Physics9C_A_HW8_Soln

# Physics9C_A_HW8_Soln - MAGNETIC FIELD AND MAGNETIC FORCES...

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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 r 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 r (a) () ˆ 1.40 T = B i r ( ) ( ) ( ) 84 4 垐垐 1.24 10 C 1.40 T q =×= − × × × × × F vB ii j i rr r & 0, ×= ×=− j ik ( ) ( )( ) ( ) 4 6.68 10 N =− × × − =− × F kk r EVALUATE: The directions of and r r are shown in Figure 27.1a. The right-hand rule gives that × r r is directed out of the paper (+ z -direction). The charge is negative so F r is opposite to ; × r r Figure 27.1a F r is in the - z direction. This agrees with the direction calculated with unit vectors. (b) EXECUTE: ˆ 1.40 T = B k r ( ) ( ) ( ) 4 q + × ×− × × F j k r , − ×= jjki ( ) ( ) ( ) ( ) 4 4 7.27 10 N −− × − + × = × + × Fjii j r EVALUATE: The directions of and r r are shown in Figure 27.1b. The direction of F r is opposite to × r r since q is negative. The direction of F r computed from the right-hand rule agrees qualitatively with the direction calculated with unit vectors. Figure 27.1b 27.3. IDENTIFY: The force F r on the particle is in the direction of the deflection of the particle. Apply the right-hand rule to the directions of v r and B r . See if your thumb is in the direction of F r , or opposite to that direction. Use sin Fq v B φ = with 90 = ° to calculate F . SET UP: The directions of v r , B r and F r are shown in Figure 27.3. EXECUTE: (a) When you apply the right-hand rule to v r and B r , your thumb points east. F r 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 r and B r are unchanged, the particle would be deflected toward the west. 27

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27-2 Chapter 27 Figure 27.3 27.5. IDENTIFY: Apply sin Fq v B φ = and solve for v . SET UP: An electron has 19 1.60 10 C q =− × . EXECUTE: 15 6 19 3 4.60 10 N 9.49 10 m s sin (1.6 10 C)(3.5 10 T)sin60 F v qB −− × == = × ×× ° EVALUATE: Only the component sin B of the magnetic field perpendicular to the velocity contributes to the force. 27.7. IDENTIFY: Apply q × F=v B rr r . SET UP: ˆ y v v= j r , with 3 3.80 10 m s y v × . 3 7.60 10 N, 0, xy FF = = and 3 5.20 10 N z F × . EXECUTE: (a) () x yz zy v Bv B q v B = . 36 3 (7.60 10 N) ([7.80 10 C)( 3.80 10 m s )] 0.256 T zxy BF q v × × × = 0 , x x z v B = which is consistent with F r as given in the problem. There is no force component along the direction of the velocity. z yx v B q v B = . 0.175T xz y q v . (b) y B is not determined. No force due to this component of B r along v r ; measurement of the force tells us nothing about . y B (c) 33 ( 0.175 T)(+7.60 10 N) ( 0.256 T)( 5.20 10 N) xx yy zz ⋅= + + = × + − × 0 . B r and F r are perpendicular (angle is 90 ) ° . EVALUATE: The force is perpendicular to both v r and B r , so vF r r is also zero. 27.15. (a) IDENTIFY: Apply Eq.(27.2) to relate the magnetic force F r to the directions of and .
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## This note was uploaded on 09/22/2011 for the course PHY 09C 9c taught by Professor Zhu during the Spring '11 term at UC Davis.

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Physics9C_A_HW8_Soln - MAGNETIC FIELD AND MAGNETIC FORCES...

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