Problem 22.31 A0.65 m copper rod with a mass of 4.00x102 kg carries a current of

Question

Problem 22.31 A0.65 m copper rod with a mass of 4.00x102 kg carries a current of 15 A in the positive x direction. Let upward be the positive y direction. Part A What is the magnitude of the minimum magnetic field needed to levitate the rod? Express your answer using two significant figures. Submit Request Answer Part B What is the direction of the minimum magnetic field needed to levitate the rod? O positive z direction O negative direction O positive y direction O negative y direction O positive z direction O negative z direction Submit Request Answer Provide Feedback

Explanation / Answer

A)

The copper rod is lying along the x-axis with current in the positive x direction

now the force on it is solely due to gravity , that is , equal to mg

Now to levitate it you just need to apply force equal to mg

The force on a Wire / Rod of length L carrying current i kept in a magnetic field B ==> F=BiL

hence F = BiL = mg

B * 15 * 0.65 = 0.04 * 9.8

B= 0.040 Tesla

B)

Now to determine the direction You can use Flemings Left hand rule

If i is towards +ve x and and force towards +y then magnetic field should be applied in -ve z direction..

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