Homework_7 - Homework 7 solutions 5.3. Consider a single...

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Unformatted text preview: Homework 7 solutions 5.3. Consider a single current loop in the x- y plane. On the axis, the magnetic scalar potential for the magnetic induction is ( ) ( ) 1 2 2 2 . 2 I z z z z z a =- + Hence the magnetic induction on the axis is ( ) ( ) ( ) 2 1 2 3 2 2 2 2 2 1 . 2 z I z B z z a z a =- + + For a loop at position z the magnetic induction on the axis is ( ) ( ) ( ) ( ) 2 1 2 3 2 2 2 2 2 1 . 2 z z z I B z z z a z z a - =- - +- + With N loops per unit length, we get the magnetic induction in the solenoid is ( ) ( ) ( ) ( ) 2 1 2 3 2 2 2 2 2 1 . 2 L z z z IN B z dz z z a z z a - =- - +- + The integral can be evaluated by noting that the integrand is a perfect differential ( ) ( ) ( ) 1 2 1 2 1 2 2 2 2 2 2 2 , 2 2 L z IN IN d z z L z z B z dz dz z a z z a z L a -- = = + + - +- + from which the result follows. 5.4. (a) The equations 0, 0, = = B B give ( ) 1 0, 0. z z B B B B z z + =- = We look for series solution for the components away from the z-axis ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 , , , . z B z a z a z a z B z b z b z b z = + + + = + + + Substituting in the above equations, we find ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 1 2 2 1 2 1 2 1 2 3 0, 2 0. a z a z a z b z b z b z a z a z a z b z b z + + + + + + + = + + +- + + = Since the coefficients of powers of are independent, we get ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 0, 1 0, 1 0, n n n n a z n a z b z a z n b z- + = + + =- + = with ( ) b z unconstrained. Eliminating ( ) , n a z we get ( ) ( ) ( ) 2 1 1 1 0. n n n b z b z +- + + = Note that ( ) a z = implies ( ) 1 0. b z = Therefore the odd b s are all zero. The above recurrence relation for n = 1 gives ( ) ( ) 2 4 0. b z b z + = Hence ( ) ( ) ( ) ( ) ( ) 2 2 2 1 , . 4 z B z b z b z b z b z = + + =- + But ( ) ( ) 0, , z B z b z = which allows us to write ( ) ( ) ( ) 2 2 2 0, 1 , 0, . 4 z z z d B z B z B z dz =- + We also have ( ) ( ) 1 1 . 2 n n a z b z n + = - + We see that the even a s are all zero. Also ( ) ( ) ( ) ( ) ( ) ( ) 1 3 3 2 1 , 2 1 1 . 4 16 a z b z a z b z b z = - = - = Hence ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 3 3 1 1 , 2 16 0, 0, 1 1 ....
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This note was uploaded on 12/02/2011 for the course PHYS 809 taught by Professor Macdonald during the Fall '11 term at University of Delaware.

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Homework_7 - Homework 7 solutions 5.3. Consider a single...

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