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Chapter 27 HW Solutions

Chapter 27 HW Solutions - PHYS 241 Ch 27 Homework Solutions...

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PHYS 241 Ch. 27 Homework Solutions 27.3.I DENTIFY : 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 F q vB φ = with 90 φ = ° to calculate F . S ET U P : The directions of v r , B r and F r are shown in Figure 27.3. E XECUTE : (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) 6 3 sin (8.50 10 C)(4.75 10 m/s)(1.25 T)sin90 0.0505 N F q vB φ - = = × × = ° E VALUATE : If the particle had negative charge and v r and B r are unchanged, the particle would be deflected toward the west. Figure 27.3 27.7. I DENTIFY : Apply q × F = v B r r r . S ET U P : ˆ y v v = j r , with 3 3.80 10 m s y v = - × . 3 7.60 10 N, 0, x y F F - = + × = and 3 5.20 10 N z F - = - × . E XECUTE : (a) ( ) x y z z y y z F q v B v B qv B = - = . 3 6 3 (7.60 10 N) ([7.80 10 C)( 3.80 10 m s )] 0.256 T z x y B F qv - - = = × × - × = - ( ) 0, y z x x z F q v B 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 x y y x y x F q v B v B qv B = - = - . 0.175 T x z y B F qv = - = - . (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) 3 3 ( 0.175 T)(+7.60 10 N) ( 0.256 T)( 5.20 10 N) x x y y z z B F B F B F - - = + + = - × + - - × B F r r 0 = B F r r . B r and F r are perpendicular (angle is 90 ) ° . E VALUATE : The force is perpendicular to both v r and B r , so v F r r is also zero. 27.11. I DENTIFY and S ET U P : B d Φ = B A r r Circular area in the xy -plane, so ( 29 2 2 2 0.0650 m 0.01327 m A r π π = = = and d A r is in the z -direction. Use Eq.(1.18) to calculate the scalar product. E XECUTE : (a) ( 29 ˆ 0.230 T ; and d = B k B A r r r are parallel ( 29 0 φ = ° so . d B dA = B A r r B is constant over the circular area so 2 3 (0.230 T)(0.01327 m ) 3.05 10 Wb B d B dA B dA BA - Φ = = = = = = × B A r r (b) The directions of and d B A r r are shown in Figure 27.11a. cos with 53.1 d B dA φ φ = = ° B A r r 1
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Figure 27.11a B and φ are constant over the circular area so cos cos cos B d B dA B dA B A φ φ φ Φ = = = = B A r r ( 29 ( 29 2 3 0.230 T cos53.1 0.01327 m 1.83 10 Wb B - Φ = ° = × (c) The directions of and d B A r r are shown in Figure 27.11b. 0 since and are perpendicular ( 90 ) d d φ = = ° B A A B r r r r 0. B d Φ = = B A r r Figure 27.11b E VALUATE : Magnetic flux is a measure of how many magnetic field lines pass through the surface. It is maximum when B r is perpendicular to the plane of the loop (part a) and is zero when B r is parallel to the plane of the loop (part c). 27.13. I DENTIFY : The total flux through the bottle is zero because it is a closed surface. S ET U P : The total flux through the bottle is the flux through the plastic plus the flux through the open cap, so the sum of these must be zero. plastic cap 0. Φ + Φ = ( 29 2 plastic cap cos cos B A B r π Φ = -Φ = - Φ = - Φ E XECUTE : Substituting the numbers gives plastic Φ = – (1.75 T)π(0.0125 m) 2 cos 25° = –7.8 × 10 –4 Wb E VALUATE : It would be impossible to calculate the flux through the plastic directly because of the
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