Đáp án đề thi giúp cho các bạn sinh viên nắm bắt được cấu trúc và cách giải đề thi, dạng đề thi chính để có kế hoạch ôn thi một cách tốt hơn. Tài liệu hữu ích cho các các bạn sinh viên đang theo học môn này và những ai quan tâm đến môn học này dùng làm tài liệu tham khảo. | SOLUTIONS KEYS AND SCORES For Questions in Final Exam of Principles of Physics 2 Edited by Phan Gia Anh Vu Date of Exam July 24th 2020 Question Answer Score 1 A. It is horizontal northward. In electric field a charge of q will be exerted by the force F qE . In this case q is negative so F is parallel but in opposite direction with E . 2 C. A D In the conducting spherical shell the inner and the outer surface will possess negative and positive charge respectively. q Applying the Gauss s law in for the Gaussian spheres that contain the 0 points A B C and D it can be found that A B B 0 C 0 . 3 D. N C E Applying the relationship between E and B c . B E B c 10 7 108 N C 4 B. mm The positions of minima are given by y L tan . The angles in which the minima occur can be found from d sin m With small angles tan sin The position of the first dark fringe is y L . d 10 6 So d L 1 1 10 4 m mm y 5 10 3 5 The magnetic field create by the current in the wire I is given by B 0 . At the position of the loop 2 a the magnetic field is perpendicular to and out of B the page as in the figure on the right . The farer a point from the wire is the weaker the magnetic field is. a If the loop moves away from the wire the magnetic flux through the loop will decrease. So the induced current in the loop is counterclockwise. b If the loop moves to the wire the magnetic flux through the loop will increase. So the induced current in the loop is clockwise. Page 3 6 Applying the right hand rule we can find the direction of the magnetic field created by the current I1 at the position of I2 as in the figure. The force exerting on the wire carrying I2 is vertical and upwards. As the result the two wires attract each other. 7 Let number the charges from 1 to 4 as in the figure. The electric fields created by the charges q1 q2 q3 q4 are E1 E2 E3 E4 respectively. 2 3 0 5 We can see that E2 E4 E1 2 E3 . So the total electric field at the
