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A long straight wire of negligible resistance is bent into V shape, its two arms making an angle `alpha` with each other and placed horizotally in a vertical, homogeneous field B. A rod of total mass m, and resistance r per unit length, is placed on V shaped conductor, at a distance `x_(0)` from its vertex A, and perpendicular to the bisector of angle `alpha` (see fegure) The rod is started off with an initial veloctiy `v_(0)` in the direction of bisector and away from vertex A. The rod is long enough not to fall off the wire during the subsquent motion, and the electrical contact between the two is good although friction between them is negligible. Choose CORRECT statements(s)A. At any position x, let velocity of wire is v then`B^(2)/(r )x_(0)^(2)tan,alpha/(2)+mV_(0)=B^(2)/(r )x^(2)tan,alpha/(2)+mv`B. maximum value of x cordinate of wire is `x_(max)=sqrt(x_(0)^(2)+(mv_(0)r )/(B^(2)tan,(alpha)/(2)))`C. As x increases, v decreaseD. Whatever is the direction of vertical magnetic field, the rod will ultimately stop, |
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Answer» Correct Answer - A::B::C::D `A=x^(2)tan,(alpha)/(2)` `SO phi =Bx^(2)tan,(alpha)/(2)` `E=B(dA)/(dt)` `I=(E)/(R )=(B)/(rl) (dA)/(dt)` Now `I =IBl=B^(2)/(r )(dA)/(dt)` `m(dv)/(dt) = -(B^(2))/(r)(dA)/(dt) implies B^(2)/(r )A_((X)) + mv =` constant |
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