A thin, uniform rod with negligible mass and length 0.220 m is attached to the f

Question

A thin, uniform rod with negligible mass and length 0.220 m is attached to the floor by a frictionless hinge at point P (the figure (Figure 1)).  A horizonal spring with force constant k=4.90N/m connects the other end of the rod to a vertical wall.  The rod is in a uniform magnetic field B=0.360T directed into the plane of the figure.  There is current I=6.60A in the rod, in the direction shown.

1)  Calculate the magnitude of the torque due to the magnetic force on the rod, for an axis at P.

2) Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque?  Explain.

3)  How much energy is stored in the spring when the rod is in equilibrium?

A thin, uniform rod with negligible mass and length 0.220 m is attached to the floor by a frictionless hinge at point P (the figure (Figure 1)). A horizonal spring with force constant k=4.90N/m connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B=0.360T directed into the plane of the figure. There is current I=6.60A in the rod, in the direction shown. 1) Calculate the magnitude of the torque due to the magnetic force on the rod, for an axis at P. Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? Explain. How much energy is stored in the spring when the rod is in equilibrium?

Explanation / Answer

F = B*I*L = 0.36*6.6*0.22 = 0.52272 N

torque due to magnetic force = F*(L/2) = 0.0574992 Nm

net torque = 0

F*(L/2) = T*L*sin53


F = T*2*sin53
T = 0.3272 N

T = k*x

x = 0.0667 m

E = 0.5*k*x^2 = 0.01089 J

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