Magnetostatics. Teacher Packet

Transcription

1 AP * PHYSICS Magnetostatics Teacher Packet AP* is a trademark of the College Entrance Examination oard. The College Entrance Examination oard was not involved in the production of this material.

2 Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on magnetostatics. Standards Magnetostatics is addressed in the topic outline of the College oard AP* Physics Course Description Guide as described below. III. Electricity and Magnetism D. Magnetic Fields 1. Forces on moving charges in magnetic fields 2. Forces on current-carrying wires in magnetic fields 3. Fields of long current-carrying wires AP Physics Exam Connections Topics relating to magnetostatics are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked over magnetostatics. These questions are available from the College oard and can be downloaded free of charge from AP Central. Free Response Questions 2008 Question Form Question Question Form Question Question Question Question 7

3 What I Absolutely Have to Know to Survive the AP* Exam Determine the magnetic force exerted on a moving charge placed in a magnetic field. Determine the magnetic force exerted on a current carrying wire placed in a magnetic field. Apply right-hand rules for finding the direction of the force on a charged particle or a current carrying wire placed in a magnetic field. Determine magnetic fields of long current carrying wires. Key Formulas and Relationships F = qvsin θ The magnetic force acting on a point charge q (Coulombs), moving with m velocity v ( ), making an angle θ with a magnetic field (Teslas). s F = qv Use when the force F is perpendicular to both the velocity v and the F = qv+ qe tot magnetic field. The total force when electric and magnetic fields act simultaneously on a charge q. mv r = q The path of a charge q of mass m with a velocity v perpendicular to a magnetic field in a circular radius r. F = ILsin θ The force a current carrying wire experiences due to its presence in a F = IL Use when the angle θ is 90. μ II F = 2π d magnetic field, I is the current in the wire, L is a vector equal in magnitude to the length of the wire, making an angle θ with the magnetic field,. Force per unit length between two infinitely long parallel current carrying 7 wires where μ is the permeability constant 4 π x 10 T m / A, 0 ( ( ) ) d is the separation between the wires and I and I are the currents in 1 2 the wires. μ0i = The magnitude of the magnetic field produced by a current carrying wire at 2π r a perpendicular distance r from the wire. E v = Electric and magnetic fields can be used together to precisely select the velocity of a charged particle.

4 Important Concepts Two north poles or two south poles near each other will exert a repulsive force on each other. This is called the law of poles: unlike poles attract, and like poles repel. A compass needle is a magnet that is free to spin until it lines up in the north-south direction. Historically, people defined the north end of a magnet as the end that points geographically north, even though the north end of a magnet is not attracted to the north end of another magnet. We define a magnetic field as the space around a magnet in which another magnet will experience a force. Magnetic field is a vector which points in the direction the north pole of a compass needle would point (north to south), and is denoted by the letter. If you sprinkle iron filings around a magnet they will line up with the magnetic field lines around the magnet. The magnetic field lines for a bar magnet, a horseshoe magnet, and the Earth are shown below. Note that magnetic fields circulate or flow out of the north pole and into the south pole. ] Fundamentally, magnetism is caused by moving charges, such as a current in a wire. Thus, a moving charge or current-carrying wire produces a magnetic field, and will experience a force if placed in an external magnetic field. We will define current as the flow of positive charge, that is, from positive to negative. THE FORCE ON A CHARGED PARTICLE MOVING IN A MAGNETIC FIELD F = qvsinθ Magnetic forces are only exerted on moving charges. A charge will experience no force due to the magnetic field if it is moving with velocity, v along the magnetic field lines at an angle of 0 or 180 relative to the magnetic field vector. The charge will experience a maximum force if the velocity vector, v is perpendicular to the magnetic field vector,. Magnetic fields are centripetal in nature and hence, 2 mv = qv. Magnetic fields change the direction of a charged r particle, but they do NOT change a charged particle s speed or kinetic energy. Magnetic fields do no work on charged particles. If a charged particle is projected into a magnetic field at an angle of 90 then the path of the circular radius of the charged particle in the magnetic field is given by

5 mv r = q If the charged particle is projected into the magnetic field at an angle other than 0 or 90, then the path is a helix. Thus, a moving charge will experience a force as it moves through a magnetic field, as long as it is crossing the magnetic field lines and not moving parallel to them. Right-hand rule for finding the force on a moving charge in a magnetic field There are many variations of the right hand rule. Follow your pheerless physics teacher s instructions as to appropriate right hand rules. Listed below is one of many variations of the right hand rule. Place your fingers in the direction of the magnetic field (north to south), your thumb in the direction of the velocity of the moving charge, and the magnetic force on the charge will come out of your palm. Example 1: Consider the positive charge moving through the magnetic field in the figure below: Since this is a moving positive charge, you would use your right hand to find the direction of the force that this charge would experience. Placing your fingers in the direction of the magnetic field and your thumb in the direction of the velocity of the charge gives the direction of the force as coming out of the page toward you.

6 Example 2: Consider the negative charge moving through the magnetic field in the figure below: In this case the magnetic field is into the page ( ), and the negative charge is moving to the left. In finding the direction of the force acting on the negative charge, you would use your left hand. Placing your fingers in the direction of the magnetic field (into the page away from you) and your thumb in the direction of the velocity of the negative charge, the force points out of your palm which is facing up toward the top of the page. Thus, this negative charge would initially turn toward the top of the page. If you prefer to use the vector cross product F = qv then the cross product yields both the magnitude and direction of the force on a positive charge moving in the magnetic field. If the charge is negative, the magnitude of the force will be the same, but the direction will be opposite that determined using the right hand rule. Since the charge, q, is a scalar it does not affect the direction of the magnetic force. The first vector is velocity, which may be represented by the length of your right fore-arm. Since cross products deal with right angles, bend the fingers of the right hand so that they form a right angle and point them in the direction of the second vector, the magnetic field. Then extend the thumb straight out and it points in the direction of the resultant force. THE FORCE ON A CURRENT CARRYING WIRE IN A MAGNETIC FIELD F = ILsinθ A current flowing through a wire produces a magnetic field, and we can specify the direction of the magnetic field. If we pass a current-carrying wire perpendicularly through a card and sprinkle iron filings on the card around the wire, the filings will form circles around the wire, lining up with the magnetic field produced by the wire. If we put little compasses on one of these circles, the needle of each compass would line up tangent to the circle and the north ends of all the compass needles would point around the wire in the same direction, as shown below. If we are looking down on the card, the north ends of the compasses are pointing around the wire in a counterclockwise direction. We can envision the magnetic field due to the current in the wire as circling the wire in a

7 counterclockwise direction. If the current were flowing downward in the wire, the magnetic field would circle the wire in a clockwise direction. FIRST RIGHT-HAND RULE FOR FINDING THE DIRECTION OF THE MAGNETIC FIELD AROUND A CURRENT-CARRYING WIRE: Place your thumb in the direction of the current (positive to negative), and your fingers will curl around in the direction of the magnetic field produced by that current. If you choose to use electron flow (flow of negative charge) instead of current flow, you can apply the same rule using your left hand. Another way to indicate that a vector is pointing into the page is by using the symbol. This symbol tells you a vector (such as force, velocity, or magnetic field) points down into the page, as if you are looking at the cross-feathers on the end of an arrow moving away from you. If you were looking at the point of an arrow coming toward you, you would see a dot ( ), and therefore a dot is the symbol for indicating a vector which is coming out of the page toward you. We could then draw the magnetic field around the wire in the figure below as pointing into the page ( ) below the wire and out of the page ( ) on top of the wire: out in THE SECOND RIGHT-HAND RULE Force on a Current-Carrying Wire in a Magnetic Field Since a current-carrying wire creates a magnetic field around itself according to the first right-hand rule, every currentcarrying wire is a source of a magnetic field. Thus, if we place a current-carrying wire in an external magnetic field, it will experience a force. The direction of the force acting on the wire is given by the second right-hand rule: Second right-hand rule for finding the force on a current-carrying wire in a magnetic field: Place your fingers in the direction of the magnetic field (north to south), your thumb in the direction of the current in a wire, and the magnetic force on the wire will come out of your palm.

8 Example 1: If we place a current-carrying wire between the poles of a horseshoe magnet, the wire will experience a force according to the second right-hand rule, as shown below. We place our fingers in the direction of the magnetic field (north to south), our thumb in the direction of the current (out of the page), and force acting on the wire is directed up to the top of the page. This wire would be pushed out of the horseshoe magnet. It is important to note that a current-carrying wire will only experience a force if it crosses the magnetic field lines. A wire which is parallel to the magnetic field lines would not experience a force. If we bent this current-carrying wire into a loop and mounted it on stirrups above a magnet, it would experience a force on both sides of the loop and spin, and we would have an electric motor. Current carrying wires are sources of magnetic fields. The magnetic field surrounding a long current carrying wire is μ0i = 2π r where I is the current in the wire and r is the perpendicular distance from the wire. Note that the magnetic field around a current carrying wire is proportional to 1 r while the electric field around a point charge is proportional to 1 2 r. In addition, magnetic fields circulate about current carrying wires while electric fields radiate outward from positive charges.

9 Question 1 (15 pts) Free Response e = -1.6 x C m = kg = 0.5 T v Plate 1 Plate 2 An electron is accelerated from rest between two charged parallel plates as shown. The electric field between the plates has a magnitude of 400 N/C and the separation between the plates is 0.20 m. The electron then passes through a small hole in Plate 2 into a region of uniform magnetic field strength which exists everywhere to the right of Plate 2. The magnetic field is directed out of the page. A. On the diagram above, i. label each plate as positively charged or negatively charged ii. clearly indicate the direction of the electric field between the plates (3 points max) 1 point for labeling each plate correctly as positive or negative 1 point for drawing the electric field lines horizontally 1 point for drawing the electric field lines from positive to negative

10 . Determine the electron s speed as it passes through the hole in plate 2. (4 points max) 1 ev = mv 2 2eV v = m V = Ed 2eEd v = = m v = s N 400 C kg m 19 ( C) ( m) Alternate solution 19 N ( C) 400 F qe C a = = = 31 m m kg 13 m a = s m v = ( ) 2 13 v = 2ad = m s m s 1 point for a statement of conservation of energy 1 point for indicating that the kinetic energy the electron acquires is the work done by the electric field between the two parallel plates 1 point for recognizing that the electric potential can be expressed in terms of V=Ed 1 point for the correct answer including correct units and reasonable number of significant digits 1 point for correct application of Newton s second law 1 point for recognizing that the net force is applied by the electric field 1 point for using the correct kinematic equation 1 point for the correct answer including correct units and reasonable number of significant digits

11 C. On the figure above, sketch the subsequent path of the electron in the magnetic field. (1 point max) 1 point for indicating that the path of the electron will be upward, circular, and counterclockwise D. Determine the radius of the path of the electron. (3 points max) F = ev 2 mv FC = r 2 mv ev = r ( kg ) mv r = = e C T m r = ( )( 0.5 ) m s 1 point for equating the magnetic force and the centripetal force 1 point for correct substitution of the mass and charge of an electron 1 point for the correct answer or answer consistent with part including correct units and reasonable number of significant digits

12 E. Determine the magnitude and direction of an electric field that would cause the electron to continue moving in a straight line while in the magnetic field. Draw arrows on the figure above representing the direction of the electric field. (4 points max) The net force in this case is zero, so the magnitudes of the electric force and magnetic force are equal to each other. ee = ev ev E = = v e 6 m E = ( 0.5T) s 6 N E = C If the electron is to follow a straight line path through the magnetic field, the electric force on the charge would need to be directed downward to counter the upward magnetic force. The electric field would be directed upward, from the bottom of the page to the top of the page. 1 point for recognizing that the electric field applies a force on the moving charge 1 point for recognizing that the electric force is equal and opposite to the magnetic force 1 point for the correct answer or answer consistent with part including correct units and reasonable number of significant digits 1 point for the correct direction of the electric field

13 Question 2 (10 pts) 1 Amp 2 m A 4 m 4 Amps The two long straight wires shown above are perpendicular, insulated from each other, and small enough so that they may be considered to be in the same plane. The wires are fixed and not free to move. Point A, in the same plane as the wires, is 2.0 meter from the wire carrying a current of 1.0 ampere and is 4.0 meters from the wire carrying a current of 4.0 amperes. A. Determine i. the magnitude and direction of the magnetic field at point A due to the 1-amp current. (2 points max) 7 Tm 4π A μ0i A = = 2πr 2π ( 2.0m) 7 = T ( ) y RHR, the magnetic field due to the 1.0 Amp current is coming out of the page at Point A. 1 point for the correct answer including correct units and reasonable number of significant digits 1 point for stating that the magnetic field is out of the page

14 ii. the magnitude and direction of the magnetic field at point A due to the 4-amp current. (2 points max) 7 Tm 4π A μ0i A = = 2πr 2π ( 4.0m) 7 = T ( ) y RHR, the magnetic field due to the 4.0 Amp current is coming out of the page at point A. 1 point for the correct answer including correct units and reasonable number of significant digits 1 point for stating that the magnetic field is coming out of the page. What is the magnitude and direction of the net magnetic field at point A due to the currents? (2 points max) 7 7 = T T A = T out of the page 1 point for the correct calculation of the net magnetic field at point A or answer consistent with part A including correct units and reasonable number of significant digits 1 point for stating that the magnetic field is coming out of the page

15 A charged particle at point A is instantaneously moving with a velocity of 2.0 x 10 6 m s to the right and experiences a force of 1.0 x 10-7 N toward the bottom of the page due to the two currents. C. State whether the charge on the particle is positive or negative. Justify your answer. (2 points max) Positive y the RHR, fingers in the direction of the magnetic field, the thumb in the direction of the particle s velocity, the force comes out of the palm toward the bottom of the page. 1 point for the correct answer or an answer consistent with part 1 point for correct use of the Right Hand Rule D. Determine the magnitude of the charge of the particle. (2 points max) F = qv 1 point for a correct application of 7 F = qv F ( N ) q = = v 6 m ( T) 1 point for the correct answer or s answer consistent with part 7 q = C including correct units and reasonable number of significant digits

16 Multiple Choice 1. An electron moves through a uniform magnetic field. The force on the electron is greatest when it A) moves perpendicular to the magnetic field at a fast velocity. ) moves perpendicular to the magnetic field at a slow velocity C) moves parallel to the magnetic field at a fast velocity D) moves parallel to the magnetic field at a slow velocity. E) does not move. A The magnitude of the force is equal to the product of the charge, speed, magnetic field, and the sine of the angle between the velocity and the magnetic field, which is θ. Questions 2-4 An electron moving with velocity m/s enters a uniform magnetic field of 0.5 T as pictured below. 2. The electron will experience a force which is initially A) into the page. ) out of the page. C) toward the top of the page. D) toward the bottom of the page. E) to the left. D We use the left-hand rule, since the electron is a negative charge, placing our fingers into the page, and the thumb to the right, in the direction of the velocity which crosses the magnetic field lines. The force, then, comes out of the palm and toward the bottom of the page.

17 3. The magnitude of the force acting on the electron is A) N ) N C) N D) N E) N D m F = qv = ( C) ( T) = s N 4. The resulting path of the electron is a circle of radius m. The work done by the magnetic field on the electron is F) J G) J H) J I) J J) zero E The magnetic force on a moving charge is always perpendicular to the magnetic field and hence no work is done by the field. The kinetic energy and the speed of the particle remain unchanged.

18 Questions 5-6 A stream of electrons are flowing from the cathode to the anode across a cathode ray tube (left to right) producing a current of 2 A. The strength of the magnetic field between the north pole and the south pole is 0.2 T. N S 5. With a north pole and south pole as shown above, which direction will the beam deflect? A) up toward the top of the page ) down towards the bottom of the page C) out of the page D) into the page E) The beam of electrons will not be deflected and will travel in a straight path. C Using the left hand rule for moving negative charges moving in a magnetic field, we can show that the force on the electron beam is out of the page toward you. According to left-hand rule, the fingers point toward the bottom of the page in the direction of the magnetic field, the thumb points toward the right of the page in the direction of the current, and the force comes out of the palm and out of the page. 6. The length of the beam of electrons in the magnetic field is 0.15 m. The force on the beam of electrons is A) N ) 0.06 N C) 0.15 N D) 2.0 N E) 6.7 N F = IL= (2 A)(0.2 T)(0.15 m) = 0.06 N

19 7. All of the following will affect the force acting on a current carrying wire, EXCEPT A) the amount of current in the wire. ) the length of the wire. C) the strength of the magnetic field. D) the amount of charge per unit time in the wire. E) the magnitude of the charges in the wire. E F = IL 8. A wire on the x-axis of a coordinate system has a current I in the +x direction as shown above. What is the direction of the magnetic field due to the wire at point P? A) to the left ) to the right C) down into the page and perpendicular to the page D) up out of the page and perpendicular to the page E) toward the bottom of the page D y the right hand rule, a current flowing to the right in the wire will produce a magnetic field which is into the page below the wire and coming out of the page above the wire.

20 9. A wire carrying a current I directed toward the bottom of the page is placed in a magnetic field as shown above. The force on the wire is directed A) into the page ) out of the page C) toward the top of the page D) toward the bottom of the page E) to the left According to right-hand rule, the fingers point to the right of the page in the direction of the magnetic field, the thumb points toward the bottom of the page in the direction of the current, and the force comes out of the palm and out of the page.

21 10. Two negatively charged particles enter a uniform magnetic field, as pictured above. The particles have the same charge but different masses. Ignoring the gravitational force, the trajectory of the less massive particle is best represented by path A) 1 ) 2 C) 3 D) 4 E) 5 E The left hand rule indicates the force on the particle is to the right. Since the particles have the same charge and the same velocity, the less massive particle is deflected more than the more massive particle since it has less inertia. mv r = q