24 HMK #6 Chp 27b

24 HMK #6 Chp 27b - Ch. 27 PHYSICS 2B Spring 2011 Homework...

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Ch. 27 PHYSICS 2B Spring 2011 Homework Solutions – Chapter 27 Additional Problems Author: Evan Grohs - egrohs@physics.ucsd.edu Ch. 27 Problems: 3, 4, 7, 14, 15, 24, 27, 30, 32, 38, 42, 45, 46 Page 1/7
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Ch. 27 page 2/7 M AGNETIC F IELD AND M AGNETIC F ORCES 27.3. I DENTIFY : 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 with 90 ° to calculate F . S ET U P : The directions of v , B and F are shown in Figure 27.3. E XECUTE : (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  ° E VALUATE : 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. I DENTIFY : Apply Newton’s second law, with the force being the magnetic force. S ET U P : ˆˆ ˆ  j i= k E XECUTE : mq F =a =vB  gives q m vB a and 84 3 (1.22 10 C)(3.0 10 m/s)(1.63 T) ( ) ˆ (0.330 m/s ) . 1.81 10 kg  ji a= = k E VALUATE : The acceleration is in the -direction z and is perpendicular to both v and B . 27.7. I DENTIFY : Apply q F=v B . S ET U P : ˆ y v v= j , 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 . 27
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Ch. 27 page 3/7 E XECUTE : (a) () x yz zy Fq 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 as given in the problem. There is no force component along the direction of the velocity. z xy yx v B q v B . 0.175 T xz y q v . (b) y B is not determined. No force due to this component of B along v ; 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 and F are perpendicular (angle is 90 ) . E VALUATE : The force is perpendicular to both v and B , so vF is also zero. 27.14. I DENTIFY : When B is uniform across the surface, cos B BA    BA . S ET U P : A is normal to the surface and is directed outward from the enclosed volume. For surface abcd , ˆ A A= i . For surface befc , ˆ A k . For surface aefd , cos 3/5 and the flux is positive. E XECUTE : (a) 0 . B abcd  (b) ( ) (0.128 T)(0.300 m)(0.300 m) 0.0115 Wb. B befc (c) 3 5 ( ) cos (0.128 T)(0.500 m)(0.300 m) 0.0115 Wb. B aefd BA (d) The net flux through the rest of the surfaces is zero since they are parallel to the x -axis. The total flux is the sum of all parts above, which is zero.
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24 HMK #6 Chp 27b - Ch. 27 PHYSICS 2B Spring 2011 Homework...

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