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Various questions related to magnetic fields, including the forces on negative and positive charges, the effect of electric and magnetic fields on particle motion, and the deflection of charged particles in velocity selectors. It also discusses the interaction between current-carrying wires and magnetic fields.
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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.
CTM- 4 A proton (charge q = +e), moving to the right with an initial speed v, passes into a region where there is a vertical electric field pointing down and a horizontal magnetic field pointing into the page. Is it possible that the forces from the E- and B-fields cancel and the proton moves straight through undeflected. (Ignore gravity in this problem.) A) Yes, the forces can cancel and the particle moves straight through B) No, the forces cannot perfectly cancel; the particle path will be bent.
The E- and B-fields of the velocity selector are adjusted so that protons with a certain speed v pass through undeflected. Now electrons with the same speed are shot into the velocity selector (with same E- and B-fields as before). Do the electrons also pass through undeflected? A) Yes, the electrons go straight through B) No, the electrons are bent Answers: Question 1. Yes, the forces can cancel and the proton moves straight through. The electric field force is down and has magnitude eE (since F E = q E , and q = +e). The magnetic field force is up (by the right-hand rule) and has magnitude evB. The two forces exactly cancel when eE = evB, or v = E /B. Question 2. Yes, the electrons go straight through. Electrons have the same size charge as the proton, just opposite sign. So both the electric force and the magnetic forces have the same magnitude as before, only the directions are switched. The forces still cancel.
FE = eE FB = evB proton
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.
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.
CTM- 8 4 parallel wires each carry a current I. 3 of the wires carry current out the page, 1 carries current into the page, as shown. What is the direction of the B-field at the center of the square? x y (out) (in) (out) (out) Hint:
Answer: Each of the four currents (labeled 1, 2, 3, 4 below) creates a B-field at the cent of the square. The net B-field is the vector sum of these 4 fields and it points to the upper right. x y 2 (out) 4 (in) 1 (out) 3 (out)
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 v^2 / 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 = (2R)/T. So the period T = (2R) / v. Using R = (mv)/(qB), we get T = (2m)/(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.
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)
CTM- 13 A row of positive charges sites stationary on the ground. A person with a gauss- meter(which measures the magnetic field) is running to the right along the row of charges, at the same height as the charges and in front of them (in the diagram below). What is the direction of the B-field which the observer measures? Answer: From the person's point of view, in the person's reference frame, the person is at rest and the (+) charges are moving to the left, so there is a current to the left. The B-field produced at the person's position is up.
