 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017/9/d/solution

(0, y)

of the middle wire, the horizontal attractions towards the two fixed outer wires cancel. Each attraction has magnitude

μ_{0} I^{2} / (2 π d^{2} + y^{2})

per unit length. Projecting them in the

y

direction gives

μ_{m} \overset{y}{¨} = - \frac{μ _{0} I ^{2}}{π} \frac{y}{d ^{2} + y ^{2}} ≃ - \frac{μ _{0} I ^{2}}{π d ^{2}} y (∣ y ∣ ≪ d) .

Neglecting dissipative effects and other transverse forces, this is simple harmonic motion. Release from rest at

y_{0}

gives

y (t) = y_{0} cos Ω t, Ω = \frac{μ _{0} I ^{2}}{π μ _{m} d ^{2}} = 2.00 s^{- 1}, T = \frac{2 π}{Ω} = 3.14 s .

The inverse-angular-frequency timescale is

Ω^{- 1} = 0.500 s

; it differs from the oscillation period by

2 π

. The assumed displacement plane matters: small displacements towards one outer wire along

x

are unstable, rather than oscillatory.

This result concerns the specified magnetic motion with gravity absent from, or compensated in, the

y

direction. The source diagram fixes coordinate directions, not their orientation relative to gravity. If

y

is physically vertical with upward positive and the wire is unsupported, add

- μ_{m} g

to the force per unit length. The maximum restoring magnetic force is only

μ_{0} I^{2} / (2 π d) = 2.0 \times 1 0^{- 5} N m^{- 1}

, compared with weight

9.8 \times 1 0^{- 3} N m^{- 1}

; that different setup would fall and could not oscillate about the original plane.

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