A 1.08 kg copper rod rests on two horizontal rails 0.84 m apart and carries a...

A 1.08 kg copper rod rests on two horizontal rails 0.84 m apart and carries a current of 60 A from one rail to the other. The coefficient of static friction between rod and rails is 0.53. What is the smallest magnetic field (not necessarily vertical ) that would cause the rod to slide?

A) What is the angle of B from the vertical? (deg)

B) What is the magnitude of B?

Expert Solution

A. magnetic force is F = i L B

mg = i L B

force of friction is μ_s ( mg - i L B )

F_s - F = 0

iL B cos θ = μ_s ( mg - iLB sinθ)

B = μs mg / iL ( cos θ + μs sinθ)

θ = tan^-1 (μs)

= tan^-1 ( 0.53)

= 27.9 degrees

B.The smallest value of the magnetic field is

B_min = 0.53 * 1.08*(9.8) / 60 * 0.84 ( cos 27.9 + 0.53 sin 27.9)

= 0.0983 T

genius_generous answered 7 months ago

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