A 0.62 kg copper rod rests on two horizontal rails 0.52 m apart and carries a cu

Question

A 0.62 kg copper rod rests on two horizontal rails 0.52 m apart and carries a current of 40 A from one rail to the other. The coefficient of static friction between rod and rails is 0.50. What is the smallest magnetic field (not necessarily vertical ) that would cause the rod to slide? What is the angle of B from the vertical? (deg)

Magnetic field? I got 0.152 T but it won't take T as a unit in Loncapa.

For angle of B? I got 26.57 deg but it won't take deg as a unit. I believe these answers are right but can you work through them just to make sure? And I don't understand why it won't take the deg or T

Explanation / Answer

The forces along x and y direction are:

Fx = i L Bx ;

Bx = B cos(theta) ; By = B sin(theta)

F(n) = mg - i L By

Ff = us (mg - i L By)

Fx - Ff = 0

Fx = Ff

i L Bx = us (mg - i L By)

i L B cos(theta) = us (mg - i L B sin(theta))

B = us mg/ [i L ( cos(theta) + us sin(theta))]

theta = tan^-1(us) = tan^-1(0.5) = 26.57 deg

B = 0.5 x 0.62 x 9.81/[40 x 0.52 (cos26.57 + 0.5 x sin26.57)] = 0.233 T

Hence, B = 0.131 T

Theta = 26.57 deg

[unit will be Tesla(T) only]

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