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Unformatted text preview: pan (zp695) midterm 03 review chiu (56565) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 0.0 points The electric field is measured all over a cu bical surface, and the pattern of field detected is shown in the above figure. On the right side of the cube and the bottom of the cube, the electric field has the value 336 , 336 , N/C . On the top of the cube and the left side of the cube, the electric field is zero. On half of the front and back faces, the electric field has the value 336 , 336 , N/C ; on the other half of the front and back faces, the electric field is zero. One edge of the cube is d = 0 . 55 m long. (Hint: Notice that he electric field vec tors are perpendicular to the diagonal area of the cube) Select the correct choices for the unit vector of 7 E Ia. E = i j 2 Ib. E = i + j 2 Ic. E = i + j Outgoing unit normal for the right side shaded area denoted by n right is: IIa. n right = i IIb. n right = i IIc. n right = i 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 E = i j 2 Hence, the answer is Ia. The outgoing normal vector for the right side shaded area is n right = i . Hence, the correct choice is IIb. 002 (part 2 of 2) 0.0 points Find the electric charge within the cube. Correct answer: 1 . 799 10 9 C. Explanation: Let : 5 = 8 . 85 10 12 J / V 2 m , E x = 336 N/C , and d = 0 . 55 m . The total outgoing flux by symmetry equals twice the outgoing flux through the shaded area at the right side. cube = 2 right cube = 2 E i j 2 ( d 2 i ) pan (zp695) midterm 03 review chiu (56565) 2 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 flux enclosed is given by cube = 2 E x d 2 Based on Gauss law, the chage enclosed is Q = 5 cube = 5 (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 a uniform magnetic field applied by some currentcarrying 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 field made by the coils? Correct answer: 1 . 63 T. Explanation: let : e = 1 . 6 10 19 C , m P m n = 1 . 7 10 27 kg , and T = 82 ns = 8 . 2 10 8 s . Given the period, = 2 T . The cyclotron frequency is = q B m . Solving, B = 2 m q T = 2 (2 m N + 2 m P ) 2 e T = 4 m P e T = 4 (1 . 7 10 27 kg) (1 . 6 10 19 C)(8 . 2 10 8 s) = 1 . 63 T ....
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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