10. A thin, uniform rod with negligible mass and length 0.2m is attached to the

Question

10. A thin, uniform rod with negligible mass and length 0.2m is attached to the floor by a frictionless hinge at point P as seen in the figure below. A horizontal spring with spring constant k 4.8N /m connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic field B-.34T as seen in the figure and a current I-65A flows through the rod in the direction shown At the moment illustrated in the figure a 3530. (a) Calculate the torque due to the magnetic force on the rod with respect to an axis passing perpendicularly to the plane of the figure through P. (b) When the rod is in equilibrium and makes an angle a 53 with the floor, is the spring stretched or compressed? (c) How much energy is stored in the spring when the rod is in equilibrium (See next page for sketch

Explanation / Answer

magnetic force Fb = I*L*B

torque = Fb*L/2

torque = I*L^*B/2 = 6.5*0.2^2*0.34/2 = 0.0442 Nm

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(b)

stretched

torque due to spring = torque due to magnetic force

torque due to spring = k*x*L*sinalpha = 0.0442

4.8*x*0.2*sin53 = 0.0442

x = 0.0576 m

energy stored U = (1/2)*k*x^2 = (1/2)*4.8*0.0576^2 = 0.00796 J

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