{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW9 - lakha(ml29327 HW#9 Jaafari(1444 This print-out should...

This preview shows pages 1–2. Sign up to view the full content.

lakha (ml29327) – HW#9 – Jaafari – (1444) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points The magnetic field over a certain range is given by vector B = B x ˆ ı + B y ˆ , where B x = 7 T and B y = 6 T. An electron moves into the field with a velocity vectorv = v x ˆ ı + v y ˆ + v z ˆ k , where v x = 5 m / s, v y = 8 m / s and v z = 8 m / s. The charge on the electron is - 1 . 602 × 10 19 C. What is the ˆ ı component of the force ex- erted on the electron by the magnetic field? Correct answer: 7 . 6896 × 10 18 N. Explanation: Let : v x = 5 m / s , v y = 8 m / s , v z = 8 m / s , q = - 1 . 602 × 10 19 C , B x = 7 T , and B y = 6 T . The force exerted on the electron by the mag- netic field is given by vector F = qvectorv × vector B = q vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ˆ ı ˆ ˆ k v x v y v z B x B y 0 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = q [ - v z B y ˆ ı + v z B x ˆ + ( v x B y - v y B x ) ˆ k ] = ( - 1 . 602 × 10 19 C) {- (8 m / s) (6 T)ˆ ı + (8 m / s) (7 T)ˆ + [(5 m / s) (6 T) - (8 m / s) (7 T)] ˆ k } = (7 . 6896 × 10

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.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern