i need help with part C Learning Goal:To practice Problem-Solving Strategy 28.2 Ampere's Law.A solid cylindrical conductor is supported by insulating disks on the axis of aconducting tube with outer radius R. = 7.75 cm and inner radius R, = 3.15 cm .(Figure 1) The central conductor and the conducting tube carry equal currents ofI = 3.35 A in opposite directions. The currents are distributed uniformly over thecross sections of each conductor. What is the value of the magnetic field at adistance r = 5.03 cm from the axis of the conducting tube?SubmitPrevious AnswersCorrectThe magnetic field lines are concentric circles around the axis of the current carrying conductors. For this reason the choice of a circular integraconductors will greatly simply your calculations.EXECUTE the solution as followsPart CWhat is the value of the magnetic field at a distance r = 5.03 cm from the axis of the conducting tube? Recall that uo = 4T × 10¬7 T · m/A.Express your answer numerically in teslas.• View Available Hint(s)Figure1 of 1ΑΣφ?B =TSubmitEVALUATE your answerRaR5Part D Complete previous part(s)

Answered: Image /qna-images/answer/9cd89769-6171-43cb-908d-d5321b002f74.jpg

Learning Goal: To practice Problem-Solving Strategy 28.2 Ampere's Law. A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius R, = 7.75 cm and inner radius R, = 3.15 cm (Eigure 1) The central conductor and the conducting tube carry equal currents of I = 3.35 A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 5.03 cm from the axis of the conducting tube? Submit Previous Answers v Correct The magnetic field lines are concentric circles around the axis of the current carrying conductors. For this reason the choice of a circular inte conductors will greatly simply your calculations. EXECUTE the solution as follows Part C What is the value of the magnetic field at a distance r = 5.03 cm from the axis of the conducting tube? Recall that to = 4 x 10-7 T ·m/A. Express your answer numerically in teslas. > View Available Hint(s) Figure 1 of 1 > ? B= T Submit EVALUATE your answer Part D Complete previous part(s)

Learning Goal: To practice Problem-Solving Strategy 28.2 Ampere's Law. A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius R, = 7.75 cm and inner radius R, = 3.15 cm (Eigure 1) The central conductor and the conducting tube carry equal currents of I = 3.35 A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 5.03 cm from the axis of the conducting tube? Submit Previous Answers v Correct The magnetic field lines are concentric circles around the axis of the current carrying conductors. For this reason the choice of a circular inte conductors will greatly simply your calculations. EXECUTE the solution as follows Part C What is the value of the magnetic field at a distance r = 5.03 cm from the axis of the conducting tube? Recall that to = 4 x 10-7 T ·m/A. Express your answer numerically in teslas. > View Available Hint(s) Figure 1 of 1 > ? B= T Submit EVALUATE your answer Part D Complete previous part(s)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Learning Goal: To practice Problem-Solving Strategy 28.2 Ampere's Law. A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 7.15 cm and inner radius R = 5.05 cm. (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.15 A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 5.41 cm from the axis of the conducting tube? Figure R₂ Ra 1 of 1 The magnetic field lines are concentric circles around the axis of the ing → integration path dl that is coaxial with the conductors will greatly simply your calculations. EXECUTE the solution as follows Part C What is the value of the magnetic field at a distance r = 5.41 cm from the axis of the conducting tube? Recall that μ = 4π × 10-7 T·m/A. Express your answer numerically in teslas. ► View Available Hint(s) VE ΑΣΦ B= 7.9510-6 Submit Previous…
Learning Goal: To practice Problem-Solving Strategy 28.2 Ampere's Law. A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 7.15 cm and inner radius R = 5.05 cm. (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.15 A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 5.41 cm from the axis of the conducting tube? Figure R₂ r Ra 1 of 1 = Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: Ra = 7.15 cm, R₁ 5.05 cm, r = 5.41 cm, and r > Ra. Rank from largest to smallest. To rank items as equivalent, overlap them. Largest BarRRa The figure shows a horizontal solid conductor inside a horizontal thick conducting tube. The tube is a hollow cylinder with its inner and outer radii R subscript b and R…
1. Fill in the table for this circuit. Show your work and then check your answer with PHET V R P R1 R2 R3 TOTAL 14 V 780 2 IR R3 3.3 k2 R2 1.5 k2
pls see the attached picture, thank you!A. Theoretical Development. 1. (a) Design an ammeter with three scales: 0-200 µA, 0-2 mA, and 0-10 mA. Use a current sensor rated at 200 μA, 100 mV. What resistance does the instrument insert in the circuit at each scale? (b) Using the same current sensor as in (a), design a voltmeter with two scales: 0-10 V and 0- 100 V. What is the maximum current the instrument draws from the circuit for each scale?
t Jumper Cables and Ohm's Law 3 of 20 > Review | Constants Part B Learning Goal: To learn to apply the concept of current density and microscopic Ohm's law. What electric field E' would have been required to create current of 30.0 A in an aluminum wire of the same diameter? Express your answer in V/m. Use three significant figures. A "gauge 8" jumper cable has a diameter d of 0,326 cm. The cable carries a current I of 30.0 A. The electric field E in the cable is 0.062 V/m • View Available Hint(s) Below is a list of some common metals with their resistivities. ? Resistivity p (N. m) Material E' = V/m copper 1.72 x 10-8 Submit silver 1.47 x 10-8 aluminum 2.75 x 10 8 Part C Complete previous part(s) gold 2.44 x 10-8 Part D Complete previous part(s) steel 20 x 10-8 Next > Provide Feedback
Direction: Answer the following and show your solution. REFERENCE: a. Watch the video provided discussing what is current resistivity and resistance. i. Electric current - https://youtu.be/HXOok3mfMLM ii. Resistance - https://youtu.be/NfcgA1axPLo iii. Resistivity - https://youtu.be/nN3bb0aK1GM 1. The main purpose of a battery in a car or truck is to run the electric starter motor, which starts the engine. The operation of starting the vehicle requires a large current to be supplied by the battery. Once the engine starts, a device called an alternator takes over supplying the electric power required for running the vehicle and for charging the battery. (a) What is the average current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow from the battery?
II. Problem Solving. Answer the following problems. Show your complete solution for each problem on the space provided. 1. When a lightning strikes from the cloud to the ground, current as high as 25,000 amperes can occur and last for about 40µs. How much charge is transferred from the cloud to the ground during such a strike? 2. An aluminum cube has sides of length 1.80 m. What is the resistance between the two opposite faces of the cube?
Student Porta X A Classwork for x L 11-physics-1 x Danna solves O The 8-Year-O x b Success Confi X G electricity ph oogle.com/file/d/1leMb1M1fOZOQİXLCxpnmYwykAyqQcjSo/view Practice Problems 1. Determine the current and potential difference through each resistor in Figure 11.38. R = 10.0 2 AV 10.0 V R3 = 100 N R = 200 N %D R2 = 30.0 2 ww- R = 75.0 N %3D e here to search wwww
8. Interpret the circuit below to determine total resistance. + V3 + Show your work... 9 V V2 9 V ww R1 10 22 ww R2 25 Ω ww R3 50 Q
If three resistors connected in series to a battery. Suppose R1 carries a current of 4.5A, R2 has a resistance of 22.50, and R3 dissipates 95W of power. What is the voltage across R3 and Resistance of R3 ? 1. Voltage (in V) across R3 %3D and 2. Resistance (in 2) of R3 =
Problem solving. (Show your solution) 1. Find the equivalent resistance of the combination of resistors shown. R, == 5.0 2 R= 12.0 n Rg 30.0 2 R= 6.0 2 Rg = 6.0 2
! Required information A 274-V emf is connected to the terminals A and B. A B 15.092 R₁ A R₂ Here, V = 274V, R₁ = 15 £2, and R2 = 1992. What is the current in the 15- resistor?