Figure P27.70 shows a 1.0-m, 0.350-kg horizontal metal rod attached to two ropes

Question

Figure P27.70 shows a 1.0-m, 0.350-kg horizontal metal rod attached to two ropes, each of which makes an angle of 30 degree with the horizontal. Each rope then drapes over a pulley and is attached to a vertical rod of mass m. Assume the pulleys have negligible mass. There is a 0.25-T magnetic Field in the region of the rod, parallel to the xy plane, oriented 45 degree from the long axis of the horizontal rod. When the horizontal rod is made to carry a current I, suddenly the rod moves upward until the ropes holding it make an angle of 5.0 degree with the horizontal. What was the value of I?

Explanation / Answer

Initially,

on horizontal rod in vertical direction,

Tsin30 + Tsin30 = 0.350x9.81

T = 3.43 N

and T = mg So, T will be same 3.43 N in final equilibrium.

F = ILBsin@

@ is angle between length vector and B vector.

@ = 45 degrees

magnetic force Fm = I x 1 x 0.25 x sin45 = 0.177I

Now balancing force on rod in vertical direction,

Tsin5 + Tsin5 + 0.177I = Mg

2*3.43*sin5 + 0.177I = 0.350 x 9.81

I = 32.41 A

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