A particle with a positive charge of 3 C moves upward at a speed of 10 m/s. It passes simultaneously through a magnetic field of 0.2 T directed into the page and an electric field of 2 N/C directed to the right. How is the motion of the particle affected? Answering this question is a matter of calculating the force exerted by the magnetic field and the force exerted by the electric field, and then adding them together. The force exerted by the magnetic field is: Using the right-hand-rule, we find that this force is directed to the left. The force exerted by the electric field is: This force is directed to the right. In sum, we have one force of 6 N pushing the particle to the left and one force of 6 N pushing the particle to the right. The net force on the particle is zero, so it continues toward the top of the page with a constant velocity of 10 m/s. Magnetic Force on Current-Carrying Wires Since an electric current is just a bunch of moving charges, wires carrying current will be subject to a force when in a magnetic field. When dealing with a current in a wire, we obviously can’t use units of q and v. However, qv can equally be expressed in terms of Il, where I is the current in a wire, and l is the length, in meters, of the wire—both qv and Il are expressed in units of C · m/s. So we can reformulate the equation for the magnitude of a magnetic force in order to apply it to a current-carrying wire: In this formulation, is the angle the wire makes with the magnetic field. We determine the direction of the force by using the right-hand rule between the direction of the current and that of the magnetic field. 251 EXAMPLE In the figure above, a magnetic field of T is applied locally to one part of an electric circuit with a 5 resistor and a voltage of 30 V. The length of wire to which the magnetic field is applied is 2 m. What is the magnetic force acting on that stretch of wire? We are only interested in the stretch of wire on the right, where the current flows in a downward direction. The direction of current is perpendicular to the magnetic field, which is directed into the page, so we know the magnetic force will have a magnitude of F = IlB, and will be directed to the right. We have been told the magnetic field strength and the length of the wire, but we need to calculate the current in the wire. We know the circuit has a voltage of 30 V and a resistance of 5 , so calculating the current is just a matter of applying Ohm’s Law: Now that we know the current, we can simply plug numbers into the equation for the force of a magnetic field on a current-carrying wire: The Magnetic Field Due to a Current So far we have discussed the effect a magnetic field has on a moving charge, but we have not discussed the reverse: the fact that a moving charge, or current, can generate a magnetic field. There’s no time like the present, so let’s get to it. The magnetic field created by a single moving charge is actually quite complicated, and is not covered by SAT II Physics. However, the magnetic field created by a long straight wire carrying a current, I, is relatively simple, and is fair game for SAT II Physics. The magnetic field strength is given by: 252 The constant is called the permeability of free space, and in a vacuum it has a value of about N/A 2 . For SAT II Physics, it’s not important to memorize this equation exactly. It’s more important to note that the strength of the magnetic field is proportional to the strength of the current and is weaker the farther it is from the wire. The direction of the magnetic field lines are determined by an alternate version of the right-hand rule: if you held the wire with your thumb pointing in the direction of the current, the magnetic field would make a circular path around the wire, in the direction that your fingers curl. EXAMPLE Two parallel long straight wires carrying a current I stand a distance r apart. What force does one wire exert on the other? Consider the magnetic field created by the bottom wire as it affects the top wire. According to the right-hand rule, the magnetic field will point out of the page, and will have a strength of B = ( I)/(2πr). The force exerted by the bottom wire on the top wire is F = IlB. If we substitute in for B the equation we derived above, we find the force per unit length is: Using the right-hand rule once more, we find that the force pulls the top wire down toward the bottom wire. We can apply the same equations to find that the top wire pulls the bottom wire up. In other words, the two wires generate magnetic fields that pull one another toward each 253 other. Interestingly, the fact that each wire exerts an opposite force on the other is further evidence of Newton’s Third Law. Key Formulas Magnetic Force on a Moving Charge Magnitude of the Magnetic Force on a Moving Charge Radius of the Circle Described by a Charged Particle Moving Perpendicula r to a Magnetic Field Magnetic Force on a Current Magnetic Field Created by a Current Practice Questions 254 1. . The pointer on a compass is the north pole of a small magnet. If a compass were placed next to a bar magnet, as shown above, in what direction would the pointer point? (A) (B) (C) (D) (E) 2. . A positively charged particle in a uniform magnetic field moves in a circular path in the clockwise direction, parallel to the plane of the page. In what direction do the magnetic field lines point? (A) Out of the page (B) Into the page (C) To the left (D) To the right (E) In a clockwise pattern parallel to the plane of the page 3. . What should one do to maximize the magnitude of the magnetic force acting on a charged particle moving in a magnetic field? I. Maximize the strength of the magnetic field II. Minimize the particle’s velocity III. Ensure that the particle is moving in the same direction as the magnetic field lines (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, and III 255 4. . What is the magnetic force experienced by a negatively charged particle of 1.0 C that is moving upward at a velocity of 2.0 10 3 m/s in a magnetic field of strength 4.0 10 –4 T, directed into the page? (A) 0.8 N to the left (B) 0.8 N to the right (C) 2.0 10 –7 N to the left (D) 2.0 10 –7 N to the right (E) 5.0 10 6 N to the left 5. . A charged particle is moving in a circular orbit in a magnetic field. If the strength of the magnetic field doubles, how does the radius of the particle’s orbit change? (A) It is quartered (B) It is halved (C) It is unchanged (D) It is doubled (E) It is quadrupled 6. . Which of the following is not a possible trajectory of a charged particle in a uniform magnetic field? (A) (B) (C) (D) (E) 256 7. . A positively charged particle of 2.0 C moves upward into an area where both a magnetic field and an electric field are acting. The magnetic field has a magnitude of 4.0 10 –4 T and the electric field has a magnitude of 0.1 N/C. At what velocity must the particle be moving if it is not deflected when it enters this area? (A) 4.0 10 –3 m/s (B) 125 m/s (C) 250 m/s (D) 500 m/s (E) The particle will be deflected to the left regardless of its velocity 8. . A current-carrying wire in a magnetic field is subject to a magnetic force. If the current in the wire is doubled, what happens to the magnetic force acting on the wire? (A) It is quartered (B) It is halved (C) It is unchanged (D) It is doubled (E) It is quadrupled 257 9. . Two wires carry current in opposite directions. Which of the following graphs represents the magnetic force acting on each wire? (A) (B) (C) (D) (E) There is no net force acting on either wire 10. . A current-carrying wire passes through a uniform magnetic field, as shown above. At which point is the magnetic field the strongest? (A) A (B) B (C) C (D) D (E) The magnetic field strength is uniform throughout Explanations 258 1. B To solve this problem, it is helpful to remember how the magnetic field lines around a bar magnet look: The arrows of the magnetic field lines show the direction toward which a north magnetic pole would be attracted. Since the compass needle is a south magnetic pole, it’s attracted in the opposite direction of the field lines. Note that the correct answer is B, and not E. The magnet points along the magnetic field lines, and not straight at the north pole of the magnet. 2. A This question demands that we apply the right-hand rule backward. Force, velocity, and magnetic strength are related by the formula . Since the particle is positively charged, q is positive, and the F vector will point in the same direction as the vector. Let’s imagine the particle at the six o’clock position. That means the particle is moving to the left, so stretch your fingers in the leftward direction. It’s moving under the influence of a centripetal magnetic force that pulls it in a circle. This force is directed toward the center of the circle, so point your thumb upward toward the center of the imaginary clock face. To do this, you’ll have to have your palm facing up, and you’ll find that when you curl your fingers around, they point out of the plane of the page. That’s the direction of the magnetic field lines. 3. A The magnetic force experienced by a moving particle is given by the formula . Since F is proportional to the cross product of v and B, we can maximize F by maximizing v and B, and by ensuring that v and B are perpendicular to one another. According to these requirements, only statement I will maximize the magnetic force: both statements II and III will serve to minimize the magnetic force. 4. B Magnetic force is related to charge, velocity, and magnetic field strength by the formula . Since the velocity vector and the magnetic field strength vector are perpendicular, we can calculate the magnitude of the magnetic force quite easily: 259 The minus sign in the answer signifies the fact that we are dealing with a negatively charged particle. That means that the force is in the opposite direction of the vector. We can determine the direction of this vector using the right-hand rule: point your fingers upward in the direction of the v vector and curl them downward in the direction of the B vector; your thumb will be pointing to the left. Since we’re dealing with a negatively charged particle, it will experience a force directed to the right. 5. B If the particle is moving in a circular orbit, its velocity is perpendicular to the magnetic field lines, and so the magnetic force acting on the particle has a magnitude given by the equation F = qvB. Since this force pulls the particle in a circular orbit, we can also describe the force with the formula for centripetal force: F = mv 2 /r. By equating these two formulas, we can get an expression for orbital radius, r, in terms of magnetic field strength, B: Since magnetic field strength is inversely proportional to orbital radius, doubling the magnetic field strength means halving the orbital radius. 6. D When a charged particle moves in the direction of the magnetic field lines, it experiences no magnetic force, and so continues in a straight line, as depicted in A and B. When a charged particle moves perpendicular to the magnetic field lines, it moves in a circle, as depicted in C. When a charged particle has a trajectory that is neither perfectly parallel nor perfectly perpendicular to the magnetic field lines, it moves in a helix pattern, as depicted in E. However, there are no circumstances in which a particle that remains in a uniform magnetic field goes from a curved trajectory to a straight trajectory, as in D. 7. C The electric field will pull the charged particle to the left with a force of magnitude F = qE. The magnetic field will exert a force of magnitude F = qvB. The direction of this force can be determined using the right-hand rule: extend your fingers upward in the direction of the velocity vector, then point them out of the page in the direction of the magnetic field vector. You will find your thumb is pointing to the right, and so a positively charged particle will experience a magnetic force to the right. 260 [...]... determined from the information given here 26 9 2 A bar of length 2 cm slides along metal rails at a speed of 1 cm/s The bar and rails are in a magnetic field of 2 T, pointing out of the page What is the induced emf in the bar and rails? (A) V (B) V (C) V (D) V (E) V 3 A wire in the shape of an equilateral triangle with sides of length 1.00 m sits in a magnetic field of 2. 00 T, pointing to the right What... current will change direction and flow clockwise It doesn’t matter whether the magnet or the loop is moving, so long as one is moving relative to the other 26 7 Applications Electromagnetic induction is important to humans because it is useful SAT II Physics has been known to ask questions about real-world applications of electromagnetic induction The two most common applications are the electric generator... be strongest at point C Electromagnetic Induction 26 1 CHARGES MOVING IN A MAGNETIC FIELD create an electric field, just as charges moving in an electric field create a magnetic field This is called electromagnetic induction Induction provides the basis of everyday technology like transformers on power lines and electric generators On average, SAT II Physics asks only one question about electromagnetic... pointing to the right What is the magnitude of the magnetic flux through the triangle? (A) 0 Wb (B) 1.00 Wb (C) 1. 73 Wb (D) 2. 00 Wb (E) 3. 46 Wb 4 A device that transforms mechanical energy into electrical energy is called a: (A) Transformer (B) Inductor (C) Motor (D) Galvanometer (E) Generator 27 0 5 A wire carrying 5.0 V is applied to a transformer The primary coil has 5 turns and the secondary coil has... the string oscillates up and down once every half second The frequency is just the reciprocal of the period: f = 1/T = 2 Hz DETERMINING WAVELENGTH: The maximum positive displacement of the mass’s oscillation signifies a wave crest Since each crest is 1 .25 m apart, the wavelength, , is 1 .25 m DETERMINING WAVE SPEED: Given the frequency and the wavelength, we can also calculate the wave speed: m/s Phase... area, with a magnitude equal to the area in question If we imagine flux graphically, it is a measure of the number and length of flux lines passing through a certain area 26 3 The unit of flux is the weber (Wb), where 1 Wb = 1 T · m2 Changing Magnetic Flux As we will see shortly, is more important than : our interest is in how flux changes, not in its fixed value The formula for magnetic flux suggests... of time, t: 27 2 In this equation, A is the amplitude, f is the frequency, and T is the period of the oscillation It is useful to think of each of these quantities in terms of a graph plotting the mass’s displacement over time The graph shows us an object moving back and forth withina distance of 1 m from its equilibrium position It reaches its equilibrium position of x = 0 at t = 0, t = 2, and t = 4... beach, the air molecules transmitting sound, etc., the medium through which these waves are transmitted 2 The medium itself is not propagated: For the “wave” to work, each person in the stadium only needs to stand up and sit back down The “wave” travels around the stadium, but the people do not 27 3 Think of waves as a means of transmitting energy over a distance One object can transmit energy to another... and wavelength, of a particular wave, we can calculate the wave speed, v: EXAMPLE 27 4 Ernst attaches a stretched string to a mass that oscillates up and down once every half second, sending waves out across the string He notices that each time the mass reaches the maximum positive displacement of its oscillation, the last wave crest has just reached a bead attached to the string 1 .25 m away What are... Law by introducing a minus sign: EXAMPLE 26 5 The square in the previous example, with sides of length 2 m and in a magnetic field of strength 10 T, is rotated by 60º in the course of 4 s What is the induced emf in the square? In what direction does the current flow? We established in the previous example that the change in flux as the square is rotated is 20 Wb Knowing that it takes 4 seconds to . the particle’s velocity III. Ensure that the particle is moving in the same direction as the magnetic field lines (A) I only (B) I and II only (C) I and III only (D) II and III only (E) I, II, . magnetic field strength is given by: 25 2 The constant is called the permeability of free space, and in a vacuum it has a value of about N/A 2 . For SAT II Physics, it’s not important to memorize. and is not covered by SAT II Physics. However, the magnetic field created by a long straight wire carrying a current, I, is relatively simple, and is fair game for SAT II Physics. The magnetic