21 MAGNETIC FORCES AND MAGNETIC FIELDS


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1 CHAPTER 1 MAGNETIC FORCES AND MAGNETIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 (d) RightHand Rule No 1 gives the direction of the magnetic force as x for both drawings A and B In drawing C, the velocity is parallel to the magnetic field, so the magnetic force is zero (b) Using RightHand Rule No 1 (see Section 1), we find that the direction of the magnetic force on a positively charged particle is to the west Reversing this direction because the particle is a negative electron, we see that the magnetic force acting on it points to the east 3 (a) Using RightHand Rule No 1 (see Section 1), we find that the direction of the magnetic force on a positively charged particle is straight down toward the bottom of the screen 4 B = T, south 5 (c) The electric force points out of the screen, in the direction of the electric field An application of Right Hand Rule No 1 shows that the magnetic force also points out of the screen, parallel to the electric force When two forces have the same direction, the magnitude of their sum has the largest possible value 6 (e) In this situation, the centripetal force, F c = mv /r (Equation 53), is provided by the magnetic force, F = qvb sin 900 (Equation 11), so mv /r = qvb sin 900 Thus, q mv / rb, and the charge magnitude q is inversely proportional to the radius r Since the radius of curve 1 is smaller than that of curve, and the radius of curve is smaller than that of curve 3, we conclude that q 1 is larger than q, which is larger than q 3 7 (a) The magnetic force that acts on the electron in regions 1 and is always perpendicular to its path, so the force does no work According to the workenergy theorem, Equation 63, the kinetic energy, and hence speed, of the electron does not change when no work is done 8 (d) According to Equation 1, the radius r of the circular path is given by r mv / qb Since v, q, and B are the same for the proton and the electron, the moremassive proton travels on the circle with the greater radius The centripetal force F c acting on the proton must point toward the center of the circle In this case, the centripetal force is provided by the magnetic force F According to RightHand Rule No 1, the direction of F is related to the velocity v and the magnetic field B An application of this rule shows that the proton must travel counterclockwise around the circle in order that the magnetic force point toward the center of the circle 9 r proton /r electron = (c) When, for example, a particle moves perpendicular to a magnetic field, the field exerts a force that causes the particle to move on a circular path Any object moving on a circular path experiences a centripetal acceleration 11 F = 30 N, along the y axis
2 1 (e) The magnetic field is directed from the north pole to the south pole (Section 11) According to Right Hand Rule No 1 (Section 15), the magnetic force in drawing 1 points north 13 (c) There is no net force No force is exerted on the top and bottom wires, because the current is either in the same or opposite direction as the magnetic field According to RightHand Rule No 1 (Section 15), the left side of the loop experiences a force that is directed into the screen, and the right side experiences a force that is directed out of the screen (toward the reader) The two forces have the same magnitude, so the net force is zero The two forces on the left and right sides, however, do exert a net torque on the loop with respect to the axis 14 (d) According to RightHand Rule No 1 (Section 15), all four sides of the loop are subject to forces that are directed perpendicularly toward the opposite side of the square In addition, the forces have the same magnitude, so the net force is zero A torque consists of a force and a lever arm For the axis of rotation through the center of the loop, the lever arm for each of the four forces is zero, so the net torque is also zero 15 N = 86 turns 16 (a) RightHand Rule No (Section 17) indicates that the magnetic field from the top wire in points into the screen and that from the bottom wire points out of the screen Thus, the net magnetic field in is zero Also, the magnetic field from the horizontal wire in 4 points into the screen and that from the vertical wire points out of the screen Thus, the net magnetic field in 4 is also zero 17 (b) Two wires attract each other when the currents are in the same direction and repel each other when the currents are in the opposite direction (see Section 17) Wire B is attracted to A and repelled by C, but the forces reinforce one another Therefore, the net force has a magnitude of F BA + F BC, where F BA and F BC are the magnitudes of the forces exerted on wire B by A and on wire B by C However, F BA = F BC, since the wires A and C are equidistance from B Therefore, the net force on wire B has a magnitude of F BA The net force exerted on wire A is less than this, because wire A is attracted to B and repelled by C, the forces partially canceling The net force expected on wire C is also less than that on A It is repelled by both A and B, but A is twice as far away as B 18 (a) The magnetic field in the region inside a solenoid is constant, both in magnitude and in direction (see Section 17) 19 B = T, out of the screen 0 (d) According to Ampere s law, I is the net current passing through the surface bounded by the path The net current is 3 A + 4 A 5 A = A
3 CHAPTER ELECTROMAGNETIC INDUCTION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 35 m/s (e) The work done by the hand equals the energy dissipated in the bulb The energy dissipated in the bulb equals the power used by the bulb times the time Since the time is the same in each case, more work is done when the power used is greater The power, however, is the voltage squared divided by the resistance of the bulb, according to Equation 06c, so that a smaller resistance corresponds to a greater power Thus, more work is done when the resistance of the bulb is smaller 3 (c) The magnetic flux Ф that passes through a surface is BAcos (Equation ), where B is the magnitude of the magnetic field, A is the area of the surface, and is the angle between the field and the normal to the surface Knowing Ф and A, we can calculate Bcos / A, which is the component of the field parallel to the normal or perpendicular to the surface 4 (b) The magnetic flux Ф that passes through a surface is BAcos (Equation ), where B is the magnitude of the magnetic field, A is the area of the surface, and is the angle between the field and the normal to the surface It has the greatest value when the field strikes the surface perpendicularly 0 and a value of zero when the field is parallel to a surface 90 The field is more nearly perpendicular to face 1 0 than to face 3 70 and is parallel to face 5 (d) Faraday s law of electromagnetic induction states that the average emf induced in a coil of N loops is N / t (Equation 3), where Ф is the change in magnetic flux through one loop and t is the time interval during which the change occurs Reducing the time interval t during which the field magnitude increases means that the rate of change of the flux will increase, which will increase (not reduce) the induced emf 6 3 V 7 (c) According to Faraday s law, the magnitude of the induced emf is the magnitude of the change in magnetic flux divided by the time interval over which the change occurs (see Equation 3) In each case the field is perpendicular to the coil, and the initial flux is zero since the coil is outside the field region Therefore, the changes in flux are as follows: BL, BL (see Equation ) The A B C corresponding time intervals are ta tb t, tc t Dividing gives the following results for the magnitudes of the emfs: A BL, B L B, B L BL C t t t t 8 (a) An induced current appears only when there is an induced emf to drive it around the loop According to Faraday s law, an induced emf exists only when the magnetic flux through the loop changes as time passes Here, however, there is no magnetic flux through the loop The magnetic field lines produced by the current are circular and centered on the wire, with the planes of the circles perpendicular to the wire Therefore, the magnetic field is always parallel to the plane of the loop as the loop falls and never penetrates the loop In
4 other words, no magnetic flux passes through the loop No magnetic flux, no induced emf, no induced current 9 (d) When the switch is closed, current begins to flow counterclockwise in the larger coil, and the field that it creates appears inside the smaller coil Using RHR reveals that this field points out of the screen toward you According to Lenz s law, the induced current in the smaller coil flows in such a direction that it creates an induced field that opposes the growth of the field from the larger coil Thus, the induced field must point into the screen away from you Using RHR reveals that the induced current must, then, flow clockwise The induced current exists only for the short period following the closing of the switch, when the field from the larger coil is growing from zero to its equilibrium value Once the field from the larger coil reaches its equilibrium value and ceases to change, the induced current in the smaller coil becomes zero 10 (b) The peak emf is proportional to the area A of the coil, according to Equation 4 Thus, we need to consider the areas of the coils The length of the wire is L and is the same for each of the coil shapes For the circle, the circumference is πr = L, so that the area is the area is A square A circle L L r For the square, 4 L L For the rectangle, the perimeter is (D + D) = L, so that the area is 4 16 L L L Arectangle The circle has the largest area, while the rectangle has the smallest area, corresponding to answer b cm 1 (d) The back emf is proportional to the motor speed, so it decreases when the speed decreases The current V I drawn by the motor is given by Equation 5 as I, where V is the voltage at the socket, is the R back emf, and R is the resistance of the motor coil As decreases, I increases 13 (c) According to Equation 7, the mutual inductance is St M If the time interval is cut in half and I the change in the primary current is doubled, while the induced emf remains the same, the mutual inductance must be reduced by a factor of four 14 (b) The energy stored in an inductor is given by Equation 10 as inductors store the same amount of energy, we have I1 L L 1414 I L L / 1 15 (e) According to Equation 8, we have 1 1 L1 I 1 L I P Energy Thus, 1 LI Since the two LI N Since Ф is the same for each coil, the number of turns is proportional to the product LI of the inductance and the current For the coils specified in the table, this product is (LI) A = L 0, (LI) B = L 0 /, (LI) C = 4L 0 16 (c) The current in the primary is proportional to the current in the secondary according to Equation 13: I I N / N The current in the secondary is the secondary voltage divided by the resistance, according P S S P to Ohm s law Thus, when the resistance increases, the current in the secondary decreases and so does the
5 current in the primary The wall socket delivers to the primary the same power that the secondary delivers to the resistance, assuming that no power is lost within the transformer The power delivered to the resistance is given by Equation 015c as the square of the secondary voltage divided by the resistance When the resistance increases, the power decreases Hence, the power delivered to the primary by the wall socket also decreases W 18 (a) The current in resistor (without the transformer) is the same as the current in resistor 1 (with the transformer) In either event, the current I is I = V/R, where V is the voltage across the resistance R Since the transformer is a stepup transformer, the voltage applied across resistor is smaller than the voltage applied across resistor 1 The smaller voltage across resistor can lead to the same current as does the greater voltage across resistor 1 only if R is less than R 1
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