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- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW20.pdf

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HW20 Due: 11:59pm on Tuesday, October 20, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] Force on Moving Charges in a Magnetic Field Learning Goal: To understand the force on a charge moving in a magnetic field. Magnets exert forces on other magnets even though they are separated by some distance. Usually the force on a magnet (or piece of magnetized matter) is pictured as the interaction of that magnet with the magnetic field at its location (the field being generated by other magnets or currents). More fundamentally, the force arises from the interaction of individual moving charges within a magnet with the local magnetic field. This force is written , where is the force, is the individual charge (which can be negative), is its velocity, and is the local magnetic field. This force is nonintuitive, as it involves the vector product (or cross product) of the vectors and . In the following questions we assume that the coordinate system being used has the conventional arrangement of the axes, such that it satisfies , where , , and are the unit vectors along the respective axes. Let's go through the right-hand rule. Starting with the generic vector cross-product equation point your forefinger of your right hand in the direction of , and point your middle finger in the direction of . Your thumb will then be pointing in the direction of . Part A Consider the specific example of a positive charge moving in the +x direction with the local magnetic field in the +y direction. In which direction is the magnetic force acting on the particle? Express your answer using unit vectors (e.g., - ). (Recall that is written x_unit.) ANSWER: Direction of = Correct Part B Now consider the example of a positive charge moving in the +x direction with the local magnetic field in the +z direction. In which direction is the magnetic force acting on the particle? Express your answer using unit vectors. ANSWER: Direction of = Correct Part C Now consider the example of a positive charge moving in the xy plane with velocity (i.e., with magnitude at angle with respect to the x axis). If the local magnetic field is in the +z direction, what is the direction of the magnetic force acting on the particle? Hint C.1 Finding the cross product The direction can be found by any of the usual means of finding the cross product: 1. Use the determinant expression for the cross product. (See your math or physics text.) 2. Use the general definition , where any term with the three directions in the normal order of xyz or any cyclical permutation (e.g., yzx or zxy) has a positive sign, and terms with the other order (xzy, zyx, or yxz) have a negative sign. Express the direction of the force in terms of , as a linear combination of unit vectors, , , and . ANSWER: Direction of = Correct Part D First find the magnitude of the force on a positive charge in the case that the velocity (of magnitude ) and the magnetic field (of magnitude ) are perpendicular. Express your answer in terms of , , , and other quantities given in the problem statement. ANSWER: = Correct Part E Now consider the example of a positive charge moving in the -z direction with speed with the local magnetic field of magnitude in the +z direction. Find , the magnitude of the magnetic force acting on the particle. Express your answer in terms of , , , and other quantities given in the problem statement. ANSWER: = 0 Correct There is no magnetic force on a charge moving parallel or antiparallel to the magnetic field. Equivalently, the magnetic force is proportional to the component of velocity perpendicular to the magnetic field. Part F Now consider the case in which the positive charge is moving in the yz plane with a speed at an angle with the z axis as shown (with the magnetic field still in the +z direction with magnitude ). Find the magnetic force on the charge. Hint F.1 Direction of force Find the direction of the force . Note that it must be perpendicular to both and . Express the direction of the force using unit vectors. ANSWER: Direction of = Correct Hint F.2 Relevant component of velocity Of course, this problem can be solved by simply applying a rule for finding the vector product. However, another useful way to think about it is to realize that only the component of velocity perpendicular to the field generates any force. Find , the component of velocity perpendicular to . Express your answer in terms of , , and unit vectors. ANSWER: = Answer not displayed Express the magnetic force in terms of given variables like , , , , and unit vectors. ANSWER: = Correct Charged Particles Moving in a Magnetic Field Ranking Task Five equal-mass particles (A–E) enter a region of uniform magnetic field directed into the page. They follow the trajectories illustrated in the figure. Part A Which particle (if any) is neutral? Hint A.1 Neutral particles Since the magnitude of the magnetic force acting on a particle is given by , a neutral particle (with ) will not experience a magnetic force. ANSWER: particle A particle B particle C particle D particle E none Correct Part B Which particle (if any) is negatively charged? Hint B.1Find the direction of the magnetic force Hint not displayed ANSWER: particle A particle B particle C particle D particle E none Correct Part C Rank the particles on the basis of their speed. Hint C.1 Determining velocity based on particle trajectories A charged particle moving in a uniform magnetic field follows a circular trajectory. By Newton's second law, the magnetic force acting on the particle must be equal to the product of its mass and acceleration: . In our scenario, the velocity and field vectors are perpendicular, so . Also, since the particle moves along a circular path, the acceleration must equal the expression for centripetal acceleration: . This can be solved for velocity to yield . Thus, the speed of a particle can be determined by measuring the radius of its circular path in a known magnetic field, assuming that you also know the charge and mass of the particle. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Correct Part D Rank the particles A, B, C, and E on the basis of their speed. Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: ANSWER: View Correct Part E Now assume that particles A, B, C, and E all have the same magnitude of electric charge. Rank the particles A, B, C, and E on the basis of their speed. Hint E.1Charged particle trajectories in magnetic fields Hint not displayed Rank from largest to smallest. To rank items as equivalent, overlap them. ANSWER: View Correct For Part D of "Determining the Velocity of a Charged Particle", if you do it in the most straightforward manner, it is essential that you keep all numbers to full precision in your calculator until you plug into your final expression for the dot product of force and velocity. GBA Determining the Velocity of a Charged Particle A particle with a charge of 5.50 is moving in a uniform magnetic field of 1.30 ) . The magnetic force on the particle is measured to be 3.30×10 −7 7.60×10 −7 . Part A Are there components of the velocity that cannot be determined by measuring the force? Hint A.1 Magnetic force on a moving charged particle Recall the following formula: . If you know , does uniquely define ? ANSWER: yes no Correct Part B Calculate the x component of the velocity of the particle. Hint B.1 Relation between and Which component of the force depends on the x component of the velocity? ANSWER: x y Answer not displayed Express your answer in meters per second to three significant figures. ANSWER: = -106 Correct Part C Calculate the y component of the velocity of the particle. Hint C.1 Relation between and Hint not displayed Express your answer in meters per second to three significant figures. ANSWER: = -46.2 Correct Part D Calculate the scalar product . Work the problem out symbolically first, then plug in numbers after you've simplified the symbolic expression. Hint D.1 Formula for dot product The dot product of two vectors and is given by . Express your answer in watts to three significant figures. ANSWER: 0 Correct Part E What is the angle between and ? Hint E.1Another dot product formula Recall that , where is the angle between and . Express your answer in degrees to three significant figures. ANSWER: 90 Correct Notice that the dot product of the velocity and the force is zero. This will always be the case. Since , must be perpendicular to both and . This result is important because it implies that magnetic fields can only change the direction of a charged particle's velocity, not its speed. Charge Moving in a Cyclotron Orbit Learning Goal: To understand why charged particles move in circles perpendicular to a magnetic field and why the frequency is an invariant. A particle of charge and mass moves in a region of space where there is a uniform magnetic field (i.e., a magnetic field of magnitude in the +z direction). In this problem, neglect any forces on the particle other than the magnetic force. Part A At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). If is positive, what is the direction of the force on the particle due to the magnetic field? Hint A.1 The right-hand rule for magnetic force A charged particle moving through a region of magnetic field experiences a magnetic force. This force is directed perpendicular to both the velocity vector and the magnetic field vector at the point of interaction. The requirement that the force be perpendicular to both of the other vectors specifies the direction of the force to within an algebraic sign. This algebraic sign is determined by the right-hand rule. To employ the right-hand rule: 1. Spread your right thumb and index finger apart by 90 degrees. 2. Bend your middle finger so that it is perpendicular to your thumb and index finger. 3. Orient your hand so that your thumb points in the direction of the velocity and your index finger in the direction of the magnetic field. If the charge is positive, your middle finger is now pointing in the direction of the force as shown in the figure. If the charge is negative, the force is in the direction opposite your middle finger. ANSWER: x direction x direction y direction y direction z direction z direction Correct Part B This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ____________. ANSWER: have a component along the direction of motion remain perpendicular to the direction of motion have a component against the direction of motion first have a component along the direction of motion; then against it; then along it; etc. Correct Part C The fact that the magnetic field generates a force perpendicular to the instantaneous velocity of the particle has implications for the work that the field does on the particle. As a consequence, if only the magnetic field acts on the particle, its kinetic energy will ____________. ANSWER: increase over time decrease over time remain constant oscillate Correct Part D The particle moves in a plane perpendicular to the magnetic field direction as shown in the figure. What is , the angular frequency of the circular motion? Hint D.1 How to approach the problem Hint not displayed Hint D.2 Determine the magnetic force Hint not displayed Hint D.3 Determine the acceleration of the particle Hint not displayed Hint D.4 Express the angular speed in terms of the linear speed Hint not displayed Express in terms of , , and . ANSWER: = Correct Note that this result for the frequency does not depend on the radius of the circle. Although it appeared in the equations of force and motion, it canceled out. This implies that the frequency (but not the linear speed) of the particle is invariant with orbit size. The first particle accelerator built, the cyclotron, was based on the fact that the frequency of a charged particle orbiting in a uniform field is independent of the radius. In the cyclotron, radio frequency voltage is applied across a gap between the two sides of the conducting vacuum chamber in which the protons circulate owing to an external magnetic field. Particles in phase with this voltage are accelerated each time they cross the gap (because the field reverses while they make half a circle) and reach energies of millions of electron volts after several thousand round trips. Mass Spectrometer J. J. Thomson is best known for his discoveries about the nature of cathode rays. Another important contribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio of mass to (positive) charge of an ion may be accurately determined in a mass spectrometer. In essence, the spectrometer consists of two regions: one that accelerates the ion through a potential and a second that measures its radius of curvature in a perpendicular magnetic field. The ion begins at potential and is accelerated toward zero potential. When the particle exits the region with the electric field it will have obtained a speed . Part A With what speed does the ion exit the acceleration region? Hint A.1 Suggested general method Hint not displayed Hint A.2 Initial energy Hint not displayed Hint A.3 Final energy Hint not displayed Find the speed in terms of , , , and any constants. ANSWER: = Correct Part B After being accelerated, the particle enters a uniform magnetic field of strength and travels in a circle of radius (determined by observing where it hits on a screen--as shown in the figure). The results of this experiment allow one to find in terms of the experimentally measured quantities such as the particle radius, the magnetic field, and the applied voltage. What is ? Hint B.1Cyclotron frequency Hint not displayed Hint B.2Relationship of and Hint not displayed Hint B.3Putting it all together Hint not displayed Express in terms of , , , and any necessary constants. ANSWER: = Correct By sending atoms of various elements through a mass spectrometer, Thomson's student, Francis Aston, discovered that some elements actually contained atoms with several different masses. Atoms of the same element with different masses can only be explained by the existence of a third subatomic particle in addition to protons and electrons: the neutron. Score Summary: Your score on this assignment is 98.9%. You received 39.54 out of a possible total of 40 points. clockwork MasteringPhysics: Assignment Print View

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