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Heimadæmi 6 Due: 11:00pm on Thursday, February 25, 2016 You will receive no credit for items you complete after the assignment is due. Grading Policy Problem 27.68 The rectangular loop shown in the figure is pivoted about the yaxis and carries a current of 15.0A in the direction indicated. Part A If the loop is in a uniform magnetic field with magnitude 0.48T in the + xdirection, find the magnitude of the torque required to hold the loop in the position shown. Express your answer using two significant figures. ANSWER: τ = 3.0×10−2 N ⋅ m Correct Part B What is the direction of the torque required to hold the loop in the position shown. ANSWER: +^i −^i +^ j −^ j +k ^ −k ^ Correct Part C Repeat part A for the case in which the field is in the − zdirection. Express your answer using two significant figures. ANSWER: τ = 1.7×10−2 N ⋅ m Correct Part D What is the direction of the torque required to hold the loop in the position shown. ANSWER: +^i −^i +^ j −^ j +k ^ −k ^ Correct Part E For the magnetic field in part A, what torque would be required if the loop were pivoted about an axis through its center, parallel to the yaxis? Express your answer using two significant figures. ANSWER: τ = 3.0×10−2 N ⋅ m Correct Part F For the magnetic field in part C, what torque would be required if the loop were pivoted about an axis through its center, parallel to the yaxis? Express your answer using two significant figures. ANSWER: τ = 1.7×10−2 N ⋅ m Correct Problem 27.22 Part A A thin copper rod that is 1.0 m long and has a mass of 0.050 kg is in a magnetic field of 0.10 T. What minimum current in the rod is needed in order for the magnetic force to cancel the weight of the rod? ANSWER: 4.9 A 7.6 A 2.5 A 9.8 A 1.2 A Correct Exercise 27.18 An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 × 10 −27 horizontally at 35.6 km/s enters a uniform, vertical, 1.80 T magnetic field. Part A What is the diameter of the path followed by this alpha particle? Express your answer with the appropriate units. ANSWER: = 0.821 mm d Correct Part B What effect does the magnetic field have on the speed of the particle? ANSWER: The speed remains constant. The speed increases with time. The speed decreases with time. ) traveling kg Correct Part C What is the magnitude of the acceleration of the alpha particle while it is in the magnetic field? Express your answer with the appropriate units. ANSWER: m = 3.09×1012 a s2 Correct Part D What is the direction of this acceleration? ANSWER: The acceleration is perpendicular to v and ⃗ and so is horizontal, out of the center of curvature of the ⃗ B particle’s path. The acceleration is perpendicular to v and ⃗ and so is horizontal, toward the center of curvature of the ⃗ B particle’s path. The acceleration is parallel to v and perpendicular to ⃗ and so is vertical, along the particle’s path. ⃗ B The acceleration is perpendicular to v and parallel to ⃗ and so is horizontal, along the particle’s path. ⃗ B Correct Part E Explain why the speed of the particle does not change even though an unbalanced external force acts on it. ANSWER: The unbalanced force → is parallel to so it changes the direction of but not its magnitude, which is the FB v ⃗ v ⃗ speed. The unbalanced force → is perpendicular to so it changes the direction of but not its magnitude, which v ⃗ FB v ⃗ is the speed. The unbalanced force → is parallel to so it changes the magnitude of but not its direction, which is the FB v ⃗ v ⃗ speed. The unbalanced force → is perpendicular to so it changes the magnitude of but not its direction, which FB is the speed. Correct v ⃗ v ⃗ Electromagnetic Velocity Filter When a particle with charge q moves across a magnetic field of magnitude B, it experiences a force to the side. If the proper electric field E⃗ 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 v 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 v along the positive x axis. They encounter electric field plates spaced a distance d apart that generate a uniform electric field of magnitude E 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 righthand rule, you can see that the positive z axis is directed out of the screen. ^ Choose the direction of B . Hint 1. Method for determining direction Assume a sign for the charge. Since both the electric force and magnetic force depend on q, in particular, they also depend on its sign. So the sign doesn't matter here. Apply the righthand rule to the equation for the ⃗ magnetic force, F M ⃗ = qv ⃗ × B . Hint 2. Righthand 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 crossproduct vector. ANSWER: ^i −^i ^ j −^ j ^ k −k ^ Correct 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. Express the magnetic field Bbal that will just balance the applied electric field in terms of some or all of the variables q, v , and E . Hint 1. Find the magnetic force What is FM , the magnitude of the force due to a magnetic field B⃗ (with a magnitude of B) interacting with a ⃗ charge q moving at velocity v (a speed of v )? Express FM in terms of some or all of the variables q, B, and v . ANSWER: FM = qBv Hint 2. Find the force due to the electric field What is FE , the magnitude of a force on a charge q due to an applied electric field E⃗ ? Express FE in terms of one or both of the variables q and E . ANSWER: FE = qE ANSWER: Bbal = E v Correct 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 deviceparticles 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 v 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 v . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with __________. 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. Hint 1. Use Newton's law If the velocity is "wrong" the forces won't balance and the resulting transverse force will cause a transverse acceleration. Use a = F /m to determine how this acceleration will depend on q and m. 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. ANSWER: both q and m large q large and m small q small and m large both q and m small Correct 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 q and inversely proportional to m: a= F ⃗ E +F m ⃗ M = q m ⃗ ⃗ (E + v ⃗ × B). So for maximum deviation, q should be large and m small. Exercise 27.49 The figure shows a portion of a silver ribbon with z1 = 11.0 mm and y1 = 0.20 mm , carrying a current of 140 A in the +x direction. The ribbon lies in a uniform magnetic field, in the y direction, with magnitude 0.91 T . Apply the simplified model of the Hall effect. Part A If there are 5.85×1028 free electrons per cubic meter, find the magnitude of the drift velocity of the electrons in the x direction. Express your answer using two significant figures. ANSWER: vd = 6.8×10−3 m/s Correct Part B Find the magnitude of the electric field in the zdirection due to the Hall effect. Express your answer using two significant figures. ANSWER: E = 6.2×10−3 V/m Correct Part C Find the direction of the electric field in the zdirection due to the Hall effect. ANSWER: −z +z Correct Part D Find the Hall emf. Express your answer using two significant figures. ANSWER: EHall = 6.8×10−5 V Correct Exercise 26.35 The resistance of a galvanometer coil is 30.0 Ω , and the current required for fullscale deflection is 500μA. Part A Show in a diagram how to convert the galvanometer to an ammeter reading 25.0 mA full scale. ANSWER: add a shunt resistor in parallel with the galvanometer coil add a resistor in series with the galvanometer coil Correct Part B Compute the shunt resistance. ANSWER: Rs = 0.612 Ω Correct Part C Show how to convert the galvanometer to a voltmeter reading 500 mV full scale. ANSWER: add a shunt resistor in parallel with the galvanometer coil add a resistor in series with the galvanometer coil Correct Part D Compute the series resistance. ANSWER: Rs = 970 Ω Correct Problem 26.21 Part A (a) For the circuit shown in the figure, determine the current in the 7.0Ω resistor. Express your answer using three significant figures. ANSWER: 1.55 A Correct Part B (b) For the circuit shown in the figure, determine the current in the 8.0Ω resistor. Express your answer using three significant figures. ANSWER: 1.27 A Correct Part C (c) For the circuit shown in the figure, determine the current in the 4.0Ω resistor. Express your answer using three significant figures. ANSWER: 0.284 A Correct PSS 27.1: Magnetic Forces Learning Goal: To practice ProblemSolving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v ⃗ = (3.00 × 10 magnetic field 4 ^ m/s) j . What are the magnitude and direction of the particle’s acceleration produced by a uniform ⃗ ^ ^ B = (1.63 T) i + (0.980 T) j ? ProblemSolving Strategy 27.1: Magnetic Forces IDENTIFY the relevant concepts: The righthand rule allows you to determine the magnetic force on a moving charged particle. SET UP the problem using the following steps: ⃗ ⃗ 1. Draw the velocity vector v and magnetic field B with their tails together so that you can visualize the plane in which these two vectors lie. 2. Identify the angle ϕ between the two vectors. 3. Identify the target variables. This may be the magnitude and direction of the force, the velocity, or the magnetic field. EXECUTE the solution as follows: 1. Express the magnetic force using the equation F ⃗ = F = qvB sin ϕ. ⃗ qv ⃗ × B . The magnitude of the force is given by ⃗ ⃗ ⃗ ⃗ 2. Remember thatF is perpendicular to the plane of the vectors v and B. The direction of v ⃗ × B is determined by the righthand rule. If q is negative, the force is opposite to v ⃗ × B⃗ . EVALUATE your answer: Whenever you can, solve the problem in two ways. Verify that the results agree. IDENTIFY the relevant concepts The problem asks for the acceleration of a moving charged particle. Since acceleration is related to force, you will need to determine the magnetic force acting on the particle. SET UP the problem using the following steps Part A ⃗ ⃗ Draw the velocity v and magnetic field B vectors. Since they have different units, their relative magnitudes aren't relevant. Be certain they have the correct orientations relative to the given coordinate system. The dot in the center of the image represents the particle. ^ Recall that ^i , ^ j , and k are the unit vectors in the x, y, and z directions, respectively. ANSWER: Correct The strategy points out that there are two ways to solve problems with magnetic forces. In this problem, you already have the components of the vectors, so the cross product method will be much easier. This means that ⃗ ⃗ you do not need to find the value of ϕ, the angle between v and B. Also, note that the coordinate system in the vector drawing applet is twodimensional. To make it three dimensional, add a positive z axis oriented out of the screen. EXECUTE the solution as follows Part B Find the acceleration vector for the charge. Enter the x, y, and z components of the acceleration in meters per second squared separated by commas. Hint 1. How to find cross products Recall that the cross product distributes like a regular scalar product: ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ A × (B + C ) = A × B + A × C You will also need to use the following relations for products of unit vectors: ^ ^ ^ i × j = k ^ ^ ^ j × i = −k ^ ^ ^ j ×k = i ^ ^ ^ k × j = −i ^ ^ ^ k× i = j ^ ^ ^ i × k = −j ^ Finally, remember that the cross product of any vector with itself is zero. For example, ^ j × j Hint 2. Find v ⃗ × B⃗ Calculate v ⃗ × B⃗ , in terms of its components. ⃗ . = 0 Enter the x, y, and z components of v ⃗ × B⃗ in tesla meters per second separated by commas. Hint 1. How to find cross products Recall that the cross product distributes like a regular scalar product: ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ A × (B + C ) = A × B + A × C You will also need to use the following relations for products of unit vectors: ^ ^ ^ i × j = k ^ ^ ^ j × i = −k ^ ^ ^ j ×k = i ^ ^ ^ k × j = −i ^ ^ ^ k× i = j ^ ^ ^ i × k = −j ^ Finally, remember that the cross product of any vector with itself is zero. For example, ^ j × j . = 0 ANSWER: = 0,0,−4.89×104 T ⋅ m/s ⃗ v ⃗ × B Correct ANSWER: = 0,0,0.330 m/s2 a⃗ Correct EVALUATE your answer Part C You can check your result by comparing its magnitude to the magnitude the acceleration would have if the particle's velocity had the same magnitude but it was perpendicular to the magnetic field. ⃗ ⃗ Find the value of the expression qvB/m (the magnitude of a⃗ when v is perpendicular to B), where q is the magnitude of the charge, v is the magnitude of the velocity, B is the magnitude of the magnetic field, and m is the mass of the particle. Express your answer in meters per second squared. Hint 1. Find the magnitude of the velocity ^ What is the value of v ? Remember that the magnitude of a vector v x ^i + v y ^ j + v z k is given by − − − − − − − − − − 2 2 2 √vx + vy + vz Express your answer in meters per second squared. ANSWER: v = 3.00×104 m/s2 Hint 2. Find the magnitude of the magnetic field ^ What is the value of B? Remember that the magnitude of a vector Bx ^i + By ^ j + Bz k is given by − −−−−−−−−−− √B 2 x +B 2 y +B 2 z Express your answer in teslas. ANSWER: B = 1.90 T ANSWER: = 0.385 m/s2 qvB/m Correct This quantity is of similar size to the magnitude of your answer from Part B. If you wanted to check precisely, you could find the value of ϕ and multiply the value you calculated above by sin ϕ. You would find that you had the same magnitude as the magnitude of the acceleration vector you found in Part B. Note that the magnitude of the ⃗ magnetic force, and therefore the magnitude of the particle's acceleration, is at its maximum when v is perpendicular to B⃗ , so it is not surprising that your answer to Part C is somewhat larger than the magnitude of the acceleration calculated in Part B. To check the direction of your answer from Part B, use the righthand rule. Point the fingers of your right hand ⃗ ⃗ parallel to v in your answer to Part A and then turn your wrist so that you can curl those fingers down toward B. You will find that your thumb points into the screen, which is the negative z direction. Thus, your answer from Part B has the proper direction as well as the proper magnitude. ± Determining the Velocity of a Charged Particle A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ = the particle is measured to be F ⃗ = −( 1.25 T )^ k. The magnetic force on −( 3.80×10−7 N )^i + ( 7.60×10−7 N )^ j. Part A Are there components of the velocity that cannot be determined by measuring the force? Hint 1. Magnetic force on a moving charged particle Recall the following formula: ⃗ ⃗ F = qv ⃗ × B . ⃗ B If you know , does ANSWER: yes no Correct ⃗ v ⃗ × B uniquely define ? v ⃗ Part B Calculate the x component of the velocity of the particle. Express your answer in meters per second to three significant figures. ⃗ ⃗ Hint 1. Relation between v and F Which component of the force depends on the x component of the velocity? ANSWER: x y ANSWER: vx = 119 m/s Correct Part C Calculate the y component of the velocity of the particle. Express your answer in meters per second to three significant figures. ⃗ ⃗ Hint 1. Relation between v and F Which component of the force depends on the y component of the velocity? ANSWER: x y ANSWER: vy = 59.6 m/s Correct Part D Calculate the scalar product v ⃗ ⋅ F ⃗ . Work the problem out symbolically first, then plug in numbers after you've simplified the symbolic expression. Express your answer in watts to three significant figures. Hint 1. Formula for dot product The dot product of two vectors A⃗ and B⃗ is given by ⃗ ⃗ A ⋅ B = A x Bx + A y By + A z Bz . ANSWER: 0 W Correct Part E ⃗ ⃗ What is the angle between v and F? Express your answer in degrees to three significant figures. Hint 1. Another dot product formula Recall that ⃗ ⃗ ∣∣ ⃗ ∣ ⃗ ∣ A ⋅ B = A B cos θ ∣ ∣∣ ∣ , where θ is the angle between A⃗ and B⃗ . ANSWER: 90 ∘ Correct Notice that the dot product of the velocity and the force is zero. This will always be the case. Since F ⃗ = ⃗ ⃗ qv ⃗ × B , ⃗ B ⃗ must be perpendicular to both v 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. F Exercise 27.38 A straight, vertical wire carries a current of 2.30 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has a magnitude of B = 0.594 T and is horizontal. Part A What is the magnitude of the magnetic force on a 1.00 the magnetic field direction is east? Express your answer with the appropriate units. ANSWER: F = 1.37×10−2 N section of the wire that is in this uniform magnetic field, if cm Correct Part B What is the direction of this magnetic force? ANSWER: west south north east Correct Part C What is the magnitude of the magnetic force on a 1.00 the magnetic field direction is south? section of the wire that is in this uniform magnetic field, if cm Express your answer with the appropriate units. ANSWER: F = 1.37×10−2 N Correct Part D What is the direction of this magnetic force? ANSWER: north east west south Correct Part E What is the magnitude of the magnetic force on a 1.00 the magnetic field direction is 30.0 ∘ south of west? Express your answer with the appropriate units. ANSWER: section of the wire that is in this uniform magnetic field, if cm F = 1.37×10−2 N Correct Part F What is the direction of this magnetic force? ANSWER: = 60.0 ∘ north of west ϕ Correct Score Summary: Your score on this assignment is 98.8%. You received 10.87 out of a possible total of 11 points.

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