hw7 - homework 07 LIN, KEVIN Due: Mar 2 2008, 4:00 am 1...

Info iconThis preview shows pages 1–3. 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: homework 07 LIN, KEVIN Due: Mar 2 2008, 4:00 am 1 Question 1, chap 30, sect 1. part 1 of 1 10 points Suppose a new particle is discovered, and it is found that a beam of these particles passes undeflected through crossed electric and magnetic fields, where E = 338 V / m and B = 0 . 00596 T. If the electric field is turned off, the particles move in the magnetic field in circular paths of radius r = 1 . 9 cm. Determine q m for the particles from these data. Correct answer: 5 . 00807 10 8 C / kg (toler- ance 1 %). Explanation: Let : E = 338 V / m , q e = q p = 1 . 60218 10 19 C , m e = 9 . 10939 10 31 kg , m p = 1 . 67262 10 27 kg , B = 0 . 00596 T , and r = 0 . 019 m . Since the particle passes undeflected, the electric force on it is equal to the magnetic force. When the electric field is turned off, only the magnetic force is exerted on it as the centripetal force. So we get two equations q E = B q v B q v = mv 2 r . So, the charge to mass ratio for this new particle is q m = E r B 2 = 338 V / m (0 . 019 m) (0 . 00596 T) 2 = 5 . 00807 10 8 C / kg . For comparison, the electron charge-to-mass ratio is q e m e = 1 . 60218 10 19 C 9 . 10939 10 31 kg = 1 . 75882 10 11 C / kg , and the proton charge-to-mass ratio is q p m p = 1 . 60218 10 19 C 1 . 67262 10 27 kg = 9 . 57883 10 7 C / kg . Question 2, chap 29, sect 1. part 1 of 2 10 points An electron is projected into a uniform magnetic field given by vector B = B x + B y , where B x = 4 . 7 T and B y = 1 . 4 T. The magnitude of the charge on an electron is 1 . 60218 10 19 C . x y z v = 390000 m / s electron 4 . 7 T 1 . 4 T B Find the direction of the magnetic force when the velocity of the electron is v , where v = 390000 m / s. 1. hatwide F = k correct 2. hatwide F = k 3. hatwide F = 1 2 ( ) 4. hatwide F = 5. hatwide F = 6. hatwide F = 7. hatwide F = 8. hatwide F = 1 2 ( ) homework 07 LIN, KEVIN Due: Mar 2 2008, 4:00 am 2 9. hatwide F = 1 2 ( + ) Explanation: Let : q = 1 . 60218 10 19 C , B x = 4 . 7 T , and B y = 1 . 4 T . Basic Concepts: Magnetic force on a mov- ing charge is given by vector F = qvectorv vector B . Solution: vector B = (4 . 7 T) + (1 . 4 T) v = (390000 m / s) for the electron. Find: The vector expression for the force on the electron. This solves both part 1 and part 2. We will go through two methods of doing the problem. The first is more mathematically oriented and the second uses more of a reasoning argu- ment. Method 1: The force acting on a charge q with velocity vectorv in the presence of an external magnetic field vector B is given by vector F = qvectorv vector B Taking the cross product of vectorv with vector B we obtain vector F = qvectorv vector B = q vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle...
View Full Document

Page1 / 14

hw7 - homework 07 LIN, KEVIN Due: Mar 2 2008, 4:00 am 1...

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

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