problemset2

# x x 2 y y 2 z z 2 r r 1 1 1 1 x

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Unformatted text preview: I1 I2 ˆ iy 2πd (i) I1 = I , � � −µ0 I2 ˆ ix 2πd = I2 = 0 Force =0 Length (ii) I1 = I , I2 = I Force −µ0 I 2 ˆ = iy Length 2πd (iii) I2 = −I I1 = I , Force µ0 I 2 ˆ =+ iy Length 2πd Problem 2.2 A The idea here is similar to applying the chain rule in a 1D problem d dx � 1 f (x) � � d = df � 1 f (x) �� � � � df −f (x) =2 dx f (x) →→ f (x) corresponds to |− − − � |. So, by diﬀ. f (x) we get part of the answer to the derivative of r r can just do it directly too. � →− |− − →� | = (x − x� )2 + (y − y � )2 + (z − z � )2 r r � � � � � � � � 1 1 1 1 ˆx ∂ ˆy ∂ ˆz ∂ � − →� = i → → + i ∂y |→ − →� | + i ∂z |→ − →� | −− −− ∂x |− − − � | r r |→ − − | r r r r r r So we can apply the trick above by just considering x, y, and z components separately. � ∂ − →� ∂ �� |→ − − | = r r (x − x� )2 + (y − y � )2 + (z − z � )2 ∂x ∂x x − x� = � (x − x� )2 + (y − y � )2 + (z − z � )2 x − x� = − →� → − − | |r r Similarly, ∂ ∂x ∂ → →� z − z � ∂ − →� y − y� |→ − − | = − →� and r r |− − − | = → → � . r r →−− | − −− | ∂y ∂z |r r |r r � 1 →− |− − →� r r � = ∂→ − − ∂x |− − →� | r r −− → − →� |2 |r r and so on for y and z . →− |− − →� |2 = (x − x� )2 + (y − y � )2 + (z − z � )2 r r 3 1 f (x) . But we Problem Set 2 6.641, Spring 2009 so: � � � − (x − x� )ˆ + (y − y � )iˆ + (z − z � )iˆ ix y z 1 � − →� = 3 |→ − − | r r [(x − x� )2 + (y − y � )2 + (z − z � )2 ] 2 � → −3 Denominators = |− − →� | 2 . Thus, r r � � →− →− −1 1 −( − − → � ) r r (− − → � ) r r � − →� = − →� 3 = − →� 2 → →� →−− | →−− | → − − | |− − − | |r r |r r |r...
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## This document was uploaded on 03/19/2014 for the course EECS 6.641 at MIT.

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