ragsdale (zdr8) HW7 ditmire (585) This print-out should have 8 questions. Multiple-choice questions ma continue on the net column or page find all choices efore answering. 00 0.0 points A wire carring a current 0 A has a length 0. m etween the pole faces of a magnet at an angle 60 (see the figure). The magnetic field is approimatel uniform at 0.5 T. We ignore the field eond the pole pieces. θ l What is the force on the wire? Correct answer:.9904 N. Let : = 0 A, l = 0. m, θ = 60, and = 0.5 T. we use F = l sinθ, so F = l sin θ = (0 A) (0. m) (0.5 T) sin60 =.9904 N. 00 0.0 points The magnetic force on a straight 0.45 m segment of wire carring a current of 4.5 A is 0. N. What is the magnitude of the component of the magnetic field that is perpendicular to the wire? Correct answer: 0.5086 T. The magnetic force is F m = l = F m l 0. N = (4.5 A) (0.45 m) = 0.5086 T 00 0.0 points A rectangular loop of wire hangs verticall as shown in the figure. A magnetic field is directed horizontall, perpendicular to the wire, and points out of the page at all points as represented the smol. The magnetic field is ver nearl uniform along the horizontal portion of the wire a (length is 0. m) which is near the center of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a alance which measures a downward force (in addition to the gravitational force) of 0 N when the wire carries a current 0. A. a l F What is the magnitude of the magnet field at the center of the magnet? Correct answer:.5 T. Let : l = 0.45 m, = 4.5 A, F m = 0. N. and Let : F = 0, l = 0. m, and = 0. A.
ragsdale (zdr8) HW7 ditmire (585) The magnetic forces on the two vertical sections of the wire loop point to the left and right, respectivel. The are equal ut in opposite directions, so the add up to zero. Thus the net magnetic force on the loop is that on the horizontal section a whose length is l = 0. m (and θ = 90 so sin θ = ), and = F l = 0 N (0. A) (0. m) =.5 T. 004 (part of 4) 0.0 points Two wires each carr a current in the plane and are sujected to an eternal uniform magnetic field, which is directed along the positive ais as shown in the figure. L R L wire # wire # What is the magnitude of the force on wire # due to the eternal field?. = ( L + R). = ( L + R) ( ). L + R = 4. = R L 5. = (L + R) 6. = (L + R) correct 7. = L 8. = 4 (L + R) 9. = ( L + R) 0. = R F = d s = l. For wire, l, so the magnitude of F is just F = l. For wire #, l = L + R. Thus F = (R + L). 005 (part of 4) 0.0 points What is the magnitude of the force on the curved part of wire # due to the eternal magnetic field?. F = π R. F = π ( R + L). F = R correct 4. F = R L 5. F = π R 6. F = (R + L) 7. F = π (R + L) 8. F = ( R + L) 9. F = R 0. F = 0 R θ θ Using the setup pictured aove, we can determine the force on a segment d s of the
ragsdale (zdr8) HW7 ditmire (585) curved part of wire #. df = d s = sin θ ds. ds is related to dθ via the relationship s = R θ so that ds = R dθ. Now df = R sin θ dθ, which can e integrated from θ = 0 to θ = π to get the force F on the curved part of the wire: F = R π 0 = R. sin θ d θ 006 (part of 4) 0.0 points What is the magnitude of the force on the entire length of wire # due to the eternal magnetic field?. F = (L + R) correct. F = ( L + R). F = (L + R) 4. F = ( L + R) 5. F = R L 007 (part 4 of 4) 0.0 points The direction of the force on the entire length of wire # due to the eternal magnetic field is. down.. out of the paper.. to the left. 4. to the right. 5. into the paper. correct 6. up. Appling the right hand rule, gives the direction: pointing into the paper. 008 (part of ) 0.0 points A circular current loop of radius R is placed in a horizontal plane and maintains a current. There is a constant magnetic field in the -plane, with the angle α (α < 90 ) defined with respect to -ais. The current in the loop flows counterclockwise as seen from aove., ĵ α î 6. F = ( L + R) ( ) 7. F L + R = 8. F = R 9. F = 4 (L + R) 0. F = L The force on each of the straight wire segments is F = L. Therefore the total force on wire # due to the eternal magnetic field is f the current in the loop is 0.7 A and its radius is.05 cm, what is the magnitude of the magnetic moment of the loop? Correct answer: 0.000499 A m. z ˆk F = L + L + R = (L + R). Let : = 0.7 A, R =.05 cm. and
ragsdale (zdr8) HW7 ditmire (585) 4 The magnetic dipole moment is µ = A = π R = (0.7 A) (.5) (.05 cm) = 0.000499 A m 009 (part of ) 0.0 points What is the direction of the torque vector τ?. cos αĵ + sin α î a magnetic field with a magnitude of 0.09 T in the vertical direction along the z ais as shown in the figure elow. The acceleration of gravit is 9.8. θ z i = 0.8 A 0.74 m. +ĵ. sin αĵ + cos α î 0.74 m 4. ĵ 5. î 6. sin αĵ cos α î 7. +î 8. ˆk = 0.09 T = 0.09 T Determine the angle that the plane of the coil makes with the z ais when the coil is in equilirium. Correct answer: 8.97668. 9. + ˆk correct 0. cos αĵ sin α î The torque is τ = µ = µ (ĵ) [ ( î) + (+ĵ)] and ĵ ĵ = 0 and +ĵ ( î) = ˆk, so τ = µ ˆk so the direction of the torque is +ˆk, which agrees with the answer ased on the righthand rule. 00 (part of ) 0.0 points A long piece of wire with a mass of 0.47 kg and a total length of 70. m is used to make a square coil with a side of 0.74 m. The coil is hinged along a horizontal side (the ais), carries a 0.8 A current, and is placed in Let : L = 70. m, l = 0.74 m, m = 0.47 kg, i = 0.8 A, and = 0.09 T. Look down the positive -ais at the coil (that is, from the right-hand side of the original figure).
ragsdale (zdr8) HW7 ditmire (585) 5 φ θ µ mg Let θ e the angle the plane of the loop makes with the z ais as shown. Then the angle the coil s magnetic moment µ makes with the z ais is φ = 90 θ ; e.g., sinφ = cos θ and tanφ = cotθ. The numer of turns in the loop is z L N = circumference = 70. m 4 (0.74 m) = 47. The torque aout the z-ais due to gravit is τ g = r F ( ) l = cos φ mg, where l is the length of each side of the square loop. This gravitational torque tends to rotate the loop clockwise. The torque due to the magnetic force tends to rotate the loop counterclockwise aout the z-ais and has magnitude τ m = N A sin φ. At equilirium, τ m = τ g N l mg (l cos φ) sin φ =. Thus tan φ = mg N l (0.47 kg) (9.8) = (47) (0.09 T) (0.8 A) (0.74 m) = 6.04. Since tanφ = tan(90 θ) = cotθ, the angle the loop makes with the z ais at equilirium is θ = cot (6.04) = 8.97668. 0 (part of ) 0.0 points Find the torque acting on the coil due to the magnetic force at equilirium. Correct answer: 0.099 N m. At equilirium, τ m = N l sin φ = N l cos θ = (47) (0.09 T) (0.8 A) (0.74 m) cos(8.97668 ) = 0.099 N m. 0 0.0 points A long, thin conductor carries a current of 0.9 A. At what distance from the conductor is the magnitude of the resulting magnetic field 7.6 0 5 T? Correct answer:.9696 cm. Let : µ 0 =.5664 0 6 N/A, = 7.6 0 5 T, and = 0.9 A. The magnetic field is = µ 0 π r r = µ 0 π = (.5664 0 6 N/A ) (0.9 A) π (7.6 0 5 T) = 0.09696 m =.9696 cm.
ragsdale (zdr8) HW7 ditmire (585) 6 kewords: 0 0.0 points The wire is carring a current. 80 r O Find the magnitude of the magnetic field at O due to a current-carring wire shown in the figure, where the semicircle has radius r, and the straight parts to the left and to the right etend to infinit.. = µ 0 r. = µ 0 r. = µ 0 r 4. = µ 0 r 5. = µ 0 4 r correct 6. = µ 0 π r 7. = µ 0 π r 8. = µ 0 π r the iot-savart Law, = µ 0 d s ˆr r. Consider the left straight part of the wire. The line element d s at this part, if we come in from, points towards O, i.e., in the - direction. We need to find d s ˆr to use the iot-savart Law. However, in this part of the wire, ˆr is pointing towards O as well, so d s and ˆr are parallel meaning d s ˆr = 0 for this part of the wire. t is now eas to see that the right part, having a d s antiparallel to ˆr, also gives no contriution to at O. Let us go through the semicircle C. The element d s, which is along the wire, will now e perpendicular to ˆr, which is pointing along the radius towards O. Therefore d s ˆr = ds using the fact that ˆr is a unit vector. So the iot-savart Law gives for the magnitude of the magnetic field at O: = µ 0 C ds r. Since the distance r to the element d s is constant everwhere on the semicircle C, we will e ale to pull it out of the integral. The integral is C ds r = r ds = C r L C where L C = π r is the length of the semicircle. Thus the magnitude of the magnetic field is = µ 0 r π r = µ 0 4 r. 04 (part of 5) 0.0 points The current loop ACDA carries current in the direction indicated in the figure, where the segments A and CD are concentric arcs with radii a and respectivel. D A 60 O a What is the magnitude of the resultant magnetic field at the center of curvature O due to the current segment of arc CD? C
ragsdale (zdr8) HW7 ditmire (585) 7. CD = µ 0 4. CD = µ 0 6. CD = µ 0 4. CD = µ 0 5. CD = µ 0 6. CD = µ 0 7. CD = µ 0 4 8. CD = 0 correct 9. CD = µ 0 6 asic Concepts: iot-savart law: d = µ 0 d s ˆr r Conversion from degrees to radians: 60 = π radians Solution: For current along arc CD, d s is perpendicular to ˆr. Therefore, if we are not concerned with the field s direction, the iot- Savart law gives d = µ 0 ds for current along arc CD. ntegrating over the segment ields CD = µ 0 D ds C = µ 0 ( α) = µ 0 α = µ 0 = µ 0, π where the pseudo unit of α should e radians. Radian denotes a pure numer, and is not a unit. 05 (part of 5) 0.0 points What is the direction of the resultant magnetic field at the center of curvature O due to the current segment of arc CD?. negative direction. pointing out of the paper correct. undetermined 4. positive direction 5. pointing into the paper 6. negative direction 7. positive direction Using the right-hand rule, for d s ˆr, the resultant magnetic field vector points out of the paper. 06 (part of 5) 0.0 points Find the magnitude of the magnetic field at O due to the current segment C.. C = µ 0. C = µ 0. C = µ 0 4 4. C = µ 0 5. C = µ 0 6 6. C = µ 0 6 7. C = µ 0 4 8. C = µ 0 ( a) ( a) a a ( a) a ( a) a
ragsdale (zdr8) HW7 ditmire (585) 8 9. C = 0 correct For segment C, d s is from an element on C to O, so it is antiparallel to the direction of ˆr. Therefore d s ˆr = 0 = C = 0. 07 (part 4 of 5) 0.0 points What is the direction of the resultant magnetic field at the center of curvature O due to the current loop ACDA?. negative direction.. pointing out of the paper.. positive direction. 4. negative direction. 5. pointing into the paper. correct 6. positive direction. 7. undetermined. Like Part, the magnetic field due to segment DA is also 0. Like Part, the magnitude of magnetic field due to segment A is given A = µ 0 a ds = µ 0 A a (a α) = µ 0 α a However, the magnetic field due to segment A is pointing into the paper, while the magnetic field due to segment CD is pointing out of the paper. Since a <, the contriution A is larger in magnitude than that CD. Put all the previous results aove together, we get the direction of the resultant magnetic field is pointing into the paper. 08 (part 5 of 5) 0.0 points What is the magnitude of the magnetic field due to the entire loop?. = µ 0 4 ( a ). = µ ( 0 a ). = µ ( 0 6 a ) 4. = µ ( 0 4 a ) 5. = µ ( 0 a ) 6. = µ ( 0 a ) 7. = µ ( 0 a ) correct 8. = 0 9. = µ 0 6 ( a ) ts magnitude is given = A CD = µ 0 α = µ ( 0 a ). ( a ) 09 (part of ) 0.0 points A conductor in the shape of a square of edge length l carries a clockwise current as shown in the figure elow. l What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in the wire?. = 4 µ 0 π l. = 4 µ 0 π l P
ragsdale (zdr8) HW7 ditmire (585) 9. = µ 0 π l 4. = µ 0 π l 5. = µ 0 π l 6. = µ 0 π l 7. = µ 0 π l 8. = µ 0 correct π l 9. = µ 0 π l 0. = µ 0 π l the iot-savart law, d = µ 0 d s ˆr r. Consider a thin, straight wire carring a constant current along the -ais with the - ais pointing towards the center of the square, as in the following figure. ˆr r θ P a ds O Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element d s is at a distance r from P. The direction of the field at P due to this element is out of the paper, since d s ˆr is out of the paper. n fact, all elements give a contriution directl out of the paper at P. Therefore, we have onl to determine the magnitude of the field at P. n fact, taking the origin at O and letting P e along the positive ais, with ˆk eing a unit vector pointing out of the paper, we see that d s ˆr = ˆk d s ˆr = ˆk (d sin θ). Sustituting into iot-savart law gives d = ˆk d, with d = µ 0 d sin θ r. () n order to integrate this epression, we must relate the variales θ,, and r. One approach is to epress and r in terms of θ. From the geometr in the figure and some simple differentiation, we otain the following relationship r = a sin θ = a csc θ. () Since tanθ = a from the right triangle in the figure, = a cotθ and d = a csc θ dθ. () Sustituting Eqs. and into Eq. gives d = µ 0 = µ 0 a a csc θ sin θ dθ a csc θ sin θ dθ. (4) Thus, we have reduced the epression to one involving onl the variale θ. We can now otain the total field at P integrating Eq. 4 over all elements sutending angles ranging from θ to θ as defined in the figure. This gives the magnetic field as = µ 0 θ sin θ dθ a θ = µ 0 a (cos θ cos θ ). We can appl this result to this prolem. For a segment of wire, as set up aove, the magnetic field at a point P is = µ 0 a (cosθ cos θ ).
ragsdale (zdr8) HW7 ditmire (585) 0 For a square wire loop consider the ottom segment. Using the aove general formula for this case gives one = µ ( 0 ( ) cos π l 4 cos π ) 4 = µ 0, π l where the total magnetic field is = 4 one = µ 0, π l since all four sides contriute. The direction and magnitude of the field is the same for each side. 00 (part of ) 0.0 points What is the direction of the magnetic field P at point P due to the upward current in the left-hand side of the square wire?. is up the page.. is to the left.. is in to the page. correct 4. is down the page. 5. is zero. 6. is out of the page. 7. is to the right. The direction of the magnetic field due to a current element is determined the cross product in the definition of the magnetic field = µ 0 d s ˆr r. For the present case the right hand rule gives the direction of the magnetic field as into the page or in for short. 0 0.0 points Three ver long wires are strung parallel to each other as shown in the figure elow. Each wire is at the perpendicular distance 8 cm from the other two, and each wire carries a current of magnitude 4.4 A in the directions shown in the figure. 8 cm 8 cm 8 cm z z Cross-sectional View The permeailit of free space is.566 0 6 T m/a. Find the magnitude of the net force per unit length eerted on the upper wire (wire ) the other two wires. Correct answer:.95 0 5 N/m. Let : r = 8 cm = 0.8 m = 4.4 A. and Magnetic field due to a long straight wire is = µ 0 π r, and the force per unit length etween two parallel wires is F l = µ 0 π r. There are two was to solve this prolem which are essentiall the same. The first wa is to find the net magnetic field at the upper wire from the two wires elow ( net = + ) and then find the force from F = L. The crucial step here will e to add the magnetic fields as vectors. The second wa would e to use F = L to find the net force on the upper wire from the two lower
ragsdale (zdr8) HW7 ditmire (585) wires F net = F + F, where we must e sure to add the forces as vectors. You should recognize that the two methods are formall identical. Let s do it the first wa. The magnitude magnetic field from wire is found from Ampere s law to e = µ 0 π r. Using the right hand rule the direction points up and to the left of wire as shown in figure. 0 0 60 60 60 Adding Magnetic Fields ˆk ĵ î The force is then F = L ( µ0 cos (0 ) ) = L ( ˆk î) π r ( µ0 cos (0 ) ) = L ( ĵ) π r F L = µ 0 cos(0 ) π r = (.566 0 6 T m/a) π (0.8 m) (4.4 A) cos(0 ) =.95 0 5 N/m. Notice: The direction of L is in the ˆk direction, since that is the direction of the current. After all the negative signs have cancelled, we notice that the force is in the ĵ direction. This is 90 from the positive -ais. F 0 F 0 60 F ˆk ĵ î 0 (part of ) 0.0 points Two circular loops of radius a = 0.79 cm whose planes are perpendicular to a common ais and whose centers are separated a distance l = a. The currents flow in the directions shown in the figure. 60 60 Adding Forces ts components will then e = [sin (0 )ĵ cos (0 ) î]. Similarl, points down and to the left of wire ; its components are given = [ sin (0 )ĵ cos (0 ) î]. Notice that smmetr the ĵ () component of the magnetic field vanishes. The net magnetic field is therefore net = µ 0 cos (0 ) π r î. C A left = 0 A right = 0 A l Find the magnitude of the magnetic field at point A on the ais midwa etween the loops. Correct answer: 0 T. At point A the field is zero. Note: The current elements on the two coils that are diagonall opposite through A, gives d contriutions that cancel at point A. C
ragsdale (zdr8) HW7 ditmire (585) 0 (part of ) 0.0 points Find the magnitude of the magnetic field at point C on the ais a distance l to the right of the right loop. Correct answer: 0.000867 T. Let : = 0 A, a = 0.79 cm = 0.0079 m, µ 0 = 0 7. and n the tet, the magnitude of the magnetic field along an aial point a distance z from the center of a current loop of radius a carring a current, is given = µ 0 0 a (a + ) /. At point C the field from the right hand loop is in the positive -direction, and the field from the left hand loop is in the negative - direction. Adding the fields the two loops, with = l = a/ for the right-hand loop, and = l = a for the left-hand loop, we otain = µ 0 a = µ 0 a / a + a 4 ( ) ] / a + a [ (4 ) / ( ) ] / 5 = ( 0 7 ) (0 A) (0.0079 m) = 0.000867 T. (0.6988) 04 (part of ) 0.0 points Point C is a distance l to the left of the left loop. The magnetic field at point C. points in the positive -direction, has a magnitude that is the same as the field at point C, and is in the same direction as the magnetic field at C. points in the negative -direction, has a magnitude that is the same as the field at point C, and is in the same direction as the magnetic field at C. points in the negative -direction, has a magnitude the same as the field at point C, and is in the opposite direction as the magnetic field at C 4. points in the positive -direction, has a magnitude that is different from the field at point C, and is in the same direction as the magnetic field at C 5. points in the positive -direction, has a magnitude that is different from the field at point C, and is in the opposite direction as the magnetic field at C 6. points in the negative -direction, has a magnitude that is different from the field at point C, and is in the same direction as the magnetic field at C 7. points in the negative -direction, has a magnitude that is different from the field at point C, and is in the opposite direction as the magnetic field at C 8. points in the positive -direction, has a magnitude that is the same as the field at point C, and is in the opposite direction as the magnetic field at C correct As alread stated the field at point C is positive. smmetr the magnitude of the field at point C is the same. The direction of the field at C is determined the closer loop, which appropriate direction rules, leads to a field at C that points in the negative - direction. 05 (part of 4) 0.0 points The figure elow shows a straight clindrical coaial cale of radii a,, and c in which equal, uniforml distriuted, ut antiparallel currents i eist in the two conductors.
ragsdale (zdr8) HW7 ditmire (585) i out a c simplifies to ( π r ) = µ 0 en, i in O F E D C r r r r 4 Which epression gives the magnitude of the magnetic field in the region r < c (at F)?. (r ) = µ 0 i (a r ) π r (a ). (r ) = µ 0 i r π a. (r ) = 0 4. (r ) = µ 0 i π r 5. (r ) = µ 0 i r π c correct 6. (r ) = µ 0 i (a ) π r (r ) 7. (r ) = µ 0 i (a + r ) π r (a ) 8. (r ) = µ 0 i π r 9. (r ) = µ 0 i r π 0. (r ) = µ 0 i (r ) π r (a ) Ampere s Law states that the line integral d l around an closed path equals µ 0, where is the total stead current passing through an surface ounded the closed path. Considering the smmetr of this prolem, we choose a circular path, so Ampere s Law where r is the radius of the circle and en is the current enclosed. For r < c, = µ 0 en π r = = µ 0 (i π r π c π r ( r ) µ 0 i c π r = µ 0 i r π c. 06 (part of 4) 0.0 points Which epression gives the magnitude of the magnetic field in the region c < r < (at E)?. (r ) = µ 0 i (a ) π r (r ). (r ) = µ 0 i r π. (r ) = µ 0 i (a r ) π r (a ) 4. (r ) = µ 0 i π r correct 5. (r ) = µ 0 i π r ) 6. (r ) = µ 0 i (a + r ) π r (a ) 7. (r ) = µ 0 i r π c 8. (r ) = 0 9. (r ) = µ 0 i r π a 0. (r ) = µ 0 i (r ) π r (a )
ragsdale (zdr8) HW7 ditmire (585) 4 For c < r <, = µ 0 en π r = µ 0 (i) π r = µ 0 i π r. 07 (part of 4) 0.0 points Which epression gives the magnitude of the magnetic field in the region < r < a (at D)?. (r ) = µ 0 i (a ) π r (r ). (r ) = µ 0 i π r. (r ) = µ 0 i r π c 4. (r ) = µ 0 i (r ) π r (a ) 5. (r ) = µ 0 i (a + r ) π r (a ) 6. (r ) = µ 0 i r π a 7. (r ) = µ 0 i π r 8. (r ) = µ 0 i (a r ) π r (a ) correct 9. (r ) = µ 0 i r π 0. (r ) = 0 For < r < a, = µ 0 en π r = = ( µ 0 i i π (r ) π (a ) π r ( a r ) µ 0 i a π r = µ 0 i (a r ) π r (a ). ) 08 (part 4 of 4) 0.0 points Which epression gives the magnitude of the magnetic field in the region r 4 > a (at C)?. (r 4 ) = µ 0 i (r 4 ) π r 4 (a ). (r 4 ) = µ 0 i r 4 π a. (r 4 ) = µ 0 i π r 4 4. (r 4 ) = µ 0 i r 4 π 5. (r 4 ) = µ 0 i r 4 π c 6. (r 4 ) = µ 0 i (a r 4 ) π r 4 (a ) 7. (r 4 ) = µ 0 i π r 4 8. (r 4 ) = 0 correct 9. (r 4 ) = µ 0 i (a ), π r 4 (r4 ) 0. (r 4 ) = µ 0 i (a + r4 ) π r 4 (a ) t s eas to find that when we draw a circle outside the coaial cale (r 4 > a), the total current enclosed in is zero since antiparallel currents eist in the two conductors, so there will e no magnetic field outside.