{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MathematicalMethodsSolutionsManual

# MathematicalMethodsSolutionsManual - James B Seaborn...

This preview shows pages 1–10. Sign up to view the full content.

James B. Seaborn University of Richmond , Springer SOLUTIONS MANUAL for Mathematical Methodsfor the Physical Sciences

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

View Full Document
SOLUTIONS MANUAL for Mathematical Methods for the Physical Sciences

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

View Full Document
Springer New York Berlin HeideLberg BarceLona Hong Kong London Milan Paris Singapore Tokyo
James B. Seaborn University of Richmond , Springer SOLUTIONS MANUAL for Mathematical Methods for the Physical Sciences r

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

View Full Document
Copyright © 2002 by Springer-Verlag New York. Inc. This material may be reproduced for testing or instructional purposes by people using the text. Printed in the United States of America.
1 1 A Review 1. Let a = 0.075 m, b = 0.183 m, ql = 7.37 X 10- 9 C, and q2 = -3.58 X 10- 9 C. Ex = -kq1b 3 = -1,569 N/C (a 2 + b 2 )2 E y = -kqla _ kq2 = -5 085 N/C (a2 + b2)~ a 2 ' E = JE'; + E~ = 5,322 N/C ¢ = tan- 1 ~: = 252.8° The resultant electric field E is 5,322 N/C down to the left at 73° below the horizontal. 2. First, calculate the horizontal and vertical components of E (9 X 10 9 N . m 2 /C 2 )(3 x 10- 9 C) Ex = - (0.35 m)2 _ (9 x 10 9 N . m 2 /C 2 )(5 x 10- 9 C) (0.35 m) = -799.5 N/C ((0.35 m)2 + (0.15 m)2) 3/2 (9 X 10 9 N . m 2 /C 2 )(7 x 10- 9 C) (0.15 m)2 + (9 x 10 9 N . m 2 /C2)(5 x 10- 9 C) (0.15 m) = 1,322.2 N/C ((0.35 m)2 + (0.15 m)2) 3/2 E = J E'; + E~ = 1,545 N/C and The electric field is 1,545 N/C up to the left at 58.8° above the horizontal. 3. The point where E = 0 must lie on the line passing through the two charges and at a distance d to the left of the 9-f-LC charge. E = k(16 f-LC) _ k(9 f-LC) = 0 (d + 13 mm)2 tf2 . On rearranging, we get the quadratic equation for d, 7d 2 - 234d - 1,521 = (7d + 39)(d - 39) = O. Thus, the electric field is zero at the point 39 mm to the left of the 9-f-LC charge.

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

View Full Document
1 A REVIEW 2 4. Let q1 = 8/-LCand q1 = -5/-LC. A line from the field point P to q1 makes an angle () with the horizontal line connecting the two charges. A line from P to q2 also makes an angle () with the line connecting the charges. Thus, Ex = E 1 cos () + E 2 cos () = (E 1 + E 2 ) cos () (9 X 10 9 N . m 2 /C 2 )(0.055 m) (8 0-6 C -6 C) = ----------- x 1 + 5 x 10 ((0.05 m)2 + (0.055 m)2) 3/2 = 1.567 X 10 7 N/C E y = E 1 sin() - E 2 sinB = (E 1 - E 2 )sin() (9 x 10 9 N . m 2 /C2)(0.05 m) (8 0-6 C -6 C) ---------- x 1 - 5 x 10 ((0.05 m)2 + (0.055 m)2) 3/2 = 0.329 X 10 7 N/C From these components, we find that the resultant electric field is 1.601 x 10 7 N/C up to the right at 11.8 0 above the horizontal. 5. Use Eq. (1.5) and solve for the ion density n, I 2.2 x 10- 6 Cis n= 2 = 2 (3e)7r(~) v (4.8 X 1O-19C)7rcoo~lm) (6.48 x 10 6 m/s) = 2.04 x 10 12 (ions/m 3 ) (.01 m/cm)3 = 2.04 x 10 6 ions/cm 3 . 6. The current I is 1= b.Q = [2(1.6 x 1O- 19 C)/ion](2.6 x 10 19 ions) = 0 39C/ b.t 60s .1 s. The ion current is 0.139 ampere. 7. The average beam current is ( .001 m)2 1= (1.6 X 10- 19 C/el)(4.2 x 10 12 elfm 3 ) -2- 7r(1.9 x 10 7 m/s) = 1.00 x 10- 5 A. The beam current is 1.0 x 10- 5 ampere or 10/-LA. 8. The emf is given by emf = Nb.if> = 1 835v = 85(2.6T)(.254m)2 b.t' b.t Solving for b.t, we find the field collapses to zero in 7.77 milliseconds.
10. We set 1 A REVIEW 9, The emf is emf = LH> = (2.2T)(.02m)(.03m) =.033v. 6.t .04s The average voltage induced in the loop is 33 mv. 6.<1> 750Cl~m) \.B emf = N 6.t = .155v = .00327 s . Solving for B, we get 5.1 x 10- 5 T for the earth's magnetic field.

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 ]}