solution9_pdf

# solution9_pdf - maini(nm7637 – hw9 – Shneidman...

This preview shows pages 1–2. 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: maini (nm7637) – hw9 – Shneidman – (12108) 1 This print-out should have 7 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 . 35 Mm / s in the positive x direction in a magnetic field of 1 . 65 T in the positive z direction. Correct answer: 1 . 14996 pN. Explanation: Let : q = 1 . 60218 × 10 − 19 C , vectorv = (4 . 35 Mm / s) ˆ ı = (4 . 35 × 10 6 m / s) ˆ ı , and vector B = (1 . 65 T) ˆ k . vector F = qvectorv × vector B = (1 . 60218 × 10 − 19 C) × bracketleftbig (4 . 35 × 10 6 m / s) ˆ ı bracketrightbig × bracketleftBig (1 . 65 T) ˆ k bracketrightBig × 10 12 pN 1 N =- (1 . 14996 pN) ˆ , so the magnitude is 1 . 14996 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 . 2 T and B y = 1 . 6 T. The magnitude of the charge on an electron is 1 . 60218 × 10 − 19 C . x y z v = 3 . 8 × 10 5 m / s electron 4 . 2 T 1 . 6 T B Find the direction of the magnetic force when the velocity of the electron is v ˆ , where v = 3 . 8 × 10 5 m / s....
View Full Document

## This note was uploaded on 01/31/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

### Page1 / 3

solution9_pdf - maini(nm7637 – hw9 – Shneidman...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online