A thin rod of length 0.25 m is attached to the floor by a hinge at point P. A ho

Question

A thin rod of length 0.25 m is attached to the floor by a hinge at point P. A horizontal spring with a spring constant k = 4.8 N/m connects the other end of the rod to a vertical wall. The rod is in a uniform magnetic Held of B = 0.44 T directed into the page. There is a current of 6.5 A in the rod in the direction shown. Calculate the torque due to the magnetic force on the rod, for art axis at P. Is the force clockwise or counterclockwise? When the rod is in equilibrium and makes an angle of 53 degree with the floor, is the spring stretched or compressed? By how much? How much energy is stored in the spring when the rod is in equilibrium?

Explanation / Answer

a) Magnetic force, FB = BIL = 0.44 * 6.5 * 0.25 = 0.715 N

Torque, T = FB(L/2) = 0.715 * 0.25 / 2 = 0.09 N-m (clockwise)

b) Since the torque of the magnetic force is clockwise, the spring should be stretched in equilibrium.

Let the spring be stretched by x.

Net torque = 0

=> (kx)L = 0.09

=> x = 0.09/kL = 0.09 / (4.8 * 0.25) = 0.075 m

c) Energy stored, E = kx2/2 = 4.8 * 0.0752 / 2 = 0.014 J

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