Question

1 question 5 parts

- 150 m/s 2.656 x 10-25 kg -1.28 x 10-18 C 1 0.7 m | 17 A y 021 (part 1 of 5) 10.0 points A long, straight wire is in the plane of the page and carries a current of 17 A. Point P is also in the plane of the page and is a perpendicular distance 0.7 m from the wire, as shown below. P z is upward from the paper With reference to the coordinate system established above, what is the direction of the magnetic force acting on the negative particle at point P? 1. in the positive z-axis 2. in the negative 3-axis iP 0.7 m 17 A 3. in the negative z-axis 4. in the negative z-axis 5. in the positive y-axis 6. in the positive D-axis part 3 of 5) 10.0 points Determine the magnitude of the magnetic force acting on the negative particle at point P. Answer in units of N. z is upward from the paper With reference to the coordinate system in the figure above, what is the direction of the magnetic field at point P due to the current in the wire? The permeability of free space is 4 10-'T.m/A. Gravitational effects are negligible. 1. in the positive z direction 2. in the negative y direction 3. in the negative direction 4. in the positive a direction 5. in the negative z direction 6. in the positive y direction (part 2 of 5) 10.0 points A 2.656 x 10-26 kg mass particle with nega- tive -1.28 x 10-18 C charge is initially mov- ing parallel to the wire with a 150 m/s speed when it is at point P, as shown below. (part 4 of 5) 10.0 points An electric field is applied that causes the net force on the negative particle to be zero at point P. With reference to the coordinate system established above, what is the direction of the electric field at point P that could accomplish this? 1. in the positive z-axis 2. in the positive y-axis 3. in the negative y-axis 4. in the negative -axis

Answer 1

1) Dieation of magnetic field at Point P Into the plane of paper, that IS along negative ?-direction (occording to eight hand shuns bule ) 18 al C q=-1.28X10 m=2.656x10-26 kg 150 mls (-ê" direction of magnetic force = q (Wx0) --C+(-)) =(-) (ixR) ده (Positive y. direction) (3) IF1= q VB 1.28710-1 150 tot 17 List (0.7) (9.32541022 (4) Direction of magnetic field along positively-diction F-qE 9.325410-22 = 1.28x10-18 E = 7.285410-4 NIC