{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midterm 03

# midterm 03 - Version 098/ABCAC midterm 03 Turner(59070 This...

This preview shows pages 1–3. Sign up to view the full content.

Version 098/ABCAC – midterm 03 – Turner – (59070) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A coil formed by wrapping 55 . 9 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 24 . 8 with the direction of the field. When the magnitude of the magnetic field is increased uniformly from 270 μ T to 600 μ T in 0 . 215 s, an emf of 39 . 3 mV is induced in the coil. What is the total length of the wire? 1. 308.719 2. 242.655 3. 239.128 4. 272.502 5. 171.491 6. 158.831 7. 230.688 8. 314.146 9. 221.746 10. 339.031 Correct answer: 158 . 831 m. Explanation: From Faraday’s Law, E = N d Φ B dt |E| = vextendsingle vextendsingle vextendsingle vextendsingle N d Φ B dt vextendsingle vextendsingle vextendsingle vextendsingle = N Δ B Δ t A cos θ So the area of one loop in the coil is A = E N cos θ · Δ t Δ B = 0 . 504575 m 2 and the side length of the loop should be a = A = 0 . 710335 m Finally, the length of the wire is l = (4 · a ) N = 158 . 831 m 002 10.0 points Three very long wires are strung parallel to each other as shown in the figure below. Each wire is at the perpendicular distance 56 cm from the other two, and each wire carries a current of magnitude 5 . 8 A in the directions shown in the figure. I I I 3 2 1 z x y 56 cm 56 cm 56 cm × 3 2 1 y x z Cross-sectional View The permeability of free space is 1 . 2566 × 10 6 T · m / A. Find the magnitude of the net force per unit length exerted on the upper wire (wire 3) by the other two wires. 1. 2.8145e-05 2. 2.08087e-05 3. 3.14326e-05 4. 2.56557e-05 5. 6.33417e-06 6. 4.09208e-05 7. 1.46596e-05 8. 3.21781e-05 9. 6.62384e-06 10. 2.77175e-05 Correct answer: 2 . 08087 × 10 5 N / m. Explanation: Let : r = 56 cm = 0 . 56 m and I = 5 . 8 A . Magnetic field due to a long straight wire is B = μ 0 I 2 π r ,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Version 098/ABCAC – midterm 03 – Turner – (59070) 2 and the force per unit length between two parallel wires is F = μ 0 I 1 I 2 2 π r . There are two ways to solve this problem which are essentially the same. The first way is to find the net magnetic field at the upper wire from the two wires below ( vector B net = vector B 1 + vector B 2 ) and then find the force from vector F = I vector L × vector B . The crucial step here will be to add the magnetic fields as vectors . The second way would be to use vector F = I vector L × vector B to find the net force on the upper wire from the two lower wires vector F net = vector F 1 + vector F 2 , where we must be sure to add the forces as vectors. You should recognize that the two methods are formally identical. Let’s do it the first way. The magnitude magnetic field from wire 1 is found from Ampere’s law to be B 1 = μ 0 I 2 π r . Using the right hand rule the direction points up and to the left of wire as shown in figure 2. 30 30 60 60 60 B 1 B B 2 × 3 2 1 ˆ ˆ ı ˆ k Adding Magnetic Fields 30 30 60 60 60 F 13 F F 23 × 3 2 1 ˆ ˆ ı ˆ k Adding Forces Its components will then be vector B 1 = B [sin (30 cos (30 ı ] .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

midterm 03 - Version 098/ABCAC midterm 03 Turner(59070 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online