1. What is the force due to the Earth’s magnetic field (magnitude 0.00005 T) on

Question

1. What is the force due to the Earth’s magnetic field (magnitude 0.00005 T) on a horizontal wire 12 cm long carrying one amp of current? Assume the Earth’s magnetic field is perpendicular to the wire.

I have this one, if I did it correctly it should be 6 x 10^-6 N

2. Assume the mass of the wire described in problem 1 is 2 grams. Place another wire parallel to this wire, at a distance of 2 mm below it. How much current must flow in this second wire to lift the first wire? Also, what direction must the current flow in the second wire?

I'm not sure how to start with this one, any help would be appreciated. Thanks!

Explanation / Answer

force between parallel wires = uo I1 I2 L/2pid

d is the distance between the wires = 2*10^-3 m

length of the wire = 12 cm = 0.12 m

so

6*10^-6 = (4pi*10^-7 * 1 * I2* 0.12)/(2*3.14* 2*10^-3)

current in the second wire = 0.5 AMps

Since both wires are to be attracted , Currrent has to be the in the same direction

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