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CU-Boulder PHYS 1120 - More B-field Concept Tests

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1 of 12More 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.`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 theseUse Biot-Savart 02ˆId l rdB4 r�=vvmpAnswer: out of the pagePhys1120 Concept Tests, M. Dubson ©University of Colorado at BoulderBIIR2 of 12B2-3 Which point A or B has the larger magnetic field?A B C ) The B-field is the same at A and B.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.True (A) or False (B): The magnetic field B1 2/ 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 Btotal.Phys1120 Concept Tests, M. Dubson ©University of Colorado at BoulderI B1/2 =? r x r Btotal A B I I3 of 12Answer: True.  B dB Idl rrtot z z024. The field element dB 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.Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder(in)(out)(out)(out)xyHint:B)C)D)E) None of these4 of 12x y 2(out) 4(in) 1(out) 3(out) 1 2 3 4 Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder5 of 12B2-6 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 in order from biggest to smallest deflection is..A) ABC B) CBA C) BCAD) 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.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 BoulderIiADCBE) Net force is zero.A B C6 of 12magnitude as you move further from it according to BIro2. 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) magneticfield lines around an object are partlyobscured by a screen.Is this partial field line diagram possible?A) YesB) NoAnswer: 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 BoulderB7 of 12B2-9 The imaginary loop L near a wire with current I has 4 segments labeled 1, 2, 3, and 4, as shown.What is B d��vvl for each of the segments?For 1, 2, 3, 4, the line integral is A) + + 0 –B) 0 + – 0C) + 0 – 0D) + 0 + 0E) None of theseAnswer: + 0 – 0B2-10 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?A) +15 A B) –10 A C) +10 A D) –5 A E) None of theseAnswer: –5 APhys1120 Concept Tests, M. Dubson ©University of Colorado at Boulder5 A5 A5 AI(in)12348 of 12B2-11 A long straight copper wire has radius b and carries a constant current of magnitude I. The current density of magnitude J=I/(b2) 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����������C) 3rIb���������� D) None of theseHow 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/rD) B  1/r2E) None of these/don't know.Referring to the same wire, for the new loop £ shown below, what is the value ofLB d��r rl�?A) +0 I/2B) > 0I/2C) < 0I/2Phys1120 Concept Tests, M. Dubson ©University of Colorado at Boulderb r J = constant b J = constant9 of 12Answers: Part 1: Current thru loop of radius r is rbIFHGIKJ2. Current thru smaller circle = total current I  area of smaller circlearea of larger circleIrb22Part 2: Inside the wire, B  r. Ampere's Law says22o thru o2rB d I , B(2 r) I , or B r r , B r.b�� ���� = m p = m � �������� ��r rl�Outside the wire, we have the standard formula for the B-field outside a wire: BIro2So the field vs. distance from the center of the wire looks like:Part 3: Technically the answer is –o I/2 The loop encloses half the current, so by Ampere's Law,o enc oB d I = I / 2 .�� = m m�rrl� I accidentally chose the sense of the path the wrong way, so the line integral is negative.Phys1120 Concept Tests, M.


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CU-Boulder PHYS 1120 - More B-field Concept Tests

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