The earth\'s magnetic field vector in Ottawa is horizontal to the surface of the

Question

The earth's magnetic field vector in Ottawa is horizontal to the surface of the earth and roughly points towards the north. The magnitude of the corresponding magnetic flux density is ßearth_ 0.5 × 10-4 T. A straight copper (Cu) wire of length L 2 m with a circular cross-section of area A (you do not need to know the actual area to solve this problem) carries a current densityj. The straight wire is also horizontal to the surface of the earth and is oriented perpendicularly to the magnetic field (in the east-west direction). The mass density of 5. towads areo to solve thic ath a circular crossx densit is copper is pcu -8.9x 103kg m-3. The resistivity of copper is pr 1.7x 10-8 m. a. What must J be equal to, and which direction must it flow (towards east or west) in order to levitate the wire? The gravitational acceleration g-9.8m s2. [Note that you don't actually need to know the wire's length and cross-sectional area to solve this problem... they should eventually cacelot in your calculations.] b. When the wire is floating (levitating) how much power will be dissipated per unit volume? This will be a huge amount of power, so it's not quite realistic to achieve!

Explanation / Answer

Given,

B = 0.5 x 10^-4 T ; L = 2 m ; rho-cu = 8.9 x 10^3 kg/m^3 ; rho-r = 1.7 x 10^-8 Ohm-m

a)We know that the magnetic force on the current carrying wire is given by:

Fm = I L B

In order to leviate:

I L B = mg (1)

we know that, density = mass/vol => mass = density x vol

m = rho-cu x A x L

from the defination of current density,

J = I/A => I = JA

putting the value of m and I in (1)

J A L B = rho-cu A L g

J = rho-cu x g/B

J = 8.9 x 10^3 x 9.8/(0.5 x 10^-4) = 1.744 x 10^9 A/m^2

Hence, J = 1.744 x 10^9 A/m^2

b)Power dissipated is:

P = I^2 R = (JA)^2 x rho L/A

P = J^2 A^2 rho L/A = J^2 A rho L

for Unit volume

(P/AL) = J^2 rho

P/V = (1.744 x 10^9)^2 x 1.7 x 10^-8 = 5.171 x 10^10 W/m^3

Hence, P = 5.171 x 10^10 w/m^3

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