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solution9_pdf - taherisefat (mt23852) – hw 9 – Opyrchal...

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Unformatted text preview: taherisefat (mt23852) – hw 9 – Opyrchal – (121012) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron in a vacuum is first accelerated by a voltage of 75400 V and then enters a region in which there is a uniform magnetic field of . 208 T at right angles to the direction of the electron’s motion. What is the force on the electron due to the magnetic field? Correct answer: 5 . 42732 × 10 − 12 N. Explanation: Let : V = 1 . 62859 × 10 8 m / s and B = 0 . 208 T . The kinetic energy gained after acceleration is KE = 1 2 m e v 2 = q e V , so the velocity is v = radicalbigg 2 q e V m = radicalBigg 2(1 . 60218 × 10 − 19 C)(75400 V) 9 . 10939 × 10 − 31 kg = 1 . 62859 × 10 8 m / s . Then the force on it is f = qvB = (1 . 60218 × 10 − 19 C) × (1 . 62859 × 10 8 m / s)(0 . 208 T) = 5 . 42732 × 10 − 12 N . 002 10.0 points A positively charged particle moving paral- lel to the z-axis enters a magnetic field (point- ing out of of the page), as shown in the figure below. z x v y vector B vector B + q Figure: ˆ ı is in the x-direction, ˆ is in the y-direction, and ˆ k is in the z-direction. What is the initial direction of deflection? 1. hatwide F = + ˆ k 2. hatwide F = − ˆ 3. hatwide F = − ˆ k 4. hatwide F = +ˆ ı correct 5. vector F = 0 ; no deflection 6. hatwide F = +ˆ 7. hatwide F = − ˆ ı Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv × vector B Right-hand rule for cross-products. hatwide F ≡ vector F bardbl vector F bardbl ; i.e. , a unit vector in the F direc- tion. Solution: The force is vector F = qvectorv × vector B . vector B = B (+ˆ ) , vectorv = v parenleftBig − ˆ k parenrightBig , and q > , therefore , vector F = + | q | vectorv × vector B = + | q | v B bracketleftBigparenleftBig − ˆ k parenrightBig × (+ˆ ) bracketrightBig = + | q | v B (+ˆ ı ) hatwide F = +ˆ ı . taherisefat (mt23852) – hw 9 – Opyrchal – (121012) 2 This is the sixth of eight versions of the problem. 003 10.0 points A positively charged particle moving paral- lel to the x-axis enters a magnetic field (point- ing toward the right-hand side of the page), as shown in the figure below. x y v z vector B vector B + q Figure: ˆ ı is in the x-direction, ˆ is in the y-direction, and ˆ k is in the z-direction. What is the initial direction of deflection? 1. vector F = 0 ; no deflection correct 2. hatwide F = +ˆ ı 3. hatwide F = − ˆ ı 4. hatwide F = + ˆ k 5. hatwide F = − ˆ 6. hatwide F = − ˆ k 7. hatwide F = +ˆ Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv × vector B Right-hand rule for cross-products....
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This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

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solution9_pdf - taherisefat (mt23852) – hw 9 – Opyrchal...

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