 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015/10/d/solution

At terminal velocity, the upward magnetic force balances the weight. Substitute the electric current from the preceding solution:

M g = I a B = \frac{B ^{2} a ^{2}}{R} v_{s}, M = ρ_{m} πa d_{w}^{2}, R = \frac{16 ρ _{c} a}{π d _{w}^{2}} .

The geometric factors cancel, giving

v_{s} = \frac{M g R}{B ^{2} a ^{2}} = \frac{16 ρ _{m} ρ _{c} g}{B ^{2}} = 2.39 \times 1 0^{- 2} m s^{- 1} .

Here

g = 9.81 m s^{- 2}

B = 1 T

ρ_{m} = 8960 kg m^{- 3}

and

ρ_{c} = 1.7 \times 1 0^{- 8} Ω m

. As numerical checks, the specified

a = 0.08 m

and

d_{w} = 0.001 m

give

M = 2.252 \times 1 0^{- 3} kg

R = 6.926 \times 1 0^{- 3} Ω

and

I_{s} = 0.276 A

The gravitational power lost,

M g v_{s}

, equals the Joule heating

I_{s}^{2} R

, consistent with energy conservation. This terminal velocity applies only while the top arm remains in the magnetic field; after it exits, the magnetic braking ceases, apart from any brief inductive transient.

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