midterm 03 review-solutions

# midterm 03 review-solutions - pan(zp695 midterm 03 review...

This preview shows pages 1–3. Sign up to view the full content.

pan (zp695) – midterm 03 review – chiu – (56565) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 0.0 points The electric feld is measured all over a cu- bical surFace, and the pattern oF feld detected is shown in the above fgure. On the right side oF the cube and the bottom oF the cube, the electric feld has the value ± 336 , - 336 , 0 ² N/C . On the top oF the cube and the leFt side oF the cube, the electric feld is zero. On halF oF the Front and back Faces, the electric feld has the value ± 336 , - 336 , 0 ² ;o nt h eo t h e rh a l F oF the Front and back Faces, the electric feld is zero. One edge oF the cube is d =0 . 55 m long. (Hint: Notice that he electric feld vec- tors are perpendicular to the diagonal area oF the cube) Select the correct choices For the unit vector oF 7 E Ia. ˆ E = ˆ i - ˆ j 2 Ib. ˆ E = ˆ i + ˆ j 2 Ic. ˆ E = ˆ i + ˆ j Outgoing unit normal For the right side shaded area denoted by ˆ n right is: IIa. ˆ n right = - ˆ i IIb. ˆ n right = ˆ i IIc. ˆ n right = ˆ i - ˆ j 2 1. Ib, IIb 2. Ib, IIc 3. Ib, IIa 4. Ia, IIc 5. Ic, IIc 6. Ia, IIb correct 7. Ic, IIb 8. Ic, IIa 9. Ia, IIa Explanation: By inspection, the unit vector For 7 E is ˆ E = ˆ i - ˆ j 2 Hence, the answer is Ia. The outgoing normal vector For the right side shaded area is ˆ n right = ˆ i .H e n c e ,t h e correct choice is IIb. 002 (part 2 of 2) 0.0 points ±ind the electric charge within the cube. Correct answer: 1 . 799 × 10 - 9 C. Explanation: Let : 5 0 =8 . 85 × 10 - 12 J / V 2 · m , E x =336 N/C , and d . 55 m. The total outgoing ²ux by symmetry equals twice the outgoing ²ux through the shaded area at the right side. φ cube =2 φ right φ cube E ± ˆ i - ˆ j 2 ² · ( d 2 ˆ i )

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

View Full Document
pan (zp695) – midterm 03 review – chiu – (56565) 2 φ cube = 2 Ed 2 Here, E is given by E = ± E 2 x + E 2 y = 2 E x =475 . 176 N / C hence, the total fux enclosed is given by φ cube =2 E x d 2 Based on Gauss law, the chage enclosed is Q = 5 0 φ cube = 5 0 (2 E x d 2 )=1 . 799 × 10 - 9 C 003 0.0 points An alpha particle (consisting oF two pro- tons and two neutrons) is moving in a circle at constant speed, perpendicular to au n i F o rmm a g n e t i e l da p p l i e db ys om e current-carrying coils. The alpha particle makes on clockwise revolution every 82 ns. IF the speed is small compared to the speed oF light, what is the magnitude B oF the mag- netic ±eld made by the coils? Correct answer: 1 . 63 T. Explanation: let : e =1 . 6 × 10 - 19 C , m P m n . 7 × 10 - 27 kg , and T =82ns=8 . 2 × 10 - 8 s . Given the period, ω = 2 π T .T h e c y c l o t r o n Frequency is ω = qB m .So lv ing , B = 2 πm qT = 2 π (2 m N +2 m P ) 2 eT = 4 P = 4 π (1 . 7 × 10 - 27 kg) (1 . 6 × 10 - 19 C)(8 . 2 × 10 - 8 s) . 63 T . 004 0.0 points Alongwirecarriesacurrent I 1 =1Aupward , and a rectangular loop oF height h =0 . 2m and width w .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 16

midterm 03 review-solutions - pan(zp695 midterm 03 review...

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

View Full Document
Ask a homework question - tutors are online