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A horizontal wire of length 0.1 meters carries a current of 5 amps.  What would be the...

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alexandra210696 | Student, Grade 10 | eNotes Newbie

Posted December 21, 2011 at 9:50 PM via web

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A horizontal wire of length 0.1 meters carries a current of 5 amps.  What would be the magnitude and direction of an external magnetic field which could support the wire if we know it has a linear density of 3x10^-3 kg/m?

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mwmovr40 | College Teacher | (Level 1) Associate Educator

Posted January 3, 2012 at 6:21 AM (Answer #1)

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To determine the magnetic field strength required to levitate a wire one must first know the weight of the wire. Weight is given by the equation W = mg where g is the acceleration due to gravity and m is the mass of the wire.

To get the mass of the wire we must multiply the linear density by the length of the wire: m = DL = 3x10^-3 kg/m X 0.1 m = 3x10^-4 kg. So

W = 3X10^-4 kg X (-9.8 m/s^2) = -0.0029 N.

The magnetic force must be equal and opposite to this: F = 0.0029 N.

To get the magnetic field which will create this force will require this equation:

F = ILB where I is the current in the wire, L is the lenght of the wire, and B is the external magnetic field creating the Force.

So B = F/(IL) = 0.0029N/ (5ampX0.1m) = 0.006 T

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