Please help! Example 24.11 A DC power line near the equator runs east-west. At t

Question

Please help!

Example 24.11 A DC power line near the equator runs east-west. At this location, the earth's magnetic field is parallel to the ground, points north, and has a magnitude of 50 ??. A 400 m length of heavy cable that spans the distance between two towers has a mass of 1000 kg. What direction and magnitude of current would be necessary to offset the force of gravity and "levitate" the wire? The power line will actually carry current that is much less than this; 850 A is a typical value.) [4.9x10% A]

Explanation / Answer

Magnetic force: F = B*I*L

Here B = magnetic field = 50 uT

I = current

L = length of the wire = 400 m

To offset the magnetic force , force of gravity must be equal and opposite to magnetic force.

So, mg = BIL

So, I = mg/BL

= 1000*9.8/(50*10^-6*400)

= 4.9*10^5 A <----- answer

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