CH 19 - Magnetism-solutions

CH 19 - Magnetism-solutions - This print-out should have 8...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron is accelerated by a 6 . 8 kV poten- tial difference. The charge on an electron is 1 . 60218 × 10- 19 C and its mass is 9 . 10939 × 10- 31 kg. How strong a magnetic field must be expe- rienced by the electron if its path is a circle of radius 5 . 8 cm? Correct answer: 0 . 00479436 T. Explanation: Let : r = 5 . 8 cm = 0 . 058 m , V = 6 . 8 kV = 6800 V , m = 9 . 10939 × 10- 31 kg , and q = 1 . 60218 × 10- 19 C . From Newton’s second law, F = q v B = mv 2 r v = q B r m . The kinetic energy is K = 1 2 mv 2 = q 2 B 2 r 2 2 m = q V , so the magnitude of the magnetic field is B = 1 r 2 V m q = 1 . 058 m × 2 (6800 V) (9 . 10939 × 10- 31 kg) (1 . 60218 × 10- 19 C) = . 00479436 T . Dimensional analysis : The voltage must be converted from kV to V and the radius from cm to m. 002 10.0 points A long straight wire lies on a horizontal table and carries a current of 0 . 92 μ A. A proton moves parallel to the wire (opposite the cur- rent) with a constant velocity of 15200 m / s at a distance d above the wire....
View Full Document

This note was uploaded on 10/03/2010 for the course PH 102 taught by Professor Staff during the Fall '08 term at Alabama.

Page1 / 3

CH 19 - Magnetism-solutions - This print-out should have 8...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online