28 - SOURCES OF MAGNETIC FIELD 28 28.12. IDENTIFY: A...

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S OURCES OF M AGNETIC F IELD 28.12. IDENTIFY: A current segment creates a magnetic field. SET UP: The law of Biot and Savart gives 0 2 sin 4 Idl dB r  . Both fields are into the page, so their magnitudes add. EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives 7 2 2.50 cm (12.0 A)(0.00150 m) 4π 10 T m/A 8.00 cm (0.0800 m) dB     = 8.79 10 8 T The field from the 24.0-A segment is twice this value, so the total field is 2.64 10 7 T, into the page. EVALUATE: The rest of each wire also produces field at P . We have calculated just the field from the two segments that are indicated in the problem. 28.22. IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine points where the fields of the two wires cancel. (a) SET UP: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires. Consider a point a distance x from the wire carrying 2 tot 75.0 A. IB will be zero where 12 . BB EXECUTE: 0 1 0 2 2 (0.400 m ) 2 II xx   2 1 1 2 (0.400 m ) ; 25.0 A, 75.0 A I x I x I I tot 0.300 m; 0 xB  along a line 0.300 m from the wire carrying 75.0 A and 0.100 m from the wire carrying current 25.0 A. (b) SET UP: Let the wire with 1 25.0 A I be 0.400 m above the wire with 2 75.0 A. I The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires. But to have must be closer to wire #1 since , so can have tot 0 B only at points above both wires. Consider a point a distance x from the wire carrying 1 tot 25.0 A. will be zero where . EXECUTE: 0 1 0 2 2 2 (0.400 m ) 21 (0.400 m ); 0.200 m I x I x x tot 0 B along a line 0.200 m from the wire carrying current 25.0 A and 0.600 m from the wire carrying current 2 75.0 A. I EVALUATE: For parts (a) and (b) the locations of zero field are in different regions. In each case the points of zero field are closer to the wire that has the smaller current. 28.28. IDENTIFY: Apply Eq.(28.11) for the force from each wire. SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. 28
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EXECUTE: On the top wire 22 00 11 , 2 2 4 F I I L d d d      upward. On the middle wire, the magnetic forces cancel so the net force is zero. On the bottom wire , 2 2 4 F I I L d d d downward.
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28 - SOURCES OF MAGNETIC FIELD 28 28.12. IDENTIFY: A...

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