27_InstructorSolutions

# 27_InstructorSolutio - MAGNETIC FIELD AND MAGNETIC FORCES 27 27.1 IDENTIFY and SET UP Apply Eq(27.2 to calculate F Use the cross products of unit

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27-1 M AGNETIC F IELD AND M AGNETIC F ORCES 27.1. IDENTIFY and SET UP: Apply Eq.(27.2) to calculate . F ! Use the cross products of unit vectors from Section 1.10. EXECUTE: ( ) ( ) 44 && 4.19 10 m/s 3.85 10 m/s =+ × +− × vi j ! (a) () & 1.40 T = B i ! ( ) ( ) ( ) 84 4 1.24 10 C 1.40 T q =×= − × × × × × F vB ii j i !! ! && & 0, ×= ×=− j ik ( ) ( )( ) ( ) 4 6.68 10 N =− × × − =− × F kk ! EVALUATE: The directions of and ! ! are shown in Figure 27.1a. The right-hand rule gives that × ! ! is directed out of the paper (+ z -direction). The charge is negative so F ! is opposite to ; × ! ! Figure 27.1a F ! is in the - z direction. This agrees with the direction calculated with unit vectors. (b) EXECUTE: & 1.40 T = B k ! ( ) ( ) ( ) 4 q + × ×− × × F j k ! &&& , − ×= jjki ( ) ( ) ( ) ( ) 4 4 & 7.27 10 N −− × − + × = × + × Fjii j ! EVALUATE: The directions of and ! ! are shown in Figure 27.1b. The direction of F ! is opposite to × ! ! since q is negative. The direction of F ! computed from the right-hand rule agrees qualitatively with the direction calculated with unit vectors. Figure 27.1b 27.2. IDENTIFY: The net force must be zero, so the magnetic and gravity forces must be equal in magnitude and opposite in direction. SET UP: The gravity force is downward so the force from the magnetic field must be upward. The charge±s velocity and the forces are shown in Figure 27.2. Since the charge is negative, the magnetic force is opposite to the right-hand rule direction. The minimum magnetic field is when the field is perpendicular to v ! . The force is also perpendicular to B ! , so B ! is either eastward or westward. EXECUTE: If B ! is eastward, the right-hand rule direction is into the page and B F ! is out of the page, as required. Therefore, B ! is eastward. sin mg q vB φ = . 90 = ° and 32 48 (0.195 10 kg)(9.80 m/s ) 1.91 T (4.00 10 m/s)(2.50 10 C) mg B vq × == = ×× . 27

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27-2 Chapter 27 EVALUATE: The magnetic field could also have a component along the north-south direction, that would not contribute to the force, but then the field wouldn&t have minimum magnitude. Figure 27.2 27.3. IDENTIFY: The force F ! on the particle is in the direction of the deflection of the particle. Apply the right-hand rule to the directions of v ! and B ! . See if your thumb is in the direction of F ! , or opposite to that direction. Use sin Fq v B φ = with 90 = ° to calculate F . SET UP: The directions of v ! , B ! and F ! are shown in Figure 27.3. EXECUTE: (a) When you apply the right-hand rule to v ! and B ! , your thumb points east. F ! is in this direction, so the charge is positive. (b) 63 sin (8.50 10 C)(4.75 10 m/s)(1.25 T)sin90 0.0505 N v B == × × = ° EVALUATE: If the particle had negative charge and v ! and B ! are unchanged, the particle would be deflected toward the west. Figure 27.3 27.4. IDENTIFY: Apply Newton&s second law, with the force being the magnetic force. SET UP: ±± ± ×− j i= k EXECUTE: mq × F =a =vB !! gives q m × = vB a ! ! ! and 84 3 ±± (1.22 10 C)(3.0 10 m/s)(1.63 T) ( ) ± (0.330 m/s ) .
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## This note was uploaded on 03/14/2008 for the course MSE 250 taught by Professor Jabbour during the Spring '08 term at University of Arizona- Tucson.

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27_InstructorSolutio - MAGNETIC FIELD AND MAGNETIC FORCES 27 27.1 IDENTIFY and SET UP Apply Eq(27.2 to calculate F Use the cross products of unit

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