PHY303L HW7

# PHY303L HW7 - thwing(jrt2329 – Homework 07 – Yao...

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Unformatted text preview: thwing (jrt2329) – Homework 07 – Yao – (59110) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron is in a uniform magnetic field B that is directed out of 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. correct 2. toward the left 3. toward the bottom of the page. 4. toward the right 5. out of the page. 6. into the page. 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 top of the page , using right hand rule for hatwide v × hatwide B , and reversing the direction due to the negative charge on the electron. 002 10.0 points A negatively charged particle moving at 45 ◦ angles to both the x-axis and y-axis enters a magnetic field (pointing into 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 2. hatwide F = 1 √ 2 parenleftBig + ˆ k − ˆ ı parenrightBig 3. hatwide F = 1 √ 2 ( − ˆ − ˆ ı ) 4. hatwide F = 1 √ 2 parenleftBig − ˆ k − ˆ ı parenrightBig 5. hatwide F = 1 √ 2 (+ˆ − ˆ ı ) 6. hatwide F = 1 √ 2 parenleftBig − ˆ k +ˆ ı parenrightBig 7. hatwide F = 1 √ 2 parenleftBig − ˆ + ˆ k parenrightBig 8. hatwide F = 1 √ 2 ( − ˆ +ˆ ı ) correct 9. hatwide F = 1 √ 2 parenleftBig − ˆ − ˆ k parenrightBig 10. hatwide F = 1 √ 2 parenleftBig +ˆ − ˆ k parenrightBig 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- thwing (jrt2329) – Homework 07 – Yao – (59110) 2 tion. Solution: The force is vector F = qvectorv × vector B . vector B = B parenleftBig − ˆ k parenrightBig , vectorv = 1 √ 2 v (+ˆ ı +ˆ ) , and q < , therefore , vector F = −| q | vectorv × vector B = −| q | 1 √ 2 v B bracketleftBig (+ˆ ı +ˆ ) × parenleftBig − ˆ k parenrightBigbracketrightBig = −| q | 1 √ 2 v B ( − ˆ +ˆ ı ) hatwide F = 1 √ 2 ( − ˆ +ˆ ı ) . This is the third of eight versions of the problem....
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PHY303L HW7 - thwing(jrt2329 – Homework 07 – Yao...

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