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Homework_7

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

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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 ( ) ( ) 0 1 2 2 2 . 2 I z z z z z a μ Φ = - + Hence the magnetic induction on the axis is ( ) ( ) ( ) 2 0 1 2 3 2 2 2 2 2 1 . 2 z I z B z z a z a μ = - + + For a loop at position z the magnetic induction on the axis is ( ) ( ) ( ) ( ) 2 0 1 2 3 2 2 2 2 2 1 . 2 z z z I B z z z a z z a μ - = - - + - + With N loops per unit length, we get the magnetic induction in the solenoid is ( ) ( ) ( ) ( ) 2 0 1 2 3 2 2 2 2 2 0 1 . 2 L z z z IN B z dz z z a z z a μ - = - - + - + The integral can be evaluated by noting that the integrand is a perfect differential ( ) ( ) ( ) 0 0 1 2 1 2 1 2 2 2 2 2 2 2 0 , 2 2 L z IN IN d z z L z z B z dz dz z a z z a z L a μ μ ′ - - = = + + - + - + 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 ρ ρ ρ ρ ρ ρ + = - =

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We look for series solution for the components away from the z -axis ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 1 2 2 0 1 2 , , , . z B z a z a z a z B z b z b z b z ρ ρ ρ ρ ρ ρ ρ = + + + = + + + Substituting in the above equations, we find ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 0 1 2 0 1 2 2 0 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 ρ ρ ρ ρ ρ ρ ρ ρ + + + + + + + = + + + - + + = Since the coefficients of powers of ρ are independent, we get ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 1 0, 1 0, 1 0, n n n n a z n a z b z a z n b z - + = + + = - + = with ( ) 0 b z unconstrained. Eliminating ( ) , n a z we get ( ) ( ) ( ) 2 1 1 1 0. n n n b z b z + - ′′ + + = Note that ( ) 0 0 a z = implies ( ) 1 0. b z = Therefore the odd b ’s are all zero. The above recurrence relation for n = 1 gives ( ) ( ) 2 0 4 0. b z b z ′′ + = Hence ( ) ( ) ( ) ( ) ( ) 2 2 0 2 0 0 1 , . 4 z B z b z b z b z b z ρ ρ ρ ′′ = + + = - + But ( ) ( ) 0 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 ρ ρ = - + 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 0 3 3 2 0 1 , 2 1 1 . 4 16 a z b z a z b z b z = - = - = Hence ( ) ( ) ( ) ( ) ( ) ( ) 3 3 0 0 3 3 3 1 1 , 2 16 0, 0, 1 1 . 2 16 z z B z b z b z B z B z z z ρ ρ ρ ρ ρ ρ = - + + = - + + Another approach is to note that we can use a scalar potential that satisfies Laplace’s equation. If we know B z on the axis, we can find the potential on the axis and hence the potential everywhere. Working in cylindrical polar coordinates the potential satisfies 2 2 1 0. z ρ ρ ρ ρ ∂Φ ∂ Φ + = The solution is a linear superposition of terms of form ( ) 0 , kz J k e ρ - (assuming z > 0). Formally ( ) ( ) ( ) 0 0 , , kz z A k J k e dk ρ ρ - Φ = where ( ) ( ) 0 0, . kz z A k e dk - Φ = Since ( ) 0 J k ρ has a series that contains only even powers of its argument, we can show, by making use of 2 2 2 , n kz n kz n e k e z - - = that the solution for the potential can also be written as ( ) ( ) 0 , 0, . z J z z ρ ρ Φ = Φ

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

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