P30_054 - the dipole is perpendicular to the axis of the...

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54. (a) By imagining that each of the segments bg and cf (which are shown in the Fgure as having no cur- rent) actually has a pair of currents, where both currents are of the same magnitude ( i ) but opposite direction (so that the pair e±ectively cancels in the Fnal sum), one can justify the superposition. (b) The dipole moment of path abcdefgha is ±µ = ±µ bcfgb + ±µ abgha + ±µ cdefc =( ia 2 )( ˆ j ˆ i+ ˆ i) = ia 2 ˆ j =( 6 . 0 A)(0 . 10 m) 2 ˆ j=6 . 0 × 10 2 A · m 2 ˆ j . (c) Since both points are far from the cube we can use the dipole approximation. ²or ( x, y, z )= (0 , 5 . 0m , 0) ± B (0 , 5 . 0m , 0) µ 0 2 π ±µ y 3 = (1 . 26 × 10 6 T · m / A)(6 . 0 × 10 2 m 2 · A) ˆ j 2 π (5 . 0m) 3 =9 . 6 × 10 11 T ˆ j . ²or ( x, y, z )=(5 . 0m , 0 , 0), note that the line joining the end point of interest and the location of the dipole is perpendicular to the axis of the dipole. You can check easily that if an electric dipole
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Unformatted text preview: the dipole is perpendicular to the axis of the dipole. You can check easily that if an electric dipole is used, the Feld would be E (1 / 4 )( p/x 3 ), which is half of the magnitude of E for a point on the y axis the same distance from the dipole. By analogy, in our case B is also half the value or B (0 , 5 . 0 m , 0), i.e., B (5 . 0 m , , 0) = 1 2 B (0 , 5 . 0 m , 0) = 1 2 (9 . 6 10 11 T) = 4 . 8 10 11 T . Just like the electric dipole case, B (5 . 0 m , , 0) points in the negative y direction....
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