HW #6-solutions

HW #6-solutions - nanni(arn437 – HW#6 – Erskine...

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Unformatted text preview: nanni (arn437) – HW #6 – Erskine – (56905) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A duck flying due north at 42 m / s passes over Atlanta, where the Earth’s magnetic field is 8 . 9 × 10 − 5 T in a direction 74 ◦ below the horizontal line running north and south. If the duck has a net positive charge of 1 . 2 × 10 − 8 C, what is the magnitude of the magnetic force acting on it? Correct answer: 4 . 31184 × 10 − 11 N. Explanation: Let : Q = 1 . 2 × 10 − 8 C , v = 42 m / s , and B = 8 . 9 × 10 − 5 T . The Lorentz force acting on the duck is | vector F | = | qvectorv × vector B | = q v B sin θ = (1 . 2 × 10 − 8 C) (42 m / s) (8 . 9 × 10 − 5 T) × sin (74 ◦ ) = 4 . 31184 × 10 − 11 N . (The Lorentz force is the name given to the force acting on a charged particle moving in a magnetic field.) 002 10.0 points An electron is in a uniform magnetic field B that is directed into the plane of the page, as shown. v e − B B B B When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed 1. toward the top of the page. 2. toward the right 3. toward the left 4. into the page. 5. out of the page. 6. toward the bottom of the page. correct Explanation: The force on the electron is vector F = q vectorv × vector B = − e vectorv × vector B. The direction of the force is thus hatwide F = − hatwide v × hatwide B , pointing toward the bottom of the page , us- ing right hand rule for hatwide v × hatwide B , and reversing the direction due to the negative charge on the electron. 003 10.0 points A solenoid of radius 3 . 4 cm is made of a long piece of wire of radius 1 . 655 mm, length 59 . 5 m and resistivity 1 . 611 × 10 − 8 Ω · m. The solenoid is closely wound with a single wire thickness. Find the magnetic field at the center of the solenoid if the wire is connected to a battery with an emf of 30 . 87 V. The permeability of free space is 1 . 25664 × 10 − 6 T · m / A. Correct answer: 0 . 0969925 T. Explanation: Let : ℓ = 59 . 5 m , r = 1 . 655 mm = 0 . 001655 m , R = 3 . 4 cm = 0 . 034 m , μ = 1 . 25664 × 10 − 6 T · m / A , ρ = 1 . 611 × 10 − 8 Ω · m , and E = 30 . 87 V . nanni (arn437) – HW #6 – Erskine – (56905) 2 The resistance of the wire is R Ω = ρ ℓ π r 2 , so it carries current I = E R Ω = E π r 2 ρ ℓ . With a single layer of windings, the wire length is wrapped around the circumference of the solenoid, with n = ℓ 2 π R turns and B = n μ I = ℓ 2 π R μ E π r 2 ρ ℓ = μ E r 2 2 ρ R = (1 . 25664 × 10 − 6 T · m / A)(30 . 87 V) 2 (1 . 611 × 10 − 8 Ω · m) (0 . 034 m) × (0 . 001655 m) 2 = . 0969925 T ....
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This note was uploaded on 11/02/2011 for the course PHYSICS 317L taught by Professor Erskine during the Fall '11 term at University of Texas.

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HW #6-solutions - nanni(arn437 – HW#6 – Erskine...

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