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# solution9_pdf - hussain(wh4892 hw 9 Opyrchal(121104 This...

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hussain (wh4892) – hw 9 – Opyrchal – (121104) 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 59600 V and then enters a region in which there is a uniform magnetic field of 0 . 619 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: 1 . 43599 × 10 11 N. Explanation: Let : V = 1 . 44793 × 10 8 m / s and B = 0 . 619 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)(59600 V) 9 . 10939 × 10 31 kg = 1 . 44793 × 10 8 m / s . Then the force on it is f = qvB = (1 . 60218 × 10 19 C) × (1 . 44793 × 10 8 m / s)(0 . 619 T) = 1 . 43599 × 10 11 N . 002 10.0 points A negatively charged particle moving paral- lel to the z -axis enters a magnetic field (point- ing into 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 ı 4. vector F = 0 ; no deflection 5. hatwide F = +ˆ 6. hatwide F = ˆ ı correct 7. hatwide F = ˆ k 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 < 0 , 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 .

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hussain (wh4892) – hw 9 – Opyrchal – (121104) 2 This is the third 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 left-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. hatwide F = ˆ ı 2. hatwide F = ˆ 3. hatwide F = +ˆ 4. hatwide F = ˆ k 5. hatwide F = +ˆ ı 6. vector F = 0 ; no deflection correct 7. hatwide F = + ˆ k 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 .
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