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# First Notes M 362K

UT

GPA 2.94

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## Popular in Probability I

## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Tomi Olubeko on Thursday January 7, 2016. The Class Notes belongs to M 362K at University of Texas at Austin taught by in Fall 2016. Since its upload, it has received 22 views. For similar materials see Probability I in Mathematics (M) at University of Texas at Austin.

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Date Created: 01/07/16

olubeko (oeo227) – HW7 – ditmire – (58565) 1 This print-out should have 28 questions. 6. Into the page Multiple-choice questions may continue on the next column or page – ﬁnd all choices 7. To the right before answering. 003 (part 3 of 6) 10.0 points 001 (part 1 of 6) 10.0 points What is the speed of the charged particle as it enters the region of the magnetic ﬁeld? A particle of mass 1.2948 × 10 −25 kg and Answer in units of m/s −19 charge of magnitude 4.8 × 10 C is acceler- ated from rest in the plane of the page through 004 (part 4 of 6) 10.0 points a potential diﬀerence of 429 V between two What is the magnitude of the force exerted on parallel plates as shown. The particle is in- the charged particle as it enters the region of jected through a hole in the right-hand plate the magnetic ﬁeld B ? into a region of space containing a uniform Answer in units of N magnetic ﬁeld of magnitude 0.478 T oriented perpendicular to the plane of the page. The 005 (part 5 of 6) 10.0 points particle curves in a semicircular path and What is the distance from the point of injec- strikes a detector. tion to the detector? q Region of 006 (part 6 of 6) 10.0 points m Magnetic hole Field B What is the work done by the magnetic ﬁeld on the charged particle during the semicircu- lar trip? E −16 1. W = 2.0592 × 10 J −16 2. W = 1.31093 × 10 J What is the sign of the charge of the parti- cle? Neglect relativistic eﬀects. 3. W = 0 J −16 1. The charge q cannot be determined. 4. W = −1.0296 × 10 J 5. W = −1.31093 × 10 −16 J 2. The charge q is negative (−). −16 3. The charge q is positive (+). 6. W = 4.1184 × 10 J −16 002 (part 2 of 6) 10.0 points 7. W = −4.1184 × 10 J Which way does the magnetic ﬁeld point? 8. W = 1.0296 × 10 −16 J 1. Toward the bottom of page −16 9. W = −2.0592 × 10 J 2. Cannot be determined 007 10.0 points 3. To the left A square loop of wire carries a current and is located in a uniform magnetic ﬁeld. The left side of the loop is aligned and 4. Toward the top of page attached to a ﬁxed axis (dashed line in ﬁgure). 5. Out of the page olubeko (oeo227) – HW7 – ditmire – (58565) 2 q B ← axis of rotation 5. T = B = 0.37 T m mB 6. T = 2π q → y q B 4 A 7. T = 2π m 5 x .6 m 0 8. T = mB → q 0.66 m 9. T = π q B B = 0.37 T m 2m 10. T = When the plane of the loop is parallel to the q B magnetic ﬁeld in the position shown, what is the magnitude of the torque exerted on the 009 (part 1 of 4) 10.0 points loop about the axis of rotation, which is alonA device (source) emits a bunch of charged the left side of the square as indicated by the dashed line in the ﬁgure? ions (particles) with a range of velocities (see Answer in units of Nm ﬁgure). Some of these ions pass through the left slit and enter Region I in which there is a vertical uniform electric ﬁeld (inthe −ˆ 008 10.0 points A particle with a positive charge q and massdirection) and a 0.3 T uniform magnetic ﬁeld (aligned with the ±k-direction) as shown in m is undergoing circular motion with speed the ﬁgure by the shaded area. v. At t = 0, the particle is moving along the negative x axis in the plane perpendicular ±1300 V Region of ▯ to the magnetic ﬁeld B, which points in the q Magnetic positive z direction as shown in the ﬁgure Field below. m 0.3 T y 45 cm y 1.1 cm B▯ v x z x Region I Region II z ı is in the direction +x (to the ri is, ˆ Find the period of the circular motion; i.in the direction +y (up the page), and k is in the time takes for the particle to complete the direction +z (out of the page). revolution. In order for an ion to pass through both 2mB slits on a straight line, which of the following 1. T = q conditions must be true for the forces on the 2π m ion? (E is the electric force vector Bnd F is 2. T = the magnetic force vector.) q B m 1 3. T = q B 1. FE = FB 2 4. T = 2q B ▯ ▯ m 2. FE = 2FB olubeko (oeo227) – HW7 – ditmire – (58565) 3 3. F = −F▯ rod hangs from two wires (in the same plane E B as the rod) in a uniform vertical magnetic 4. FE = − 1 FB ﬁeld as in the ﬁgure. 2 The acceleration of gravity is 9.8 m/s . 5. FE = 0, F▯B ▯= 0 6. F ▯= 0, F▯ = 0 E B θ 7. FE = −2F▯ B B B 9 mm 4 m 9 ▯ ▯ v 8 m 8. FE ⊥ F B so it is impossible. b 7. A / 9. F = F ← s2 E B 010 (part 2 of 4) 10.0 points If the wires make an angle of 64 with the In which direction (relative to the coordinatevertical when in equilibrium (v = 0 m/s), system shown above) should the magnetic determine the strength of the magnetic ﬁeld. ﬁeld point in order for negatively charged ions Answer in units of T to move along the path shown by the dotted line in the diagram? 014 10.0 points A straight wire segment 4 m long makes an 1. ▯B▯ = 0; direction undetermined angle of 90 with a uniform magnetic ﬁeld of 0.33 T. ▯ 2. B = −k Find the magnitude of the force on the wire ▯B▯ if it carries a current of 3.3 A. Answer in units of N B ˆ 3. ▯ = +k ▯B▯ 015 (part 1 of 2) 10.0 points 011 (part 3 of 4) 10.0 points A circular current loop of radius R is placed In Region I, the electric potential between thin a horizontal plane and maintains a current plates is 1300 V, the distance between the I. There is a constant magnetic ﬁeld B in plates is 1.1 cm, and the magnetic ﬁeld in the xy-plane, with the angle α(0 < α < 90 ) both Regions I and II is 0.3 T. deﬁned with respect to y-axis. The current in What is the speed of a singly charged ion the loop ﬂows clockwise as seen from above. that passes through both slits and makes it y,ˆ x into Region II? α ˆı B Answer in units of m/s 012 (part 4 of 4) 10.0 points The ions that make it into Region II are ob- I served to be deﬂected downward and then follow a circular path with a radius of 0.45 m. k What is the mass o−18he ions? The charge z on each ion is 9.9 × 10 C. What is the direction of the torque vector Answer in units of kg ▯τ exerted on the current loop by the ﬁeld? 013 10.0 points 1.τ = +ˆı A metal rod having a mass per unit length of 0.0654 kg/m carries a current of 7.6 A. The olubeko (oeo227) – HW7 – ditmire – (58565) 4 2. τ = −kˆ ˆ 1.2 A 3. τ = ı√+ k 2 4. τ = −ˆ 0.48 m0.32 m ˆ 5. τ = k√−ˆı 2 P 6. τ = ı + sinα What is the magnetic moment of this cur- rent loop? 7. τ = k −ˆ sinα Answer in units of A · m 8. τ = +kˆ 018 10.0 points The two wires of length 2 m are 3 mm apart and carry a current of 16 A dc. 9. τ = +ˆ Calculate the force between these wires. 10. τ = −ˆı Answer in units of N 016 (part 2 of 2) 10.0 points 019 (part 1 of 2) 10.0 points What is the magnitude of the torque vector? Imagine a very long, uniform wire that has a linear mass density of 0.0052 kg/m and that 2 π encircles the Earth at its equator. 1. ▯τ▯ = I R B sin 2 + α π Assume the Earth’s magnetic dipole mo- 2. ▯τ▯ = I π R B sin − α ment is aligned with the Earth’s rotational 2 axis. The Earth’s magnetic ﬁeld is cylindri- 2 3. ▯τ▯ = I R B sinα cally symmetric (like an ideal bar ma2netic). The acceleration of gravity is 9.8 m/s and 4. ▯τ| = I π R B sinπ the magnetic ﬁeld of the earth is 2 × 1T. 2 What is the magnitude of the current in 2 5. ▯τ▯ = I 2π R B sinα the wire that keeps it levitated just above the ground? 6. ▯τ▯ = I π R B sinα Answer in units of A π 7. ▯τ▯ = I R B sin − α 020 (part 2 of 2) 10.0 points 2 The current in the wire goes in the 2 π 8. ▯τ▯ = I 2π R B sin 2 + α π 1. opposite direction as the Earth’s spinning 9. ▯τ▯ = I 2π R B sin − α motion (East to West). 2 2 π 10. ▯τ▯ = I π R B sin + α 2. same direction as the Earth’s spinning 2 motion (West to East). 017 10.0 points A wire loop consists of two semicircles con- 021 (part 1 of 4) 10.0 points nected by straight segments. The inner and A current I ﬂows in the directions indicated outer radii are 0.32 m and 0.48 m, respec- by the arrows in the ﬁgure along a wire bent tively. A current of 1.2 A ﬂows in this loop. to form concentric semicircles. olubeko (oeo227) – HW7 – ditmire – (58565) 5 5. B = 0 b 6. B = π µ0I I b 4b 7. B = π µ0I b b 4 b a 8. B = 4π µ0I P b b What is the magnitude of the component of 9. B = µ0I ▯ b 4b the magnetic ﬁeld (▯a ▯ ≡ Ba) at P due to the current in the inner semicircle (r = a)?10. B = µ0I b 2b 4π µ0I 1. Ba= 023 (part 3 of 4) 10.0 points a µ0I What is the magnitude of the magnetic ﬁeld 2. Ba= (▯B▯ ≡ B) at P due to the current in the a entire loop? µ0I 3. Ba= 2a 1. B = µ0I 1 + 1 π µ0I a 2π a b 4. Ba= 8 2. B = µ0I 1 − 1 µ0I 2 a b 5. Ba= π a 3. B = µ0I 1 + 1 π µ0I 2 a b 6. Ba= 4a µ0I 1 1 µ0I 4. B = 4π a + b 7. Ba= 4a µ0I 1 1 5. B = 4 a − b 8. Ba= 0 µ0I 1 1 4µ 0 6. B = + 9. Ba= a 4 a b 7. B = 0 µ0I 10. Ba= 8a µ0I 1 1 022 (part 2 of 4) 10.0 points 8. B = 2π a − b What is the magnitude of the component of µ0I 1 1 the magnetic ﬁeld (▯b ▯ ≡ Bb) at P due to 9. B = 4π a − b the current in the outer semicircle (r = b)? 024 (part 4 of 4) 10.0 points 1. B = 4µ 0 What is the direction of the magnetic ﬁeld at b b P due to the current in the entire loop? 2. B = µ0I b b 1. Into the page 3. B = µ0I 2. Out of the page b 8b 4. B = µ0I 3. The direction cannot be determined using b π b simple techniques. olubeko (oeo227) – HW7 – ditmire – (58565) 6 move. 4. The magnitude is 0, thus the direction is undetermined. b b 025 10.0 points 2.7 V . m Consider two radial legs extending to inﬁnity 5 and a circular arc carrying a current I as shown below. y 0.29 m When the switch is closed, the two parallel I 7 π metal rods 12 1. will move toward each other. I r I 2. will heat up, and remain motionless. O I x 3. will change in a way not determined by the information given. What is the magnitude of the magnetic ﬁeld B at the origin O due to the current through O 4. will move away from each other. this path? 5 µ 0 5. both will move toward the right. 1. B O 36 r 7 µ 0 6. both will move toward the left. 2. B O 36 r 027 (part 2 of 3) 10.0 points 5 µ 0 3. B O The magnetic ﬁeld at the right-most rod, due 32 r to the current in the rod immediately to its 7 µ 0 4. B O left, is 48 r 5 µ I 1. in the plane of the paper pointing left. 5. B O 0 28 r 7 µ I 2. not determined from information given. 6. B O 0 44 r 3 µ I 3. out of the plane of the paper. 7. B = 0 O 22 r 4. in the plane of the paper pointing right. 8. B O 0 5. into the plane of the paper. 3 µ 0 9. B O 14 r 028 (part 3 of 3) 10.0 points 7 µ I 10. B O 0 The battery is 2.7 V, each metal rod has a 40 r resistance R = 2.9 Ω, a length of 5.2 m, and a mass of 0.0018 kg. The rods are separated by 026 (part 1 of 3) 10.0 points Two identical parallel sections of metal rods 0.29 m. Find the magnitude of the acceleration of are connected parallel to a battery as shown. each rod. Neglect the resistance of the part of The two sections of metal rods are free to the circuit not speciﬁcally mentioned below. olubeko (oeo227) – HW7 – ditmire – (58565) 7 2 Answer in units of m/s

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