Solutions, keys and scores for questions in final exam of principles of Physics 2 - ĐH Sư phạm Kỹ thuật

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SOLUTIONS, KEYS AND SCORES For Questions in Final Exam of Principles of Physics Edited by: Phan Gia Anh Vu Date of Exam: July 24th 2020 Answer Question A It is horizontal, northward Score 0.5   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 0.5 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  D 45.0 N/C E Applying the relationship between E and B: c  B    0.5  E  B  c  1.50  107  3.00  108  45.0 N/C 0.5 B 0.100 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 0.5  106   10 m  0.100 mm y  103 The magnetic field create by the current in the wire I is given by B  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 So d  L    1 Page 0.5 0.5 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 0.5 0.5 × Let number the charges from to as in the figure The electric fields created by the     charges q1, q2, q3, q4 are E1 , E2 , E3 , E4 , respectively    0,5  We can see that E2   E4 ; E1  E3 So the total electric field at the square center is:       E  E1  E2  E3  E4  3E3  The magnitude of E3 is: E3  9  10 6  10   4.32  10 kq a 2        0,5 12 6 Nm /C  0.05         2 So the magnitude of E is: E  3E3  1.30  10 Nm /C 2 Or: E  9.19  10 iˆ  ˆj N.m /C  0,5  b) If a fifth charge is located at O, it will be exerted by the force:   F   qE   6.00  10 6 9.19  10 iˆ  ˆj  551 iˆ  ˆj N (the red arrow 0,5       in the above figure) The magnitude of this force is: F  780 N  a) The current I1 creates a magnetic field B1 that is perpendicular to the page and directs into the page In order to have the zero total magnetic field at O,  the current I2 must create the magnetic field B2 so O   that: B2   B1 0,5 Thus, the current I2 has to be in opposite direction of the current I1 and have the magnitude that: I  I I     or   r1 r2  r2 r1 r2  3.75 A Thus: I  I1  r1 12 B   I1 0,5 So I2 is counterclockwise and has the magnitude of: 3.75 A b) If the direction of I2 remains unchanged but its magnitude is doubled, then the 0,5 total magnetic field at O is nonzero Now, B1  B2 , so the total field has the   same direction with B2 (out of the page) The magnitude of B is: I2  I   4  10 7       r1 r2    I1  3.75   5     2.62  10 T 0.09   0.12 0,5 a) From the condition of constructively reflection on the thin film: nt     0,5  m Page It can be found that:   nt m The first three longest wavelengths are corresponding with the three smallest value of m  : 1, and        1.38  5.00  10 7  1.38  10  m  1.38 μm  1.38  5.00  10  2   0.690 μm 2  1.38  5.00  10  3   0.460 μm 1  b) From these wavelengths, there are wavelengths in the visible spectrum   0.690 μm; 0.460 μm Page 0,5 0,5 0,5 ...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... direction of the current I1 and have the magnitude that: I  I I     or   r1 r2  r2 r1 r2  3.75 A Thus: I  I1  r1 12 B   I1 0,5 So I2 is counterclockwise and has the magnitude of: ... and directs into the page In order to have the zero total magnetic field at O,  the current I2 must create the magnetic field B2 so O   that: B2   B1 0,5 Thus, the current I2 has to be in