Magnetic levitation A metal rod of length w = 12 cm and mass m = 60 grams has me

Question

Magnetic levitation
A metal rod of length w = 12 cm and mass m = 60 grams has metal loops at both ends, which go around two metal poles (see the figure).

The rod is in good electrical contact with the poles but can slide freely up and down. The metal poles are connected by wires to a battery, and a current I = 6 amperes flows through the rod. A magnet supplies a large uniform magnetic field B in the region of the rod, large enough that you can neglect the magnetic fields due to the 6 ampere current. The large magnetic field is oriented to have the maximum effect.

The rod floats at rest h = 3 cm above the table. What is the magnetic field in the region of the rod? Assume +x is to the right, +y is up, and +z is out of the page.

Explanation / Answer

Here,

Bar is in equilibrium.

so, Fnet = 0

Fnet = Fmag + Fgrav = 0

|Fmag| = |Fgrav|

B*I*L = M*g

magnetic field in the region of the rod is,

B =  M*g / I*L

B = 0.060 * 9.8 / 6*0.12

B = 0.816 T

Direction of B is -z.

B = <0 , 0 , -0.816> T

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