solution9_pdf

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online