IB PHYSICS HL REVIEW PACKET: MAGNETISM

Transcription

1 NAME IB PHYSICS HL REVIEW PACKET: MAGNETISM 1. This question is about electromagnetic induction. In 1831 Michael Faraday demonstrated three ways of inducing an electric current in a ring of copper. One way is to move a bar magnet through the stationary copper ring. (a) Describe briefly a way that a current may be induced in the copper ring using a stationary bar magnet You are given the following apparatus: copper ring, battery, variable resistor, lengths of insulated copper wire with connecting terminals at each end. Describe how you would use all of this apparatus to induce a current in the copper ring. copper ring In the diagram below, a magnetic field links a circular copper ring. The field is uniform over the area of the ring and its strength is increasing in magnitude at a steady rate. (4) magnetic field (c) (i) State Faraday s law of electromagnetic induction as it applies to this situation Draw on the diagram, an arrow to show the direction of the induced current in the copper ring. Explain how you determined the direction of the induced current

2 (iii) The radius of the copper ring is 0.1 m and its resistance is Ω. The field strength is increasing at rate of T s 1. Calculate the value of the induced current in the copper ring (Total 13 marks). This question is about the force between current-carrying wires. Diagram 1 below shows two long, parallel vertical wires each carrying equal currents in the same direction. The wires pass through a horizontal sheet of card. Diagram shows a plan view of the wires looking down onto the card. eye sheet of card diagram 1 diagram (a) (i) Draw on diagram 1 the direction of the force acting on each wire. Draw on diagram the magnetic field pattern due to the currents in the wire. The card is removed and one of the two wires is free to move. Describe and explain the changes in the velocity and in acceleration of the moveable wire (Total 7 marks)

3 3. This question is about electric motors and generators. (a) The diagram below is a representation of a simple dc electric motor. The armature consists of a single rectangular coil and rotates between the poles of a permanent magnet. The connections between the coil and the battery B are not shown. The split-ring is labelled C. armature C B (i) (iii) On the diagram above, draw connections from the battery B to the split-ring so that the coil will rotate continuously in one direction. On the diagram above, draw arrows to show the direction of the forces acting on the coil when connection to the battery is made and the coil is in the position shown in the diagram. Describe how these connections enable the coil to rotate continuously in one direction. In an experiment to measure the efficiency of a small dc electric motor, the motor is clamped to the edge of a bench. The motor is used to raise a small weight that is attached to a pulley wheel by cotton thread. The pulley wheel is rotated by the motor. The thread wraps around the pulley wheel, so raising the weight. axel motor pulley wheel cotton thread Side view weight End-on-view The time taken for the motor to raise the weight through a certain height is measured. It is assumed that the weight accelerates uniformly whilst being raised. The weight of the cotton thread is negligible. (i) Draw a labelled, free-body force diagram of the forces acting on the accelerating weight. (c) weight The weight has a mass of 15 g and it takes. s to raise it from rest through a height of 0.84 m. Calculate the tension in the thread as the weight is being raised. (Acceleration of free fall g = 10 m s.) In a second experiment, the current is adjusted so that the weight of mass 15 g is raised at constant speed. The motor is connected to a 6.0 V supply and it now takes the motor 3.4 s to raise the weight through 0.84 m. (i) Suggest how it might be determined that the weight is being raised at constant speed. 3 (4)

4 Determine the power delivered to the weight by the motor. (Acceleration of free fall g = 10 m s.) (iii) The current in the motor is 45 ma. Estimate the efficiency of the motor. (d) (e) (f) It is suggested that the efficiency E of the motor is related to the current I in the motor by the expression E = ki n where k and n are constants. State and explain what graph you would plot in order to determine a value of n Explain why (i) as the coil rotates, an emf is induced in the coil; the faster the coil rotates, the greater the value of the induced emf; (iii) the induced emf is not constant even when the speed of rotation of the coil is constant. An emf is also induced in a coil that is rotated mechanically in a magnetic field. This is the principle of a simple alternating current generator. For a particular generator, the graph below shows the variation with time t of the induced (generated) emf E. 3 E / V t / s (i) 3 On the graph above, label with the letter P, one point that corresponds to a time when the coil is parallel to the magnetic field. 4

5 4. This question is about magnetic fields. (a) Using the diagram below, draw the magnetic field pattern of the Earth. North Earth (c) State what other object produces a magnetic field pattern similar to that of the Earth.... A long vertical wire passes through a sheet of cardboard that is held horizontal. A small compass is placed at the point P and the needle points in the direction shown. cardboard sheet P direction of compass needle A current is passed through the wire and the compass needle now points in a direction that makes an angle of 30 to its original direction as shown below. direction of compass needle with current in wire cardboard sheet P 30 original direction of compass needle (i) Draw an arrow on the wire to show the direction of current in the wire. Explain why it is in the direction that you have drawn. The magnetic field strength at point P due to the current in the wire is BW and the strength of the horizontal component of the Earth s magnetic field is BE. Deduce, by drawing a suitable vector diagram, that BE = BW tan 60. (Total 7 marks) 5

6 5. This question is about motion of a charged particle in a magnetic field. A charged particle is projected from point X with speed v at right angles to a uniform magnetic field. The magnetic field is directed out of the plane of the page. The particle moves along a circle of radius R and centre C as shown in the diagram below. region of magnetic field out of plane of page Y v R C X charged particle (a) On the diagram above, draw arrows to represent the magnetic force on the particle at position X and at position Y. State and explain whether (i) the charge is positive or negative; work is done by the magnetic force. (c) A second identical charged particle is projected at position X with a speed v in a direction opposite to that of the first particle. On the diagram above, draw the path followed by this particle. (Total 6 marks) 6

7 6. Electromagnetic induction (a) State Faraday s law of electromagnetic induction A long straight wire carries a constant current. A rectangular loop of conducting wire is placed near the wire such that the wire is on the plane of the loop. The loop is then moved at constant speed away from the wire as shown in the diagram below. wire current loop direction of motion of loop (i) (iii) Explain why an emf is induced in the loop. On the diagram above, draw an arrow to indicate the direction of the current induced in the loop. Explain your answer. Energy is dissipated in the wire of the loop. Explain how the movement of the loop gives rise to energy dissipation. (Total 8 marks) 7

8 7. This question is about induced e.m.f. A small area A is in a region of uniform magnetic field of strength B. The field makes an angle to the normal to the area as shown below. Area A B normal (a) With reference to the diagram, define magnetic flux both in words and in symbols A thin copper ring encloses an area of m. The plane of the ring is normal to a uniform magnetic field. The magnetic field strength increases at a constant rate of T s 1. Calculate the e.m.f. induced in the ring (Total 4 marks) 8

9 MARK SCHEME 1. (a) move the ring over the end of the magnet / OWTTE; 1 ie magnet stationary, ring moved. diagram showing wire wrapped around part of the ring; and appropriate connections to battery and variable resistor; as the current is changed by altering the value of the resistance; a current is induced in the ring; 4 Mark diagram and description together look for any sensible description of the production of transformer induced currents. (c) (i) the emf induced in the ring; is equal / proportional to the rate of change of magnetic flux linking the ring; clockwise; Lenz s law: induced current is such as to oppose the change / OWTTE; current in this direction induces a field in the opposite direction to the changing field / OWTTE; 3 (iii) area = 3.14 (1.) 10 = m ; rate of flux change = m = emf = V; 5 ( ) current = = 5.4 ma; (a) (i) ; 1 [13] general shape: at least one circle around each wire and one loop around both wires; appropriate spacing of lines: increasing separation with distance from wires; correct direction of field; 3 velocity increases; acceleration increases; because the force is getting larger the closer the wires get together; 3 Watch for ecf if force is drawn in wrong direction in (a) (i) ie velocity increases, acceleration decreases, force gets smaller. 9

10 3. (a) (i) showing connection via brushes, 1 two correct forces; 1 (iii) when the split ring is in contact with the brushes the current in the coil will always be in the same direction / OWTTE; some statement to the effect that this will be so even after the coil has rotated through 180 and sides of coil are reversed; when the split ring is not in contact with the brushes, the momentum of the coil will keep it rotating / OWTTE; 3 (i) tension in thread; weight (of object) / mg; tension length > weight length; 3 s t a ; = ms. ; T mg = ma; T = m (g + a) = = 0.16N; 4 (c) (i) measure the time it takes the object to go successive distances of say 10 cm / any realistic length given or implied; if the times are equal then speed is constant / OWTTE; increase in potential energy = = 0.13J; rate of working = power input = W ; (iii) power input to motor = VI = = 0.7 W; P P in out Eff or 14%; (d) lg(e) against lg(i); lg(e) = lg(k) + nlg(i); slope / gradient = n; 3 10

11 (e) (i) a magnetic flux links the coil; as the coil rotates the flux linkage changes with time; therefore, from Faraday s law an emf will be induced; 3 the faster the speed of rotation, the greater the flux change; Faraday s law states that the emf is equal / proportional to the rate of change of flux; (iii) the amount of flux linking the coil changes with the angle that the coil makes with the magnetic field / OWTTE; 1 (f) (i) any maximum / minimum value of V; 1 4. (a) overall correct shape with no field lines touching; direction of field; bar magnet / solenoid; 1 Do not accept just magnet. (c) (i) upwards the direction of the compass needle is the resultant of two fields / OWTTE; the field must be into the plane of the (exam) paper to produce a resultant field in the direction shown / OWTTE; Award [1] for upwards because of the right hand rule / OWTTE. vector addition with correct values of two angles shown 30, 60 or 90 ; W from diagrams BE = BW tan 60 or E ; B B tan (a) (i) two arrows directed towards the centre of the circular path, within 5.0 cm of the centre. 1 [7] (i) negative by stating any rule for the direction of the magnetic force; 1 the work done is zero; since the force is at all times normal to the velocity; (c) a curved path starting at X and in the right direction ie counterclockwise; R circular path of radius ; Allow diameter 3-4 cm and be generous with how round the circle is. [6] 11

12 6. Electromagnetic induction (a) the induced emf (in a loop) is proportional to the rate of change of the magnetic flux linkage (in the loop); 1 (i) as the loop is moved away the magnetic field through the loop is getting smaller; hence the magnetic flux through it is changing with time; (iii) Accept also answers based on considerations of induced emf in each section of loop eg emf is induced in upper and lower section of loop; no emf induced in sides / emf of same sign; emf in upper section larger hence current in loop; the direction of the current is clockwise; because in this way the magnetic field created by the induced current is in the same direction as the external magnetic field thus opposing the change in flux; Accept any other reasonable formulation based on Lenz s law but not bald answer without explanation or incorrect explanation. work is being performed on the loop by the agent pushing the loop; against the attractive magnetic force between the loop and the wire; so that the loop moves at constant speed; 3 [8] 7. (a) product of normal component of magnetic field strength and area that it links / OWTTE; = BA cos θ rate of change of flux = ( ) = (Wb s 1 ); recognize that e.m.f. rate of change of flux = V; Ignore any sign. [4] 1