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# 9 - kavo(ak22862 HW09 Janow(12121 This print-out should...

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kavo (ak22862) – HW09 – Janow – (12121) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the magnitude of the magnetic force on a proton moving with velocity 4 . 38 Mm / s in the positive x direction in a magnetic field of 1 . 8 T in the positive z direction. Correct answer: 1 . 26316 pN. Explanation: Let : q = 1 . 60218 × 10 19 C , vectorv = (4 . 38 Mm / s) ˆ ı = (4 . 38 × 10 6 m / s) ˆ ı , and vector B = (1 . 8 T) ˆ k . vector F = qvectorv × vector B = (1 . 60218 × 10 19 C) × bracketleftbig (4 . 38 × 10 6 m / s) ˆ ı bracketrightbig × bracketleftBig (1 . 8 T) ˆ k bracketrightBig × 10 12 pN 1 N = (1 . 26316 pN) ˆ  , so the magnitude is 1 . 26316 pN . 002 (part 1 of 2) 10.0 points An electron is projected into a uniform mag- netic field given by vector B = B x ˆ ı + B y ˆ , where B x = 4 . 7 T and B y = 1 . 9 T. The magnitude of the charge on an electron is 1 . 60218 × 10 19 C . x y z v = 3 . 9 × 10 5 m / s electron 4 . 7 T 1 . 9 T B Find the direction of the magnetic force when the velocity of the electron is v ˆ , where v = 3 . 9 × 10 5 m / s. 1. hatwide F = 1 2 ı + ˆ ) 2. hatwide F = 1 2 ı ˆ ) 3. hatwide F = ˆ 4. hatwide F = ˆ ı 5. hatwide F = 1 2 ˆ ı ) 6. hatwide F = ˆ k 7. hatwide F = ˆ k correct 8. hatwide F = ˆ ı 9. hatwide F = ˆ Explanation: Let : q = 1 . 60218 × 10 19 C , B x = 4 . 7 T , and B y = 1 . 9 T . Basic Concepts: Magnetic force on a mov- ing charge is given by vector F = qvectorv × vector B . Solution: vector B = (4 . 7 T) ˆ ı + (1 . 9 T) ˆ v = (3 . 9 × 10 5 m / s) ˆ for the electron. Find: The vector expression for the force on the electron. This solves both part 1 and part 2. We will go through two methods of doing the problem. The first is more mathematically oriented and the second uses more of a reasoning argu- ment. Method 1: The force acting on a charge q with velocity vectorv in the presence of an external magnetic field vector B is given by vector F = qvectorv × vector B

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kavo (ak22862) – HW09 – Janow – (12121) 2 Taking the cross product of vectorv with vector B we obtain vector F = qvectorv × vector B = q vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ˆ ı ˆ ˆ k 0 v 0 B x B y 0 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = q braceleftBig [( B y )(0) ( B x )( v )] ˆ k [(0)(0) ( B x )(0)] ˆ + [( v )(0) ( B x )(0)] ˆ ı bracerightBig = q B x v ˆ k = ( 1 . 60218 × 10 19 C)(4 . 7 T) × (3 . 9 × 10 5 m / s) ˆ k = (2 . 93679 × 10 13 N) ˆ k , and the direction is + ˆ k . Method 2: The other method is to real- ize that the only component of the magnetic field which affects the electron is the compo- nent perpendicular to its velocity. Therefore, F = q | vectorv × vector B | = q v B with the direction given by the right hand rule to be in the neg- ative ˆ k direction; but recalling to reverse the direction because the electron has a negative instead of positive charge.
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9 - kavo(ak22862 HW09 Janow(12121 This print-out should...

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