A thin uniform rod with negligible mass and length 0.200 m is attached to the fl

Question

A thin uniform rod with negligible mass and length 0.200 m is attached to the floor by a frictionless hinge at point P as shown. A horizontal spring with force constant k = 4.80 N/m connects the other end of the rod to a vertical wall. the rod is in a uniform magnetic field B = 0.340 T directed into the plane of the figure. There is current I = 6.50 An in the road in the direction shown. Calculate the torque due to the magnetic force on the rod, for an axis at P (by direct integration). Is it correct to take the total magnetic force to act at the center of gravity of the rod when calculating the torque? (i.e. re-calculate the torque as if the magnetic force simply acted at the center of mass instead of distributed all along the road.) Explain. When the rod is in equilibrium and makes an angle of 53.0 degree with the floor, is the spring stretched or compressed?

Explanation / Answer

a) magnitude of the torque due to the magnetic force on the rod

T = F * (L/2) = I L B * (L / 2)

= (6.5 * 0.200 * 0.34 * 0.200) / (2) = 44.2 * 10-3 N.m

It is correct

b) The spring is stretched

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