A 0.65 copper rod with a mass of 0.22 carries a current of 11 in the positive x

Question

A 0.65 copper rod with a mass of 0.22 carries a current of 11 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.
=


Part B
What is the direction of the minimum magnetic field needed to levitate the rod?
What is the direction of the minimum magnetic field needed to levitate the rod?
positive direction
negative direction
positive direction
negative direction
positive direction
negative direction

Explanation / Answer

The force on a wire of length L in a magnetic field B carrying current I is F = I*L x B x is the vector cross product. The force per unit length is F/L, so F/L = I x B You are given F/L = 3.3*10^-2 N/m (I assume its N). The magnitude of the cross product is I*B*sinø, so the magnitude of the force per unit length is |F/L| = I*B*sinø B = |F/L| / (I*sinø) = 3.3*10^-2 / (6.2*sin7.5) = 4.1*10^-2 T

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