Magnetic Fields: Forces on Charges and Currents, Assignments of Physics

Download Magnetic Fields: Forces on Charges and Currents and more Assignments Physics in PDF only on Docsity! Magnetism Concept Tests. CTM-1 Pick the answer which is most nearly correct. Magnetic fields are caused by... A) stationary electric charges. B) magnetic monopoles. C) moving charges (currents) D) neutrinos E) None of these is correct. Answer: moving charges (currents) CTM-2 A negative particle and a positive particle are moving with given velocities in a constant, uniform magnetic field B, as shown. The direction of the B-field is to the right. The (+) particle is moving directly left; the (–) particle is moving directly up. The force on the negative particle due to the B-field is A: in B: out C: zero D: right E: left The force on the positive particle due to the B-field is A: in B: out C: zero D: right E: left Answer: The positive particle is moving anti-parallel to the B-field. The angle  is 180 and the force is FB=qvB sin = 0. The negative particle is moving at right angles to the field. By the right-hand rule, the direction of the force is out of the page. B CTM-5 A particle with unknown charge (but non-zero) q and moving left with speed v enters a regions where there is a uniform electric field down and a uniform magnetic field out of the page. The particle is observed to go in a straight line. The charge of the particle must be ... A: positive B: negative C: impossible to determine. Suppose the particle is a proton. If the speed of the proton is increased, it will A: still undergo no deflection B: deflect out of the plane of the page C: stay in the plane of the page and deflect upward D: stay in the plane of the page and deflect downward Answers: Question 1: Impossible to determine. The particle will go straight through undeflected if the speed v is v = E / B, regardless of the sign of the charge. (See previous Concept Test). Question 2: If the speed is increased, the particle stay in the plane of the page and deflect upward. When the speed is increased the upward magnetic force will get larger, but the downward electric force will remain unchanged. v E B CTM-6 A square loop of wire carrying current I is in a uniform magnetic field B. The loop is perpendicular to B (B out of the page). What is the direction of the net force on the wire? A: out of the page B: into the page C:  D:  E: None of these The same loop is now in a non-uniform field. The field grows stronger near the bottom of the page. The direction of the net force is ... Answers: Question 1: The net force is zero. Each side of the square loop feels a force of magnitude F = ILB (L = length of edge) directed away from the center of the loop. All the forces cancel. Question 2: In the non-uniform B-field shown, the net force is upward. The upward force on the top edge of the loop is larger than the downward force on the bottom edge of the loop. D B C A B stronger B weaker E: net force is zero B (out) I CTM-7 A magnetic compass is placed at the points A, B, and C near an electric circuit which has the following twisty shape: The deflection of the compass needle is a measure of the strength of the magnetic field. The relative deflection of the needle n order from biggest to smallest deflection is.. A) ABC B) CAB C) BCA D) ACB E) None of these. Answer: The correct order is CAB. Big currents make big magnetic fields. Near C there are two currents in the same direction which are effectively one big current. Near B, there are two equal currents going in opposite directions, which (if they were right on top of each other) would give a net current of zero. Point A is the middle case. A B C TM-10 A positively-charged particle with an initial speed vo is moving in a plane perpendicular to a uniform magnetic field (B into the page). There is a tenuous gas throughout the region which causes viscous drag and slows the particle over time. The path of the particle is A: a spiral inward B: a spiral outward C: something else If the time for the particle to complete the first revolution (once around) is 1 second, the time for the first 5 revolutions is A: > 5 s B: < 5 s C: 5 s Answers: Question 1: From Fnet = ma, we have qvB = m v2 / R or R = (mv)/(qB). Notice that R is proportional to v. So smaller speed v means smaller radius R. The particle spirals inward as it loses speed. Question 2: This is a tough one and requires you to work out a formula for the period T. The period T is the time for one complete revolution. T is related to the speed v by v = (distance)/time = (2R)/T. So the period T = (2R) / v. Using R = (mv)/(qB), we get T = (2m)/(qB). Notice that the period T is independent of the radius R or the speed v. So the time for each revolution is 1 second, and the time for 5 revolutions is 5 s. B CTM-11 A square loop of wire carrying a current I, is in a constant uniform magnetic field B. The magnetic field is perpendicular to the plane of the loop. The loop can rotate freely about the axis. The loop tends to rotate so that the near edge moves.. A) to the right B) to the left C) The net torque on the loop is zero, so it does not tend to move. Answer: Since from above the situation is as shown below. The direction of the forces are from the Right hand rule. The net torque is zero; there is no tendency to rotate. The net torque is zero. B I(in) I(out) F F B I axis Near edge Far edge B CTM-12 A permanent bar magnet is broken in half. The two pieces are interchanged, keeping their orientations fixed, as shown below. Do the pieces attract or repel? A) Attract B) Repel C) No force at all. Answer: The bar magnet can be thought of as a solenoid of current. Parallel currents attract. When the magnet is broken and rearranged, the currents on the ends are parallel so there is an attraction. They attract. (In the diagram, below we choose the right end to be North. If we had chosen the right end to be South, all the currents would be reversed, but the end currents would still be parallel. 1 2 2 1 1 2 2 1 NorthSouth South South South South North NorthNorth North