Ch27Word - Chapter 27 p. 1 CHAPTER 27 Magnetism 1. (a) The...

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Chapter 27 p. 1 CHAPTER 27 – Magnetism 1. ( a ) The maximum force will be produced when the wire and the magnetic field are perpendicular, s o w e h a v e F max = ILB , or F max / L = IB = (7.40 A)(0.90 T) = 6.7 N/m . ( b ) We find the force per unit length from F / L = IB sin 45.0° = ( F max / L ) sin 45.0° = (6.7 N/m) sin 45.0° = 4.7 N/m . 2. The force on the wire is produced by the component of the magnetic field perpendicular to the wire: F = ILB sin θ = (150 A)(240 m)(5.0 × 10 –5 T) sin 60° = 1.6 N perpendicular to the wire and to B . 3. For the maximum force the wire is perpendicular to the field, so we find the current from F = ILB ; 0.900 N = I (4.20 m)(0.0800 T), which gives I = 2.68 A . 4. The force on the wire is produced by the component of the magnetic field perpendicular to the wire: F = ILB sin 40° = (4.5 A)(1.5 m)(5.5 × 10 –5 T) sin 40° = 2.4 × 10 –4 N perpendicular to the wire and to B . 5. The maximum force will be produced when the wire and the magnetic field are perpendicular, so we have F max = ILB ; 1.18 N = (8.75 A)(0.555 m) B , which gives B = 0.243 T . 6. The force is maximum when the current and field are perpendicular: F max = ILB . When the current makes an angle with the field, the force is F = ILB sin . Thus we have F / F max = sin = 0.45, or = 27°. 7. ( a ) We see from the diagram that the magnetic field is up, so the top pole face is a south pole . ( b ) We find the current from the length of wire in the field: F = ILB ; 5.30 N = I (0.10 m)(0.15 T), which gives I = 3.5 × 10 2 A . ( c ) The new force is F = ILB sin = F sin = (5.30 N) sin 80° = 5.22 N . Note that the wire could be tipped either way. 8. We have the same upward force on the semicircular wire. The force on each of the horizontal wires will also be upward. Thus the total force is F total = F semi + 2 F wire = 2 IB 0 R + 2 ILB 0 = 2 IB 0 ( R + L ) upward . F max (into page) F (into page) I B I B F (out of page) I B F semi B 0 F wire F wire II
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Chapter 27 p. 2 9. If we consider a length L of the wire, for the balanced forces, we have mg = ρ p r 2 Lg = ILB ; ( 8 . 9 × 10 3 kg/m 3 )p(1.0 × 10 –3 m) 2 (9.80 m/s 2 ) = I (5.0 × 10 –5 T), which gives I = 5.5 × 10 3 A . 10. We find the force per unit length from F / L = I ( i × B ) = (3.0 A)[ i × (0.20 i – 0.30 j + 0.25 k )T] = (3.0 A)(– 0.30 k – 0.25 j )T = (– 0.75 j – 0.90 k ) N/m = (– 7.5 × 10 –3 j – 9.0 × 10 –3 k ) N/cm . 11. We choose the coordinate system shown in the diagram. We select a differential element of the curved wire, d ¬ = d x i + d y j , on which the force is d F = I d ¬ × B = I (d x i + d y j ) × (– B k ) = IB (d x j – d y i ). We find the resultant force by integration: F = ? d F = IB ? (d x j – d y i ) = IB (? x j – ? y i ), where ? x = x b x a , and ? y = y b y a . If we have the same current in the straight wire, the resultant force is F = I (? x i + ? y j ) × (– B k ) = IB (? x j – ? y i ), which is the same result. 12. If we select a differential length d ¬ of the wire, we see that the force on this element has a magnitude d F = IB d ¬ and will have a direction perpendicular to the wire at an angle θ below the horizontal.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch27Word - Chapter 27 p. 1 CHAPTER 27 Magnetism 1. (a) The...

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