two vectors are of equal magnitude but are opposite in direction. Therefore they will cancel when added together, leaving you only to worry about the x components. Find the x component of the magnetic field at point L due to wire 1 by using the magnitude of the vector (found in Part F) and the angle between the x axis and the magnetic field vector (found in the previous hint). Because of symmetry, the x component of the magnetic field at point L due to wire 2 is the same size. To find the net magnetic field at point L you need to add together the x components of the magnetic field at point L due to wire 1 and of the magnetic field due to wire 2. Express your answer in terms of http://session.masteringphysics.com/myct/assignments , , and appropriate constants. Page 10 of 32 Assignments 2/15/11 10:26 AM ANSWER: = Part H Point M is located a distance magnetic field from the midpoint between the two wires. Find the magnitude of the created at point M by wire 1. Express your answer in terms of , , and appropriate constants. ANSWER: = Part I Find the magnitude of the net magnetic field Hint I.1 created at point M by both wires. How to approach the problem Sketch a detailed diagram with all angles marked; draw vectors and ; then add them using the parallelogram rule. Hint I.2 Find the direction of the magnetic field due to wire 1 Which of the vectors best represents the direction of the magnetic field created at point M by wire 1 alone? Enter the number of the vector with the appropriate direction. http://session.masteringphysics.com/myct/assignments Page 11 of 32 Assignments 2/15/11 10:26 AM ANSWER: Hint I.3 2 Find the direction of the net magnetic field Which of the vectors best represents the direction of the net magnetic field created at point M by both wires ? Enter the number of the vector with the appropriate direction. ANSWER: Hint I.4 3 Angle between magnetic field due to wire 1 and the x axis Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point M and the x axis. Hint I.5 Find the angle between magnetic field due to wire 1 and the x axis Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point M and the x axis. Express your answer numerically, in degrees. ANSWER: Hint I.6 = Net magnetic field Consider the symmetry of the magnetic field at point M due to wire 1 and the magnetic field due to wire 2. You should note that the y components of these two vectors are of equal magnitude but are opposite in direction. Therefore they will cancel when added together, leaving you only to worry about the x http://session.masteringphysics.com/myct/assignments Page 12 of 32 Assignments 2/15/11 10:26 AM components. Find the x component of the magnetic field at point M due to wire 1 by using the magnitude of the vector (found in Part H) and the angle between the x axis and the magnetic field vector (found in the previous hint). Because of symmetry, the x component of the magnetic field at point M due to wire 2 is the same size. To find the net magnetic field at point M you need to add together the x components of the magnetic field at point M due to wire 1 and of the magnetic field due to wire 2. Express your answer in terms of , , and appropriate constants. ANSWER: = Part J Finally, consider point X (not shown in the diagram) located on the x axis very far away in the positive x direction. Which of the vectors best represents the direction of the magnetic field created at point X by wire 1 alone? Enter the number of the vector with the appropriate direction. ANSWER: 1 Part K Which of the vectors best represents the direction of the magnetic field created at point X by wire 2 alone? Enter the number of the vector with the appropriate direction. http://session.masteringphysics.com/myct/assignments Page 13 of 32 Assignments ANSWER: 2/15/11 10:26 AM 5 As you can see, at a very large distance, the individual magnetic fields (and the corresponding magnetic field lines) created by the wires are directed nearly opposite to each other, thus ensuring that the net magnetic field is very, very small even compared to the magnitudes of the individual magnetic fields, which are also relatively small at a large distance from the wires. Thus, at a large distance, the magnetic fields due to the two wires almost cancel each other out! (That is, if point X is very far from each wire, then the ratio is very close to zero.) Another way to think of this is as follows: If you are really far from the wires, then it's hard to tell them apart. It would seem as if the current were traveling up and down, almost along the same line, thereby appearing much the same as a single wire with almost no net current (because the up and down currents almost cancel each other), and therefore almost no magnetic field. Note that this only works for points very far from the wires; otherwsie it's easy to tell that the wires are separated and the currents don't cancel, since they are going up and down at different locations. It comes as no surprise then that one way to eliminate unnecessary magnetic fields in electric circuits is to twist together the wires carrying equal currents in opposite directions. Current Sheet Description: Find the magnetic field above a uniform current sheet using Ampère's law Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current through each wire. There are runs in the - y direction wires per unit length in the x direction. http://session.masteringphysics.com/myct/assignments Page 14 of 32 Assignments 2/15/11 10:26 AM Part A Write an expression for Use , the magnetic field a distance above the xy plane of the sheet. for the permeability of free space. Hint A.1 How to approach the problem You will need to use Ampère's law: . The first step in applying Ampère's law is to choose an appropriate Ampèrean loop. Because you are trying to find the magnetic field a distance above the sheet, a good choice for the Ampèrean loop is a rectangle of width Hint A.2 and height as shown. Find How much current is enclosed by the Ampèrean loop given in the first hint? Answer in terms of variables given in the problem introduction. ANSWER: = Hint A.3 Determine the direction of the magnetic field Above the sheet, in which direction does the magnetic field point ? (Be careful that your answer has the correct sign.) Hint A.3.1 Direction of a field from a single wire The magnetic field generated by a current running through a single wire in the - y direction cannot have any component in what direction ? http://session.masteringphysics.com/myct/assignments Page 15 of 32 Assignments 2/15/11 10:26 AM ANSWER: Hint A.3.2 Direction of the total field From the answer for the field from a single wire we know that each wire generates a magnetic field with components in the x and z directions. In this problem, the magnetic field in one of these directions generated by any wire is canceled out exactly by the magnetic field generated by the other wires. Which component cancels ? Hint A.3.2.1 A figure The figure shows the fields due to two wires on opposite sides of a point above the wire. ANSWER: Give your answer in terms of the unit vectors , , and . ANSWER: Hint A.4 Magnitude of the magnetic field http://session.masteringphysics.com/myct/assignments Page 16 of 32 Assignments 2/15/11 10:26 AM Because the magnetic field points in the direction (above the sheet) and the the sheet), the line integral in Ampère's law, direction (below , does not depend on contributions from the sides of the loop (which run in the z direction). In addition, the current enclosed by the loop does not depend on the length of the sides of the loop. This means that the quantity appears nowhere in Ampère's law for this problem, and therefore the magnitude of the magnetic field does not vary as a function of height above or below the sheet. By symmetry, the magnitude of the magnetic field also does not vary as a function of xy position. (Because the sheet is infinite, any xy point above the sheet is equivalent to every other.) Following this line of reasoning we conclude that the magnitude of the magnetic field is constant everywhere outside the sheet. Hint A.5 Evaluate What is the value of Use evaluated around the Ampèrean loop shown in the figure ? for the (constant) magnitude of the magnetic field. ANSWER: = Now equate the left - hand side of Ampère's law (which you just found) to the right - hand side (involving ) and solve for in terms of given variables. Express the magnetic field as a vector in terms of any or all of the following: the unit vectors , , and/or , , , , , and . ANSWER: = This equation is analogous to on either side of a infinitely charged sheet. The correspondence seems more obvious if you set the current per unit length . Then the magnetic field you just calculated is . The electric field, though, points along the perpendicular to the surface. Do you see why you had to pick the rean loop you used ? That is, why would any other loop not have worked. Did you notice that by using Ampère's law you could find the field by using a much simpler integral than Biot - Savart's law ? The drawback is that you may not always be able to find a convenient loop in situations where the current distribution is more complicated. http://session.masteringphysics.com/myct/assignments Page 17 of 32 Assignments 2/15/11 10:26 AM Force between Moving Charges Description: compare electric and magnetic forces between moving charges. 1 part requires Biot Savart law. Two point charges, with charges and , are each moving with speed toward the origin. At the instant shown is at position (0, ) and is at ( , 0). (Note that the signs of the charges are not given because they are not needed to determine the magnitude of the forces between the charges.) Part A What is the magnitude of the electric force between the two charges ? Hint A.1 Which law to use Apply Coulomb's law: where is the distance between the two charges. Hint A.2 Find the value of What is the value of for the given situation ? Express your answer in terms of . ANSWER: = Express in terms of , , , and . ANSWER: http://session.masteringphysics.com/myct/assignments Page 18 of 32 Assignments 2/15/11 10:26 AM ANSWER: = Part B What is the magnitude of the magnetic force on Hint B.1 due to the magnetic field caused by How to approach the problem First, find the magnetic field generated by charge magnetic force on due to the field of . Hint B.2 ? at the position of charge . Then evaluate the Magnitude of the magnetic field The Biot - Savart law, which gives the magnetic field produced by a moving charge, can be written , where is the permeability of free space and magnetic field is produced. Note we have is the vector from the charge to the point where the in the numerator, not , necessitating an extra power of in the denominator. Using this equation find the expression for the magnitude of the magnetic field experienced by charge due to charge . Hint B.2.1 Determine the cross product What is the magnitude of ? For any two vectors, , where is the angle between the vectors. Because, in this case, Substitute the appropriate value of Express your answer in terms of ANSWER: Note that is 45 degrees, for this problem, to arrive at a surprisingly simple answer. and . = is cubed in the denominator of the Biot - Savart law. Express the magnitude of the magnetic field of (at the location of ) in terms of , , , and . http://session.masteringphysics.com/myct/assignments Page 19 of 32 Assignments 2/15/11 10:26 AM ANSWER: = Hint B.3 Find the direction of the magnetic field Which of the following best describes the direction of the magnetic field from at according to the Biot - Savart law, the field must be perpendicular to both and . Ignore the effects of the sign of ANSWER: ? Remember, . (along the x axis) (along the y axis) (along the z axis into or out of the screen) Hint B.4 Computing the force You can evaluate the force exerted on a moving charge by a magnetic field using the Lorentz force law: , where is the force on the moving charge, charge, and is the magnetic field, is the velocity of the charge. Note that, as long as Express the magnitude of the magnetic force in terms of , is the charge of the moving and , , are perpendicular, , and . . ANSWER: = Part C Assuming that the charges are moving nonrelativistically ( ), what can you say about the relationship between the magnitudes of the magnetic and electrostatic forces ? Hint C.1 How to approach the problem Determine which force has a greater magnitude by finding the ratio of the electric force to the magnetic force and then applying the approximation. Recall that . ANSWER: The magnitude of the magnetic force is greater than the magnitude of the electric force. http://session.masteringphysics.com/myct/assignments Page 20 of 32 Assignments 2/15/11 10:26 AM The magnitude of the electric force is greater than the magnitude of the magnetic force. Both forces have the same magnitude. This result holds quite generally: Magnetic forces between moving charges are much smaller than electric forces as long as the speeds of the charges are nonrelativistic. ± Magnetic Force on a Current- Carrying Wire Description: ± Includes Math Remediation. Practice calculating the magnetic force (both magnitude and direction using the right - hand rule) on a straight current - carrying wire in a uniform external magnetic field. Learning Goal: To understand the magnetic force on a straight current - carrying wire in a uniform magnetic field. Magnetic fields exert forces on moving charged particles, whether those charges are moving independently or are confined to a current - carrying wire. The magnetic force on an individual moving charged particle depends on its velocity and charge . In the case of a current - carrying wire, many charged particles are simultaneously in motion, so the magnetic force depends on the total current and the length of the wire The size of the magnetic force on a straight wire of length in a uniform magnetic field with strength carrying current . is . Here is the angle between the direction of the current (along the wire) and the direction of the magnetic field. Hence refers to the component of the magnetic field that is perpendicular to the wire, Thus this equation can also be written as . . The direction of the magnetic force on the wire can be described using a "right- hand rule." This will be discussed after Part B. Part A Consider a wire of length = 0.50 = 0.30 that runs north- south on a horizontal surface. There is a current of flowing north in the wire. (The rest of the circuit, which actually delivers this current, is not shown.) The Earth's magnetic field at this location has a magnitude of 0.50 (or, in SI units, ) and points north and 38 degrees down from the horizontal, toward the ground. What is the size of the magnetic force on the wire due to the Earth's magnetic field? In considering the agreement of units, recall that . Express your answer in newtons to two significant figures. http://session.masteringphysics.com/myct/assignments Page 21 of 32 Assignments 2/15/11 10:26 AM ANSWER: Because the Earth's magnetic field is quite modest, this force is so small that it might be hard to detect. Part B Now assume that a strong, uniform magnetic field of size 0.55 pointing straight down is applied. What is the size of the magnetic force on the wire due to this applied magnetic field? Ignore the effect of the Earth's magnetic field. Hint B.1 Determining the angle theta Recall that is the angle between the magnetic field direction and the direction of the current flow in the wire. In this situation, what is the value of ANSWER: ? 0 degrees 38 degrees 90 degrees 180 degrees Express your answer in newtons to two significant figures. ANSWER: This force would be noticeable if the wire were of light weight. The direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a "right- hand rule" to help you determine the direction of the magnetic force. Straighten the fingers of your right hand and point them in the direction of the current. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field. Your thumb now points in the direction of the http://session.masteringphysics.com/myct/assignments Page 22 of 32 Assignments 2/15/11 10:26 AM magnetic force acting on the wire. Part C What is the direction of the magnetic force acting on the wire in Part B due to the applied magnetic field? ANSWER: due north due south due east due west straight up straight down Part D Which of the following situations would result in a magnetic force on the wire that points due north? Check all that apply. ANSWER: Current in the wire flows straight down; the magnetic field points due west. Current in the wire flows straight up; the magnetic field points due east. Current in the wire flows due east; the magnetic field points straight down. Current in the wire flows due west and slightly up; the magnetic field points due east. Current in the wire flows due west and slightly down; the magnetic field points straight down. As you can see, many current/magnetic field configurations can result in the same direction of magnetic force. Part E Assume that the applied magnetic field of size 0.55 is rotated so that it points horizontally due south. What is the size of the magnetic force on the wire due to the applied magnetic field now? Hint E.1 Determining the angle theta http://session.masteringphysics.com/myct/assignments Page 23 of 32 Assignments Recall that 2/15/11 10:26 AM is the angle between the magnetic field direction and the direction of the current flow in the wire. In this situation, what is the value of ANSWER: ? 0 degrees 38 degrees 90 degrees 180 degrees Express your answer in newtons to two significant figures. ANSWER: Notice that whenever the current in the wire and the magnetic field point in the same direction ( ) or in opposite directions ( ), the sine of is zero, so there is no magnetic force exerted on the wire. This is consistent with the earlier statement that it is the component of the magnetic field that is perpendicular to the direction of the current that produces the magnetic force. Also notice that for these two special values of (when the current is flowing parallel to or antiparallel to the magnetic field) the steps listed for the right - hand rule suggest a unique direction for the magnetic force. This is another clue that the magnetic force is zero. Electromagnetic Velocity Filter Description: Find the velocity of a charged particle that is undeflected in crossed electric and magnetic fields. Look at relation between mass, charge, and acceleration as charged particle traverses the fields. When a particle with charge side. If the proper electric field moves across a magnetic field of magnitude , it experiences a force to the is simultaneously applied, the electric force on the charge will be in such a direction as to cancel the magnetic force with the result that the particle will travel in a straight line. The balancing condition provides a relationship involving the velocity of the particle. In this problem you will figure out how to arrange the fields to create this balance and then determine this relationship. Part A Consider the arrangement of ion source and electric field plates shown in the figure. The ion source sends particles with velocity along the positive x axis. They encounter electric field plates spaced a distance apart that generate a uniform electric field of magnitude in the + y direction. To cancel the resulting electric force with a magnetic force, a magnetic field (not shown) must be added in which direction ? Using the right - hand rule, you can see that the positive z axis is directed out of the screen. http://session.masteringphysics.com/myct/assignments Page 24 of 32 Assignments 2/15/11 10:26 AM Hint A.1 Method for determining direction Assume a sign for the charge. Since both the electric force and magnetic force depend on , in particular, they also depend on its sign. So the sign doesn't matter here. Apply the right - hand rule to the equation for the magnetic force, . Hint A.2 Right - hand rule Curl the fingers of your right hand from the first vector to the second in the product. Your outstretched thumb then points in the direction of the cross - product vector. Choose the direction of . ANSWER: Part B Now find the magnitude of the magnetic field that will cause the charge to travel in a straight line under the combined action of electric and magnetic fields. http://session.masteringphysics.com/myct/assignments Page 25 of 32 Assignments 2/15/11 10:26 AM Hint B.1 Find the magnetic force What is , the magnitude of the force due to a magnetic field with a charge Express moving at velocity (a speed of ) interacting )? in terms of some or all of the variables ANSWER: (with a magnitude of , , and . = Hint B.2 Find the force due to the electric field What is , the magnitude of a force on a charge Express due to an applied electric field in terms of one or both of the variables ANSWER: . = Express the magnetic field all of the variables and ? , , and that will just balance the applied electric field in terms of some or . ANSWER: = Part C It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, would the velocity of a neutral particle be selected by passage through this device ? ) The explanation of this is that the mass and the charge control the resolution of the device - particles with the wrong velocity will be accelerated away from the straight line and will not pass through the exit slit. If the acceleration depends strongly on the velocity, then particles with just slightly wrong velocities will feel a substantial transverse acceleration and will not exit the selector. Because the acceleration depends on the mass and charge, these influence the sharpness (resolution) of the transmitted particles. Assume that you want a velocity selector that will allow particles of velocity to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with __________. Hint C.1 Use Newton's law http://session.masteringphysics.com/myct/assignments Page 26 of 32 Assignments 2/15/11 10:26 AM If the velocity is "wrong" the forces won't balance and the resulting transverse force will cause a transverse acceleration. Use to determine how this acceleration will depend on and . You want particles with the incorrect velocity to have the maximum possible deviation in the y direction so that they will not go through a slit placed at the right end. This means that the acceleration should be maximum. Assume that the selector is short enough so that particles that move away from the axis do not have time to come back to it. ANSWER: both and large large and small small and large both and small You want particles with the incorrect velocity to have the maximum possible deviation in the y direction so that they will not go through a slit placed at the right end. The deviation will be maximum when the acceleration is maximum. The acceleration is directly proportional to and inversely proportional to : . So for maximum deviation, should be large and small. Force between an Infinitely Long Wire and a Square Loop Description: Compute the force on a square, current - carrying loop of wire due to the magnetic field of a nearby infinite wire. (Hints mention Ampère's Law as a tool to calculate the magnetic field.) A square loop of wire with side length carries a current . The center of the loop is located a distance from an infinite wire carrying a current . The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown. http://session.masteringphysics.com/myct/assignments Page 27 of 32 Assignments 2/15/11 10:26 AM Part A What is the magnitude, Hint A.1 , of the net force on the loop ? How to approach the problem You need to find the total force as the sum of the forces on each straight segment of the wire loop. You'll save some work if you think ahead of time about which forces might cancel. Hint A.2 Determine the direction of force Which of the following diagrams correctly indicates the direction of the force on each individual line segment? Hint A.2.1 Direction of the magnetic field In the region of the loop, the magnetic field points into the plane of the paper (by the right - hand rule). Hint A.2.2 Formula for the force on a current- carrying conductor The magnetic force on a straight wire segment of length , carrying a current magnetic field with a uniform along its length, is , where is a vector along the wire in the direction of the current. ANSWER: a b c d http://session.masteringphysics.com/myct/assignments Page 28 of 32 Assignments 2/15/11 10:26 AM Determine the magnitude of force Hint A.3 Which of the following diagrams correctly indicates the relative magnitudes of the forces on the parallel wire segments ? Hint A.3.1 Find the magnetic field due to the wire What is the magnitude, the wire, , of the wire's magnetic field as a function of perpendicular distance from . Hint A.3.1.1 Ampère's law Use Ampère's law to obtain the magnetic field. Ampère's law states that , where the line integral can be done around any closed loop. Express the magnetic field magnitude in terms of , , and . ANSWER: = ANSWER: a b c d http://session.masteringphysics.com/myct/assignments Page 29 of 32 Assignments 2/15/11 10:26 AM Hint A.4 Find the force on the section of the loop closest to the wire What is the magnitude of the force on the section of the loop closest to the wire, that is, a distance from it ? Hint A.4.1 Formula for the force on a current- carrying conductor The magnetic force on a straight wire segment of length , carrying a current magnetic field with a uniform along its length, is , where is a vector along the wire in the direction of the current. Hint A.4.2 Find the magnetic field due to the wire What is the magnitude, the wire, , of the wire's magnetic field as a function of perpendicular distance from . Hint A.4.2.1 Ampère's law Use Ampère's law to obtain the magnetic field. Ampère's law states that , where the line integral can be done around any closed loop. Express the magnetic field magnitude in terms of , , and . ANSWER: = Express your answer in terms of , , , , and . ANSWER: = Hint A.5 Find the magnetic field due to the wire What is the magnitude, wire, , of the wire's magnetic field as a function of perpendicular distance from the . Hint A.5.1 Ampère's law Use Ampère's law to obtain the magnetic field. Ampère's law states that http://session.masteringphysics.com/myct/assignments Page 30 of 32 Assignments 2/15/11 10:26 AM , where the line integral can be done around any closed loop. Express the magnetic field magnitude in terms of , , and . ANSWER: = Express the force in terms of , , , , and . ANSWER: = Part B The magnetic moment of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current flowing in it ( the plane in which the current flows. Find the magnitude, ), and whose direction is perpendicular to , of the force on the loop from Part A in terms of the magnitude of its magnetic moment. Express in terms of , , , , and . ANSWER: = The direction of the net force would be reversed if the direction of the current in either the wire or the loop were reversed. The general result is that "like currents" (i.e., currents in the same direction) attract each other (or, more correctly, cause the wires to attract each other), whereas oppositely directed currents repel. Here, since the like currents were closer to each other than the unlike ones, the net force was attractive. The corresponding situation for an electric dipole is shown in the figure below. http://session.masteringphysics.com/myct/assignments Page 31 of 32 Assignments 2/15/11 10:26 AM Score Summary: Your score on this assignment is 0%. You received 0 out of a possible total of 8 points. http://session.masteringphysics.com/myct/assignments Page 32 of 32 ...
View Full Document

This note was uploaded on 12/28/2011 for the course PHYSICS 270 taught by Professor Drake during the Fall '08 term at Maryland.

Ask a homework question - tutors are online