1 of 11 More B-field Concept Tests B2-1. A current-carrying wire is in a B-field. The wire is parallel to the B-field as shown. What is the direction of the magnetic force on the wire? A) right B) left C) up D) down E) None of these. ` B I Answer: the force is zero, so no direction. B2-2. A circular loop of wire with radius R is carrying current I as shown. The B-field at the center of a circular loop is A) zero B) out of the page C) into the page D) None of these Use Biot-Savart 02ˆIdl rdB4r×=KKµπ I R Answer: out of the page Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder2 of 11 B2-3. Which point A or B has the larger magnetic field? A B C ) The B-field is the same at A and B. A B I I Answer: Case B has the larger magnetic field. Use the Biot-Savart Law to get the directions of the B-field due to the two semi-circular portions of the loop. In A the two fields oppose each other; in B they add. B2-4. A long straight wire is carrying current I. The magnetic field at point x has magnitude Btotal. I B1/2 =? r x r Btotal True (A) or False (B): The magnetic field GB12/ at point x due only to the current to the left of x points in the same direction and has 1/2 the magnitude of GBtotal. Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder3 of 11 Answer: True. GGGBdB Idl rrtot==×zzµπ024 . The field element dBG always points in the same direction, so the vectors just add like scalars (numbers). Integrating over the left half of the wire gives half the result. B2-5. 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? 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) 1 2 3 4 y (in) (out) (out) x (out) Hint:B) C) D) E) None of these Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder4 of 11 B2-6. A magnetic compass is placed at the points A, B, and C near an electric circuit which has the following twisty shape: A B C The deflection of the compass needle is a measure of the strength of the magnetic field. The relative deflection of the needle in order from biggest to smallest deflection is.. A) ABC B) CBA 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. B2-7. A rectangular loop of wire is carrying a current i in the clockwise direction and is near a long straight wire carrying a current I, as shown. What is the direction of the net force on the rectangular loop, due to the B-field from the long, straight wire. I A B i D C E) Net force is zero. Answer: The net force is up. The B-field due to the long straight wire decreases in Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder5 of 11 magnitude as you move further from it according to BIro=µπ2. So the upward force on the top of the wire loop F = ILB, is greater than the force on the bottom of the loop. Another way to see the answer: Parallel currents attract and anti-parallel currents repel. So the upper portion of the loop feels an upward force and the bottom portion of the loop feels a downward force. But the upper portion is closer to the straight wire, so there is a bigger field, a bigger force. B2-8. The (perhaps incorrect) magnetic field lines around an object are partly obscured by a screen. Is this partial field line diagram possible? B A) Yes B) No Answer: Yes. The field lines can be coming in toward the center from above and below behind the screen. This would be the case when two bar magnets have their North poles pointing toward each other. Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder6 of 11 B2-9. The imaginary loop L near a wire with current I has 4 segments labeled 1, 2, 3, and 4, as shown. What is Bd⋅∫KKA for each of the segments? For 1, 2, 3, 4, the line integral is A) + + 0 – 3 B) 0 + – 0 C) + 0 – 0 D) + 0 + 0 E) None of these 1 4 I(in) 2 Answer: + 0 – 0 B2-10. We need a sign convention for I_thru. Place the fingers of your right hand around the imaginary loop and your thumb points in the direction of positive I_thru. What is I_thru here, where all three wires have 5 A? 5 A 5 A5 A A) +15 A B) –10 A C) +10 A D) –5 A E) None of these Answer: –5 A Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder7 of 11 B2-11. A long straight copper wire has radius b and carries a constant current of magnitude I. The current density of magnitude J=I/(πb2) is uniform throughout the wire. What is the current contained in the circular loop £, with radius r < b, centered on the wire's center as shown? A) rIb⎛⎞⎟⎜⎟⎜⎟⎜⎝⎠ B) 2rIb⎛⎞⎟⎜⎟⎜⎟⎜⎝⎠ b r J = constant C) 3rIb⎛⎞⎟⎜⎟⎜⎟⎜⎝⎠ D) None of these How does the magnitude of the B-field a distance r < b from the center of the wire depend on r? A) B ∝ r B) B = constant C) B ∝ 1/r D) B ∝ 1/r2 E) None of these/don't know. Referring to the same wire, for the new loop £ shown below, what is the value of LBd⋅∫GGAv? A) +µ0 I/2 b J = constant B) > µ0I/2 C) < µ0I/2 Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder8 of 11 Answers: Part 1: Current thru loop of radius r is rbIFHGIKJ2. Current thru smaller circle = total current I ×=area of smaller circlearea of larger circleIrbππ22 Part 2: Inside the wire, B ∝ r. Ampere's Law says 22othru o2rBd I , B(2r) I , or Br r, B r.bℑ⎛⎞⎟⎜⋅ =µ π =µ ∝∝⎟⎜⎟⎟⎜⎜⎝⎠∫GGAv Outside the wire, we have the standard formula for the B-field outside a wire: BIro=µπ2 B b 0 r B ~ 1/r B ~ r So the field vs. distance from the center of the wire looks like: Part 3: Technically the answer is –µo I/2 The loop encloses half the current, so by Ampere's Law, …
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