Fall 2007 4 Em

A spiral spring has N turns and initial length x0. How does its length changes if a small current I is made to flow through it? The spring has an elastic constant k for longitudinal deformations. Assume that the spring can be treated as a perfect solenoid and that its radius R remains fixed.

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Answer

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It is convenient to think that the current I was created by some external source while the spring was kept at the original length x0. The coil was then short-circuited leaving the current flowing. Finally, the spring was allowed to expand or contract freely. In this formulation, the magnetic flux $\Phi = L I$ remains constant since the circuit has zero resistance, which simplifies the derivation.

Since the current flows in the same direction in the adjacent coils of the spring, these coils attract. Hence, the spring would shorten. Let x be the new length. To find the contraction $\bigtriangleup x = x - x_0$ we can minimize the sum of the magnetic and the elastic energy,

(1)
\begin{align} E = \frac{\Phi^2}{2 c^2 L} + \frac{k}{2}(\bigtriangleup x) ^2 \end{align}

where

(2)
\begin{align} L= \frac{4 \pi^2 N^2 R^2}{c^2 x } \end{align}

is the inductace of the spring (the derivation of this known formula is elementary). Since $\Phi$ = const, it is easy to take the derivative of E with respect to x. Equating it to zero, one finds that the mimimum energy is reached at

(3)
\begin{align} \bigtriangleup x \simeq - \frac{2 \pi^2}{c^2} \frac{N^2 R^2 I^2}{k x_0^2} \end{align}
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