HW20Due: 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. and . In the following questions , This force is nonintuitive, as it involves the vector product (or cross product) of the vectors 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 forefinger of your right hand in the direction of be pointing in the direction of . , and point your middle finger in the direction of point your . Your thumb will then 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., ANSWER: Direction of = - ). (Recall that is written x_unit.) 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 magnitude at angle moving in the xy plane with velocity (i.e., with 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, ANSWER: , , and . Direction of = Correct ANSWER: = Answer not displayed Express the magnetic force in terms of given variables like . Note that it must be perpendicular to both and . . Express your answer in terms of . ANSWER: = Correct Part E Now consider the example of a positive charge magnitude in the +z direction.Part D First find the magnitude of the force magnetic field (of magnitude on a positive charge in the case that the velocity (of magnitude ) and the ) are perpendicular. .2 Relevant component of velocity Of course. and other quantities given in the problem statement. 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 ). . . and unit vectors. and other quantities given in the problem statement. . the magnetic force is proportional to the component of velocity perpendicular to the magnetic field. .1 Direction of force . However. this problem can be solved by simply applying a rule for finding the vector product. Hint F. . another useful way to think about it is to realize that only the component of velocity perpendicular to the field generates any force. They follow the trajectories illustrated in the figure. ANSWER: . . and unit vectors. = 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. Express your answer in terms of . ANSWER: Direction of = Correct Hint F. Find the direction of the force Express the direction of the force using unit vectors. ANSWER: =0 Correct There is no magnetic force on a charge moving parallel or antiparallel to the magnetic field. the component of velocity perpendicular to . Find . Express your answer in terms of . Equivalently. the magnitude of the magnetic force acting on the particle. Find moving in the -z direction with speed with the local magnetic field of . Find the magnetic force on the charge. Rank from largest to smallest.1 Neutral particles Since the magnitude of the magnetic force acting on a particle is given by . C. Also. B. In our scenario. ANSWER: . the acceleration must equal the expression for centripetal acceleration: . so circular path. since the particle moves along a This can be solved for velocity to yield . a neutral particle (with ANSWER: ) will not experience a magnetic force. ANSWER: View Correct Part D Rank the particles A. the speed of a particle can be determined by measuring the radius of its circular path in a known magnetic field. the velocity and field vectors are perpendicular. overlap them. and E on the basis of their speed.1 Determining velocity based on particle trajectories A charged particle moving in a uniform magnetic field follows a circular trajectory. Rank from largest to smallest. the magnetic force acting on the particle must be equal to the product of its mass and acceleration: . By Newton's second law. Thus.Part A Which particle (if any) is neutral? Hint A.1 Find 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. assuming that you also know the charge and mass of the particle. overlap them. To rank items as equivalent. particle A particle B particle C particle D particle E none Correct Part B Which particle (if any) is negatively charged? Hint B. Hint C. . To rank items as equivalent. does uniquely define ? ANSWER: yes no Correct Part B Calculate the x component of the velocity of the particle. overlap them. if you do it in the most straightforward manner.1 Relation between and Hint not displayed .50 is moving in a uniform magnetic field of 3.60×10 −7 . Rank the particles A.ANSWER: View Correct Part E Now assume that particles A. Hint C.1 Relation between and Which component of the force depends on the x component of the velocity? ANSWER: x Answer not displayed y Express your answer in meters per second to three significant figures. GBA Determining the Velocity of a Charged Particle A particle with a charge of particle is measured to be Part A Are there components of the velocity that cannot be determined by measuring the force? Hint A. If you know .30×10 −7 7. C. To rank items as equivalent. ANSWER: View Correct For Part D of "Determining the Velocity of a Charged Particle". ANSWER: = -106 Correct Part C Calculate the y component of the velocity of the particle. Hint B. and E on the basis of their speed. 1.30 ) . B. and E all have the same magnitude of electric charge. Hint E.1 Charged particle trajectories in magnetic fields Hint not displayed Rank from largest to smallest. The magnetic force on the Recall the following formula: . 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. C.1 Magnetic force on a moving charged particle 5. B. If the charge is positive. then plug in numbers after you've simplified the The dot product of two vectors Part E What is the angle between Hint E. what is the direction of the force on the particle due to the magnetic field? Hint A. If positive. Bend your middle finger so that it is perpendicular to your thumb and index finger. where is the angle between and . Hint D. This result is important because it implies that magnetic fields can only change the direction of a charged particle's velocity. a magnetic field of magnitude problem. A particle of charge and mass moves in a region of space where there is a uniform magnetic field in the +z direction). Express your answer in watts to three significant figures.1 Formula for dot product and is given by . 3. neglect any forces on the particle other than the magnetic force. and ? Another dot product formula Express your answer in degrees to three significant figures. This algebraic sign is determined by the right-hand rule. Since be perpendicular to both and . To employ the right-hand rule: 1. 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. ANSWER: 90 Correct . This force is directed perpendicular to both the velocity vector and the magnetic field vector at the point of interaction. This will always be the case. Part A At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +z direction). In this (i.1 Recall that .Express your answer in meters per second to three significant figures. . 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. ANSWER: 0 Correct . The requirement that the force be perpendicular to both of the other vectors specifies the direction of the force to within an algebraic sign.. must Notice that the dot product of the velocity and the force is zero. Spread your right thumb and index finger apart by 90 degrees. ANSWER: = -46. your middle finger is now pointing in the direction of the force as shown in the figure. 2.2 Correct Part D Calculate the scalar product symbolic expression. not its speed. Work the problem out symbolically first.1 The right-hand rule for magnetic force is A charged particle moving through a region of magnetic field experiences a magnetic force.e. the angular Hint D. 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. if only the magnetic field acts on the particle.3 Determine the acceleration of the particle Hint not displayed Hint D. then against it. the direction of the force will ____________. etc.2 Determine the magnetic force Hint not displayed Hint D. 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. 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 frequency of the circular motion? .4 Express the angular speed in terms of the linear speed Hint not displayed . As a consequence.1 How to approach the problem Hint not displayed Hint D. at a later time. then along it.If the charge is negative. Therefore. its kinetic energy 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. the force is in the direction opposite your middle finger. Express in terms of . 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. and the applied voltage. Part A With what speed Hint A. the magnetic field. and . Although it appeared in the equations of force and motion. . In the cyclotron. . Mass Spectrometer J.2 Relationship of and Hint not displayed Hint B. Another important contribution of his was the invention. it canceled out. The ion begins at potential and is accelerated toward zero potential.3 Putting it all together Hint not displayed Express ANSWER: = Correct in terms of . and any necessary constants. together with one of his students.3 Final energy Hint not displayed Find the speed in terms of ANSWER: = . This implies that the frequency (but not the linear speed) of the particle is invariant with orbit size. . What is Hint B. J.1 does the ion exit the acceleration region? Suggested general method Hint not displayed Hint A. . When the particle exits the region with the electric field it will have obtained a speed . 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). Correct Part B After being accelerated. was based on the fact that the frequency of a charged particle orbiting in a uniform field is independent of the radius. and any constants. The results of this experiment allow one to find in terms of the experimentally measured quantities such as the particle radius. 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.2 Initial energy Hint not displayed Hint A. The first particle accelerator built. The ratio of mass to (positive) charge of an ion may be accurately determined in a mass spectrometer. ANSWER: = Correct Note that this result for the frequency does not depend on the radius of the circle. . . Thomson is best known for his discoveries about the nature of cathode rays. of the mass spectrometer. In essence. the cyclotron.1 ? Cyclotron frequency Hint not displayed Hint B. 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. 9%.54 out of a possible total of 40 points.By sending atoms of various elements through a mass spectrometer. 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. Francis Aston. . You received 39. discovered that some elements actually contained atoms with several different masses. Score Summary: Your score on this assignment is 98. Thomson's student.