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Name:_____________________________________________________ Section: A B Quiz 03A Solution: Magnetic Forces A moving charged particle, of charge q and with velocity vector v pointing into the x-‐y-‐plane ( = plane of the drawing), is subject to a magnetic field, with the magnetic field vector B in the x-‐y-‐plane, at an angle of +130o above the (+x)-‐direction. Find the direction of the magnetic force vector, F, acting on the particle. If F is in the x-‐y-‐plane: state the angle θ of F measured from the (+x)-‐direction, as a positive number (θ >0) if F points above the x-‐axis; else as a negative number (θ ≤0). If F is perpendicular to the x-‐y-‐plane: state either “F into x-‐y-‐plane” or “F out of x-‐y-‐plane” B B v F y-axis y-axis F v y-axis F B IL F F x-axis Parts (a), (b) x-axis Part (c) x-axis Parts (d), (e) .__________________________________________________________________________________________ (a) Assume the charged particle is a proton. F is perpendicular to B and v . Hence, F is in x-‐y-‐plane. Since proton charge q=+e > 0, vector qv points in the same direction as v: qv is into x-‐y-‐plane. Hence, by RH Rule, F points in the “red” direction in Fig. left panel: θ = 130o – 90o = + 40o __________________________________________________________________________________________ (b) Assume charged particle is an electron. F is perpendicular to B and v . Hence, F is in x-‐y-‐plane. Since electron charge q=–e < 0, vector qv points in opposite direction as v: qv is out of x-‐y-‐plane. Hence, by RH Rule, F points in the “blue” direction in Fig. left panel: θ = 130o – 270o = –140o __________________________________________________________________________________________ (c) Repeat parts (a) and (b), if the charged particle is traveling in the x-‐y-‐plane, into the (+x)-‐direction, assuming the same magnetic field vector, B: For either electron or proton, F, is perpendicular to B and v. Hence, F is perpendicular to x-‐y-‐plane. Proton: Since proton charge q=+e > 0, vector qv points in the same direction as v: qv is into (+x)-‐direction. Hence, by RH Rule, F points “out of x-‐y-‐plane”, see Fig. middle panel: Electron: Since electron charge q=–e < 0, vector qv points in opposite direction as v: qv is into (–x)-‐direction. Hence, by RH Rule, F points “into x-‐y-‐plane”, see Fig. middle panel: Name:_____________________________________________________ Section: A B __________________________________________________________________________________________ (d) Assuming the same magnetic field vector, B, as before, find the direction of the magnetic force vector, F, acting on a straight current carrying piece of wire, perpendicular to the x-‐y-‐plane, with the current I flowing out of the x-‐y-‐plane. F is perpendicular to B and IL. Also, vector IL points out of x-‐y-‐plane and B lies in x-‐y-‐plane at 130o above (+x)-‐direction. Hence, F lies in x-‐y-‐plane and, by RH Rule, F points in the “red” direction in Fig. right panel: θ = 130o – 270o = – 140o __________________________________________________________________________________________ (e) What is the strength of the force, F ≡ |F|, on the wire in part (d), if the current, I, is 5.0A, the wire is 3.0m long, and the magnetic field strength, B ≡ |B|, is 2.0T ? From F = I L × B follows: F ≡ |F| = I L B sin(θILB) Here I=5.0A, L≡|L|=3.0m, B≡|B|=2.0T and θILB is the angle enclosed between vector IL and vector B. Since B lies in the x-‐y-‐plane and IL is perpendicular to (out of) the x-‐y-‐plane, we have: θILB = angle(B, IL) = 90o ; sin(θILB) = sin(90o) = 1. Hence, F = (5.0 A) (3.0m) (2.0T) (1.0) = 30 N Name:_____________________________________________________ Section: A B Quiz 03B Solution: Magnetic Forces v02 A moving charged particle, of charge q and with velocity vector v pointing out of the x-‐y-‐plane ( = plane of the drawing), is subject to a magnetic field, with the magnetic field vector B in the x-‐y-‐plane, at an angle of +130o above the (+x)-‐direction. Find the direction of the magnetic force vector, F, acting on the particle. If F is in the x-‐y-‐plane: state the angle θ of F measured from the (+x)-‐direction, as a positive number (θ >0) if F points above the x-‐axis; else as a negative number (θ ≤0). If F is perpendicular to the x-‐y-‐plane: state either “F into x-‐y-‐plane” or “F out of x-‐y-‐plane” F B F F y-axis v B v y-axis y-axis B F IL F x-axis Parts (a), (b) x-axis Part (c) x-axis Parts (d), (e) .__________________________________________________________________________________________ (a) Assume the charged particle is an electron || F is perpendicular to B and v . Hence, F is in x-‐y-‐plane. Since electron charge q=–e < 0, vector qv points in opposite direction as v: qv is into x-‐y-‐plane. Hence, by RH Rule, F points in the “blue” direction in Fig. left panel: θ = 130o – 90o = + 40o __________________________________________________________________________________________ (b) Assume the charged particle is a proton || F is perpendicular to B and v . Hence, F is in x-‐y-‐plane. Since proton charge q=+e > 0, vector qv points in the same direction as v: qv is out of x-‐y-‐plane. Hence, by RH Rule, F points in the “red” direction in Fig. left panel: θ = 130o – 270o = – 140o __________________________________________________________________________________________ (c) Repeat parts (a) and (b), if the charged particle is traveling in the x-‐y-‐plane, into the (+y)-‐direction, assuming the same magnetic field vector, B. || Electron: Since electron charge q=–e < 0, vector qv points in opposite direction as v: qv is into (–x)-‐direction. Hence, by RH Rule, F points “into x-‐y-‐plane”, see Fig. middle panel: || Proton: Since proton charge q=+e > 0, vector qv points in the same direction as v: qv is into (+x)-‐direction. Hence, by RH Rule, F points “out of x-‐y-‐plane”, see Fig. middle panel: Name:_____________________________________________________ Section: A B __________________________________________________________________________________________ (d) Assuming the same magnetic field vector, B, as before, find the direction of the magnetic force vector, F, acting on a straight current carrying piece of wire, perpendicular to the x-‐y-‐plane, with the current I flowing into the x-‐y-‐plane. || F is perpendicular to B and IL. Also, vector IL points into x-‐y-‐plane and B lies in x-‐y-‐plane at 130o above (+x)-‐direction. Hence, F lies in x-‐y-‐plane and, by RH Rule, F points in the “red” direction in Fig. right panel: θ = 130o – 90o = + 40o __________________________________________________________________________________________ (e) What is the strength of the force, F ≡ |F|, on the wire in part (d), if the current, I, is 3.0A, the wire is 4.0m long, and the magnetic field strength, B ≡ |B|, is 2.0T ? || From F = I L × B follows: F ≡ |F| = I L B sin(θILB) Here I=3.0A, L≡|L|=4.0m, B≡|B|=2.0T and θILB is the angle enclosed between vector IL and vector B. Since B lies in the x-‐y-‐plane and IL is perpendicular to (out of) the x-‐y-‐plane, we have: θILB ≡ angle(B, IL) = 90o ; sin(θILB) = sin(90o) = 1. Hence, F = (3.0 A) (4.0m) (2.0T) (1.0) = 24 N __________________________________________________________________________________________ Name:_____________________________________________________ Section: A B Quiz 03C Solution: Charged Particles in a Magnetic Field Four different types of charged particles (ions), labeled P, Q, R, S, are mixed together in a particle beam, initially traveling straight upward at the same speed, in the absence of any force, as shown in Figure below. (Neglect gravity!) The particles then enter a non-‐zero uniform magnetic field, B, pointing out of plane of drawing, and the beam splits apart: each particle type now travels on a different semi-‐circular trajectory, with P-‐ and S-‐particles turning rightward and Q-‐ and R-‐particles turning leftward, as shown below. The particles carry charges qP , qQ , qR , qS and masses mP , mQ , mR , mS , for particles of type P, Q, R, S, respectively. vR vQ FR R Y B (out) Q FQ vS vP FP P S FS . B=0 __________________________________________________________________________________________ (a) Find the direction of a P-‐particle’s velocity vector, v, as it passes through the point marked “Y” on its semicircular trajectory. State your answer as one of these six: (A) leftward (B) rightward (C) upward (D) downward (E) out of plane of drawing (F) into plane of drawing || Recall from mechanics, PHYS1251/1211: The velocity vector v is always tangential to a particle’s trajectory and points in the direction of the particle’s motion. Since the P-‐particle turned rightward, after entry into the B-‐field, it travels clockwise along its semi-‐circular trajectory: v ≡ vP is rightward at point Y. __________________________________________________________________________________________ (b) Find the direction of the magnetic force F acting on a P-‐particle, as it passes through “Y”. State your answer for the direction as in part (a), using: (A) (B) (C) (D) downward (E) (F) . || Recall from mechanics, PHYS1251/1211: The force F acting on a particle traveling at constant speed on a (semi-‐)circle always points towards the center of that circle: it’s the “centripetal force”! For particle P at location Y, the center of that circle is directly below Y. Hence, at Y, the force vector F ≡ FP points downward. Name:_____________________________________________________ Section: A B __________________________________________________________________________________________ (c) Find the signs of the charges, qP , qQ , qR , qS , carried by type P, Q, R and S particles, respectively: qP : + or − ? qQ : + or − ? qR : + or − ? qS : + or − ? || Since B is perpendicular to the plane of drawing, the magnetic force vector (q v) × B lies always in the plane of drawing, for all particles at all points along their respective trajectories. Consider, e.g., the P-‐ particle velocity v ≡ vP and force F ≡ FP at point Y and assume, e.g., qP >0. Then, the vector qP v points in same direction as v, i.e., rightward, see part (a). By RH rule, the vector (qP v) × B points downward at Y, which is the correct direction for the force vector F, as required by part (b). On the other hand, if you assume qP <0, then the vector qP v points in the direction opposite to v, i.e., leftward. By RH rule, the vector (qP v) × B then points upward at Y, which is not the correct direction for the force vector F, as required by part (b). Hence, qP must be positive, qP >0, to get the magnetic force (qP v) × B to point in the direction required by the mechanics of circular motion, see part (b). By the same method you can find the signs of the other charges, qQ , qR , qS . _________________________________________________________________________________________ (d) Assume the magnitudes of all 4 particle charges are the same: |qP| = |qQ| =| qR|=|qS| . Rank order their masses, mP , mQ , mR , mS , from highest to lowest. Write, e.g.: mS > mR > mQ > mP if S is heaviest, R is 2nd heaviest, Q is 3rd heaviest, P is 4th heaviest. Hint: Compare their semi-‐circles! || The diameter, d, of the (semi-‐)circular trajectory for a particle of mass of m and charge magnitude |q|, traveling at speed v perpendicular to a uniform magnetic field B is: d ≡ 2r = 2mv/(|q| B) . Here, all 4 particle types have the same speed, v, and the same magnitude of charge, |q| ≡ |qP| = |qQ| = |qR|=|qS|, and are subject to the same magnetic field, B. Thus, d is directly proportional to the particle mass m: the larger d, the larger m. Now compare and rank order the semi-‐circular diameters of the 4 particle types shown in figure above: dP > dR > dQ > dS . Hence, it follows that mP > mR > mQ > mS . __________________________________________________________________________________________ (e) Assume all 4 particle masses are the same: mP = mQ = mR = mS . Rank order the magnitudes of their charges, |qP|, |qQ|, |qR|, |qS|, from lowest to highest. Write, e.g.: |qP| < |qQ| < |qR| < |qS|, if P has the lowest, Q the 2nd lowest, R the 3rd lowest, S the 4th lowest magnitude of charge. Hint: Compare their semi-‐circles! || The diameter, d, of the (semi-‐)circular trajectory for a particle of mass of m and charge magnitude |q|, traveling at speed v perpendicular to a uniform magnetic field B is: d ≡ 2r = 2mv/(|q| B) . Here, all 4 particle types have the same speed, v, and the same mass, m = mP = mQ = mR = mS , and are subject to the same magnetic field, B. Thus, d is inversely proportional to the magnitude of the particle charge |q|: the larger d, the smaller |q|. So, from the rank ordering the semi-‐circular diameters of the 4 particle types, see part (d) and figure above, dP > dR > dQ > dS , it follows that |qP| < |qR| < |qQ| < |qS| . __________________________________________________________________________________________ Name:_____________________________________________________ Section: A B Quiz 03D Solution: Charged Particles in a Magnetic Field Four different types of charged particles (ions), labeled P, Q, R, S, are mixed together in a particle beam, initially traveling straight upward at the same speed, in the absence of any force, as shown in Figure below. (Neglect gravity!) The particles then enter a non-‐zero uniform magnetic field, B, pointing out of plane of drawing, and the beam splits apart: each particle type now travels on a different semi-‐circular trajectory, with P-‐ and R-‐particles turning rightward and Q-‐ and S-‐particles turning leftward, as shown below. The particles carry charges qP , qQ , qR , qS and masses mP , mQ , mR , mS , for particles of type P, Q, R, S, respectively. vQ Y vS FQ Q S vP B (out) FS vR FP R FR P . B=0 .__________________________________________________________________________________________ (a) Find the direction of a Q-‐particle’s velocity vector, v, as it passes through the point marked “Y” on its semicircular trajectory. State your answer as one of these six: (A) leftward (B) rightward (C) upward (D) downward (E) out of plane of drawing (F) into plane of drawing || Recall from mechanics, PHYS1251/1211: The velocity vector v is always tangential to a particle’s trajectory and points in the direction of the particle’s motion. Since the P-‐particle turned rightward, after entry into the B-‐field, it travels counter-‐clockwise along its semi-‐circular trajectory: v ≡ vQ is lefttward at point Y. __________________________________________________________________________________________ (b) Find the direction of the magnetic force F acting on a Q-‐particle, as it passes through “Y”. State your answer for the direction as in part (a), using: (A) (B) (C) (D) downward (E) (F) || Recall from mechanics, PHYS1251/1211: The force F acting on a particle traveling at constant speed on a (semi-‐)circle always points towards the center of that circle: it’s the “centripetal force”! For particle Q at location Y, the center of that circle is directly below Y. Hence, at Y, the force vector F ≡ FQ points downward. Name:_____________________________________________________ Section: A B __________________________________________________________________________________________ (c) Find the signs of the charges, qP , qQ , qR , qS , carried by type P, Q, R and S particles, respectively: qP : + or − ? qQ : + or − ? qR : + or − ? qS : + or − ? || Since B is perpendicular to the plane of drawing, the magnetic force vector (q v) × B lies always in the plane of drawing, for all particles at all points along their respective trajectories. Consider, e.g., the Q-‐ particle velocity v ≡ vQ and force F≡ FQ at point Y and assume, e.g., qQ >0. Then, the vector qQ v points in the same direction as v, i.e., leftward, see part (a). By RH rule, the vector (qQ v) × B points upward at Y, which is not the correct direction for the force vector F, as required by part (b). On the other hand, if you assume qQ <0, then the vector qQ v points in the direction opposite to v, i.e., rightward. By RH rule, the vector (qQ v) × B then points downward at Y, which is the correct direction for the force vector F, as required by part (b). Hence, qQ must be negative, qQ <0, to get the magnetic force (qQ v) × B to point in the direction required by the mechanics of circular motion, see part (b). By the same method you can find the signs of the other charges, qP , qR , qS . __________________________________________________________________________________________ (d) Assume the magnitudes of all 4 particle charges are the same: |qP| = |qQ| =| qR|=|qS| . Rank order their masses, mP , mQ , mR , mS , from highest to lowest. Write, e.g.: mS > mR > mQ > mP if S is heaviest, R is 2nd heaviest, Q is 3rd heaviest, P is 4th heaviest. Hint: Compare their semi-‐circles! || The diameter, d, of the (semi-‐)circular trajectory for a particle of mass of m and charge magnitude |q|, traveling at speed v perpendicular to a uniform magnetic field B is: d ≡ 2r = 2mv/(|q| B) . Here, all 4 particle types have the same speed, v, and the same magnitude of charge, |q| ≡ |qP| = |qQ| = |qR|=|qS|, and are subject to the same magnetic field, B. Thus, d is directly proportional to the particle mass m: the larger d, the larger m. Now compare and rank order the semi-‐circular diameters of the 4 particle types shown in figure above: dP > dQ > dS > dR . Hence, it follows that mP > mQ > mS > mR . __________________________________________________________________________________________ (e) Assume all 4 particle masses are the same: mP = mQ = mR = mS . Rank order the magnitudes of their charges, |qP|, |qQ|, |qQ|, |qS|, from lowest to highest. Write, e.g.: |qP| < |qQ| < |qR| < |qS|, if P has the lowest, Q the 2nd lowest, R the 3rd lowest, S the 4th lowest magnitude of charge. Hint: Compare their semi-‐circles! || The diameter, d, of the (semi-‐)circular trajectory for a particle of mass of m and charge magnitude |q|, traveling at speed v perpendicular to a uniform magnetic field B is: d ≡ 2r = 2mv/(|q| B) . Here, all 4 particle types have the same speed, v, and the same mass, m = mP = mQ = mR = mS , and are subject to the same magnetic field, B. Thus, d is inversely proportional to the magnitude of the particle charge |q|: the larger d, the smaller |q|. So, from the rank ordering the semi-‐circular diameters of the 4 particle types, see part (d) and figure above, dP > dQ > dS > dR , it follows that |qP| < |qQ| < |qS| < |qR| . __________________________________________________________________________________________
