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familiari nf3795 Magnetic Force and Field yeazell 58395 1 This print out should have 11 questions Multiple choice questions may continue on the next column or page nd all choices before answering 001 10 0 points Consider two parallel wires where the mag nitude of the left current is 2 I0 and that of the right current is I0 Point A is midway be tween the wires and B is an equal distance on the other side of the wires A B The ratio of the magnitude of the magnetic eld at point A to that at point B is 0 2 3 4 3 5 2 1 3 1 2 3 2 4 1 2 3 4 5 6 7 8 9 BA BB BA BB BA BB BA BB BA BB BA BB BA BB BA BB BA BB BA BB Long Wire B 0 I 2 r Principle of superposition Right hand rule Solution Let s call the distance between the wires r The magnetic eld at A due to the upward current is Bup A 0 2 I0 2 r 2 2 0 I0 r The right hand rule tells us the direction is into the paper Due to the fact that A is the same distance from both wires the total magnetic eld at A is BA 2 0 I0 r 0 I0 r 3 0 I0 r Now the eld at B due to the upward current is Bup B 2 0 I0 2 3 r 2 2 0 I0 3 r again into the paper while the downward current gives Bdown B 0 I0 2 r 2 0 I0 r out of the paper So at B the net eld out of the paper is Comparing their magnitudes we nd Bup B 2 3 cid 19 BB Bdown B 0 I0 1 r cid 18 0 I0 3 r BA BB 3 0 I0 r 0 I 3 r 9 002 10 0 points 10 9 correct Explanation Basic Concepts Magnetic Field due to a Consider a long wire and a rectangular current loop familiari nf3795 Magnetic Force and Field yeazell 58395 2 A D I2 I1 B a b C Determine the magnitude and direction of the net magnetic force exerted on the rectan gular current loop due to the current I1 in the long straight wire above the loop 1 F 2 F 3 F 4 F 5 F 0 I1 I2 2 0 I1 I2 2 0 I1 I2 2 0 I1 I2a 2 0 I1 I2 2 0 I1 I2 a b right a b left a a b cid 19 down cid 18 b a up a b up a b a b cid 19 down 6 F 7 F 2 cid 18 0 I1 I2 2 0 I1 I2 2 Explanation 8 F a b down b a a b cid 21 cid 20 up correct To compute the net force on the loop we need to consider the forces on segments AB BC CD and DA The net force on the loop is the vector sum of the forces on the pieces of the loop The magnetic force on AB due to the straight wire can be calculated by using FAB I2 B Z A B d s In order to use this we need to know the magnitude and direction of the magnetic eld at each point on the wire loop We can apply the Biot Savart Law The result of this is that the magnitude of the magnetic eld due to the straight wire is B 0 I1 2 r and the direction of the magnetic eld is given by the right hand rule the eld curls around the straight wire with the eld coming out of the page above the wire and the eld going into the page below the wire We can now nd the force on the segment AB applying the right hand rule to nd the direction of the B we see that the force will cross product d s be in the up direction Since the wire along the segment AB is straight and always at a right angle to B the cross product simpli es to B ds Since the magnitude of the magnetic eld is constant along segment AB it can come out of the integral which simpli es to give us the result FAB I2 B1 I2 0 I1 2 a cid 19 cid 18 Following the same argument we see that the force on the segment CD is FCD I2 0 I1 2 a b cid 21 cid 20 and its direction is down This is because the direction of the current is now in in the opposite direction along segment CD We can do the use the same procedure for segments BC and DA but because the mag netic eld decreases with distance from the straight wire B is changing along these seg ments This means that the integrals are not as simple Using the right hand rule we see that the force on segment BC is directed to wards the right and the force on segment DA is directed towards the left Because the two segments of wire are symmetrically placed their magnitudes will be equal Since these forces on the loop have equal magnitudes but opposite directions they will cancel Look ing more closely at segments BC and DA we see that for each small portion of the seg ment BC we can nd a small portion on the segment DA such that the forces on these two portions are the same magnitude The small portions are the same distance from the familiari nf3795 Magnetic Force and Field yeazell 58395 3 straight wire because the currents are in op posite directions the forces on them will be in opposite directions Their contributions to the net force cancel The net force on the loop is then Floop FAB I2 FCD 0 I1 2 a cid 19 b a a b cid 21 I2 cid 20 cid 20 cid 18 0 I1 I2 2 0 I1 2 a b cid 21 I1 I2 B1 FAB I2 I2 I2 I2 B1 FCD I2 A D B C Here a positive force is in the up direction and a negative force is in the down direction So the direction of the net force is up 003 part 1 of 2 10 0 points A circular current loop of radius R is placed in a horizontal plane and maintains a current I There is a constant magnetic eld B in the xy plane with the angle 0 de ned with respect to y axis The current in the loop ows clockwise as seen from above y 90 x B I What is the direction of the torque vector exerted on the current loop by the eld k z 1 k sin k 2 k b 2 3 b b 4 5 b b 6 7 b k 2 8 k correct b 9 b 10 b sin Explanation b y x I B k z 0 and k so the torque is B Bx k Bx By …

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