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ASSIGN_4 - Assignment#4 PHYSICS 354/4-01 ELECTRICITY&...

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Unformatted text preview: Assignment #4, PHYSICS 354/4-01 ELECTRICITY & MAGNETISM II Mar. 2010 Assignment #4, Page 1 of 4 Solution to Assignment # 4 259 . 6.2, p Given B is uniform. The area of an arbitrary current loop (not necessarily flat) can be resolved into components parallel and perpendicular to B . The component parallel to B contributes no torque. The projected areas can further be broken down into tiny loops so that one can make use of B m N × = which is valid for a localized magnetic dipole. Choose the coordinate axes so that B is along the z- axis without loss of generality. a Id m d = is a differential magnetic moment. ∴ = × ∫ N I da da da B x y z z ( , , ) ( , , ) 0 0 . Since B is constant, [i.e. not a function of the variables of integration], it makes no difference whether we sum the areas first before taking the cross product or vice-versa. Cross first: z x z y z x z y z z y x B m y B m x B da y I B da x I B da da da I N ˆ ˆ ˆ ˆ ) , , ( ) , , (- =- = × = ∫ ∫ ∫ Sum first: z x z y z z y x z z y x B m y B m x B m m m B da da da I N ˆ ˆ ) , , ( ) , , ( ) , , ( ) , , (- = × = × = ∫ Thus N I da B I da B m B = × = × = × ∫ ∫ iff B is uniform in magnitude and direction. 259 . 6.3, p Rotate Figure 6.7 ( p. 259) 90 ° , to have both dipoles ( m 1 , m 2 ) pointing in the z-direction. (a) Use Eq. 6.2: , cos 2 1 2 β π B I R F = since only the horizontal component of B 1 contribute to a force. Here β is the angle B 1 makes with x ˆ at l d I 2 of m 2 as shown in the diagram. This formula is based on fact that the current loop m 2 has a finite R , and the B- field is no longer uniform over the loop, (direction varies although magnitude is reasonably constant). + = θ θ θ π μ θ ˆ sin ˆ cos 2 4 ) , ( 3 1 1 1 3 1 1 1 1 r m r r m r B o ( r 1 >> dimensions of m 1 ) ( ) ( ) 1 1 1 3 1 1 3 1 1 3 1 1 3 1 1 1 1 1 1 1 sin , 1 cos with 4 3 sin cos 3 4 cos sin sin cos 2 4 cos sin cos r R r z r R r m r m r m r m B B B B B o o o r x x r = ≈ = ≈ = + = + = + = θ θ π μ θ θ π μ θ θ θ θ π μ θ θ β θ θ , 6 4 4 3 2 4 1 2 1 4 1 1 2 r m m r R m R I F o o = ≈ π μ π μ π 2 2 2 R I m π = ....
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ASSIGN_4 - Assignment#4 PHYSICS 354/4-01 ELECTRICITY&...

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