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?
1 Answer | Add Yours
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
We’ve answered 328,065 questions. We can answer yours, too.Ask a question