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Physics 112 Exam 2 Summer 2015 21. Negative charge q is moving with velocity v in the magnetic field B as shown below (with vectors v and B on the surface of the page). What is the direction of the magnetic force? B A) Parallel to v B) Parallel to B C) Exactly between v and B q D) Into the page v E) Out of the page Solution: If the charge would be positive then from the right-hand rule follows that magnetic force is directed out of the page (rotate from v to B). On the negative charge the force is into the page. 22. A positively charged particle (q = + 60 μC) moves at 2.0 10 5 m/s in a direction that is 30o to a uniform magnetic field of 0.50 T. What is the magnetic force on the charge? A) B) C) D) E) 1.0 N 1.5 N 2.0 N 2.5 N 3.0 N Solution: F qvB sin θ 60 10 -6 C 2.0 105 m / s 0.50T sin 30 3.0 N 23. An electron is moving at speed v = 800 m/s. It enters a region with uniform electric field E = 400 V/m and uniform magnetic B field. Vectors E , B and v are perpendicular to each other. The electron continues on the straight path. What is the magnitude of the magnetic field B? A) B) C) D) E) 0.5 T 1.0 T 1.5 T 2.5 T 3.0 T Solution: The electron moves in a straight line so the electric and magnetic forces on the electron must cancel each other. FE FB qE qvB B E / v B 400 V m 0.5T 800 m s 1 Physics 112 Exam 2 Summer 2015 24. The force on a current-carrying wire in a magnetic field is the strongest when A) B) C) D) E) The current is parallel to the field lines The current is at a 30° angle with respect to the field lines The current is at a 45° angle with respect to the field lines The current is at a 60° angle with respect to the field lines The current is perpendicular to the field lines Solution: F IlB sin For 90 : F Fmax IlB . 25. A coil of wire has 100 turns and a cross sectional area of 0.10 m2. The coil is free to rotate and sits in a uniform 0.50 T magnetic field that points at 30° to the normal of the coil. If a 4.0A current passes through the coil, determine the torque on the coil? A) B) C) D) E) 2.0N m 5.0N m 10N m 20N m There is no torque because the current in the coil is steady Solution: The torque is given by equation NIAB sin . 100 4.0 A 0.10m2 0.50T sin 30o 10N m 26. A wire loop is in a uniform magnetic field. Current flows in the wire loop, as shown. What does the loop do? A) B) C) D) E) Moves to the right Moves up Remains motionless Rotates Moves out of the page Solution: There is no magnetic force on the top and bottom legs, since they are parallel to the B field. However, the magnetic force on the right side is into the page, and the magnetic force on the left side is out of the page. Therefore, the entire loop will tend to rotate. This is how a motor works! 2 Physics 112 Exam 2 Summer 2015 27. Two long parallel wires are 4.0 cm apart. Each wire carries the current 3.0A in the same direction. Find the magnetic field at a point that is 3.0 cm from one wire and 1.0 cm from the other wire. A) B) C) D) E) 4 10 5 T 6 10 5 T 4 10 5 T 6 10 7 T zero out of the page out of the page into the page into the page I I 3.0 cm 1.0 cm Solution: According to the right hand rule, at the point of interest, the magnetic field BL due to the left wire is directed into the page, and the field B R due to the right wire is directed out of the page. I I BL 0 , and BR 0 . 2rL 2rR The total field is B B L B R . Because rL 3.0cm rR 1.0cm , we have BL BR , and vector B R is directed out of the page. B BR BL 4 10 7 0 I 0 I 0 I 1 1 2rR 2rL 2 rR rL T m / A 3.0 A 1 1 1 1 5 2 10 5 3.0 T 4 10 T 2 2 2 1 . 0 3 . 0 1 . 0 10 3 . 0 10 28. If current in each of two long parallel wires is doubled, what happened with force of interaction between these wires? A) B) C) D) E) Double Quadruple Decrease by one-half Decrease by one-fourth Remain the same Solution: IIl F 0 1 2 2r 3 Physics 112 Exam 2 Summer 2015 29. If current in a solenoid is doubled, what happened with the field inside the solenoid? A) B) C) D) E) Double Quadruple Decrease by one-half Decrease by one-fourth Remain the same Solution: B 0 NI / l 0 nI 30. A bar magnet falls through a loop of wire with the north pole entering first. As the north pole enters the wire, the induced current will be (as viewed from above): A) B) C) D) E) Zero Clockwise Counterclockwise To top of loop A current whose direction cannot be determined from the information given v Solution: According to the Lenz’s law the induced current creates magnetic field that opposes the original change in the flax. As the north pole of the falling magnet enters the wire the down directed flux is increasing. The induced counterclockwise current creates magnetic field directed up. S B N I 31. A coil of 600 turns with area 100 cm2 is placed in a uniform magnetic field. The angle between the direction of the field and the perpendicular to the loop is 60°. The field changes at the rate of 0.010 T/s. What is the magnitude of induced emf in the coil? A) B) C) D) E) 0.01 V 0.02 V 0.03 V 0.10 V 0.20 V Solution: B BA cos NBA cos N 600 60 A 100cm 2 B 0.010T / s t ? B B N A cos t t B B N A cos t t 600 0.010T / s 100 10 4 m 2 cos 60 0.03V 4 Physics 112 Exam 2 Summer 2015 32. An airplane with a wing span of 50 m flies horizontally at a location where the downward component of the Earth's magnetic field is 6.0 × 10-5 T. Find the magnitude of the induced emf between the tips of the wings when the speed of the plane is 200 m/s. A) B) C) D) E) 0.4 V 0.5 V 0.6 V 1.2 V 1.8 V Solution: Blv 6.0 10 5 T 50m200m / s 0.60V 33. A 6 V battery is connected to the primary coil of a transformer that is designed to convert 120 V to 240 V. What is voltage drop across the secondary coil? A) B) C) D) E) 6V 12 V 120 V 240 V Zero Solution: Batteries provide DC current. Only a changing magnetic flux induces an EMF. Therefore, the voltage across the secondary coil is zero. 34. What is the direction of the induced current in the circular loop due to an increasing current that flows to the left in the straite wire as shown below? A) B) C) D) E) Clockwise Counterclockwise Alternating current No current induced Not enough information to answer Solution: The increasing current in the wire will cause an increasing field out of the page through the loop. To oppose this increase, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise. 5 Physics 112 Exam 2 Summer 2015 35. A 4.0-mH coil carries a current of 5.0 A. How much energy is stored in the coil's magnetic field? A) B) C) D) E) 2.0 mJ 10 mJ 20 mJ 50 mJ 80 mJ Solution: U 12 LI 2 U 1 2 4.0 10 3 H 5.0 A 50 10 3 J 50mJ 2 36. After how many time constants does the current in a LR circuit reach 90% of its maximum value? A) B) C) D) E) 1.2 2.3 3.4 4.5 never Solution: I 1 e t / I max 1 e t / R t / ln 1 I / I max t / ln 1 0.9 2.3 37. An inductance coil operates at 240 V and 60.0 Hz. It draws 12.8 A. What is the coil’s inductance? A) B) C) D) E) 1.0 10 2 H 2.0 10 2 H 3.0 10 2 H 4.0 10 2 H 5.0 10 2 H Solution: We find the reactance from Ohm’s law, and the inductance by Eq. 21-11b. V IX L X L V I X L 2 fL L XL 2 f V 2 fI 240 V 2 60.0 Hz 12.8 A 4.97 102 H 6 Physics 112 Exam 2 Summer 2015 38. The amplitude of magnetic field in a sinusoidal electromagnetic wave is 4.0 µT. What is the amplitude of the electric field in this wave? A) B) C) D) E) 2400 V/m 1200 V/m 600 V/m 300 V/m 150 V/m Solution: Speed of an electromagnetic wave in vacuum is given by, c = E/B. So, E cB 3 108 m / s 4.0 10 6 T 1200V / m 39. Unpolarized light passes through three successive ideal polarizers, each of whose axis makes a 30° angle with the previous one. What fraction of the light intensity is transmitted? A) B) C) D) E) 1/4 1/8 3/8 9/32 27/64 Solution: I 1 12 I 0 I 2 I 1 cos 2 30 34 I 1 83 I 0 I 3 I 2 cos 2 30 34 I 2 9 32 I0 40. An electromagnetic wave in vacuum has a frequency of 1.00 MHz. What is the wavelength of the wave? A) B) C) D) E) 1.00 cm 3.00 cm 1.00 m 3.00 m 300 m c 3.00 108 m / s Solution: 300m f 1.00 10 6 s 1 7 Physics 112 Exam 2 21 31 D) Into the page C) 0.03 V 22 32 E) 3.0 N C) 0.6 V 23 33 A) 0.5 T E) Zero 24 34 E) The current is perpendicular to the field lines A) Clockwise 25 35 C) 10N m D) 50 mJ 26 36 D) Rotates B) 2.3 27 37 A) 4 10 T out of the page E) 5.0 10 2 H 28 38 B) Quadruple B) 1200 V/m 29 39 A) Double D) 9/32 30 40 C) Counterclockwise E) 300 m 5 Summer 2015 8