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In the figure shown an infinitely long wire carries a current I_(1) and another uniform rigid wire ACB (bent at C at right angle) of mass m carries a current I_(2).It is hinged at corner C.Find the angular acceleration of this wire ACB just after release.Is the direction of rotation clockwise or anticlockwise.(Assume gravity is absent) [Take ln^(2)=~0.7] |
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Answer» `dtau=(DF).(2L-x)` in clockwise sense. `dF=(I_(2)dx)mu_(0)/(2pi) I_(1)/x` `therefore` Net torqeu on wire `AC` about `C` is `tau_(1)UNDERSET(L)overset(2L)int(mu_(0))/(2pi)(I_(1)I_(2)(2L-x))/xdx=(mu_(0))/(2pi)I_(1)I_(2)L(ln 4-1)` `=(0.4mu_(0)I_(1)I_(2)L)/(2pi)`(clockwise direction) Magnetic field at each point on segment `BC` due to infinite wire is uniform. `therefore` Net torque or wire `BC` about `C` is `tau_(2)=(mu_(0)/(2pi)I_(1)/(2L))I_(2) LxxL/2=mu_(0)/(8pi)I_(1)I_(2)L`(anticlockwise direction) `because tau_(1) gt tau_(2) rArr "net torque" tau=tau_(1)-tau_(2)=mu_(0)/(8pi)(0.6)I_(1)I_(2)L`(clockwise direction) moment of inertia of `L` shaped rod about `C` is `I=(mL^(2))/3` angular ACCELERATION `alpha=tau/I=(0.6mu_(0)I_(1)I_(2)L)/(8pi)xx3/(mL^(2))=(9mu_(0)I_(1)I_(2))/(40pimL)` 
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