What magnetic field strength will levitate the 2.0 g wire in the figure(Figure 1)?

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Part A

What magnetic field strength will levitate the 2.0 g wire in the figure(Figure 1)? Assume that I = 1.6 A and d = 11 cm.

Express your answer to two significant figures and include the appropriate units

Part B

What is the direction of the magnetic field?

Expert Solution

Answer:

Given, mass of the wire m = 2.0 g, current I = 1.6 A and d = 11 cm

(a) Magnetic force on a wire is F = IL x B.

In order to levitate the given force, the magnetic force has to be in the upward direction. Using Fleming's left hand rule, magnetic field B must have a component out of the page. The component of B parallel to the does not affect the force and is not determined. If the component of B which is perpendicular to the current, then B will point straight out of the page with a magnitude

B = F/IL = mg /IL = (0.002 kg) (9.8 m/s²) / (1.6 A) (0.11 m) = 0.111 T.

(b) Out of the screen.

genius_generous answered 2 years ago

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