InterviewSolution
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InterviewSolution
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A squareframe carryinga current I = 0.90 A is located in the same plane as a long straight wire carrrying a current I_0 = 5.0 A. The frome side has a length a = 8.0 cm. The axis of the frame passing through the mid-points of the opposite sides is parallel to the wire and is separated from it by the distance which is eta = 15 times greater than the side of the frame. Find : |
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Answer» Solution :(a) Force of attraction between parallel currents `F_(1) = (mu_0)/(4pi) ( 2I I_0)/((2eta -1 )(a)/(2)` Similar force of repulsion between antiparallel currents `F_(2) = (mu_0)/(4pi) (4I I_0)/((2 eta + 1)a)` Net force of repulsion between the square frame and the long straight wire `F = F_1 - F_2 = (2 mu_0)/(pi a) (I I_0)/(2 eta^2 -1)` Putting values,`F = 0.05 mu N.` b) Work perfomed in turning the frame through `180^@` `W = int_0^(pi) tau d theta,` But `d tau = bar(dM) XX bar (B)(r)` `dM = IdA - = 2a dr n B(r) = (mu_0)/(4 pi) (2I_0)/(r) n`. n = unit normal to the frame into the paper `therefore d tau = dM B sin theta` `theta` = angle between B(r) and dA during the rotation process. `thereforetau = int d tau (mu_0)/(4 pi). 2I I_(0) sin theta int_( eta a - a//2)^(eta a+a//2) (dr//r)` `= (mu_0)/(4pi) 2 I I_0a sin theta log_e (2eta + 1)/(2 eta-1) therefore` Workdone ` W = int_(0)^(pi) taud theta = (mu_0)/(4pi)2 I I_0 a LOG ""(2eta +1)/(2 eta - 1) sin theta d theta` ` = (mu_0)/(pi) I I_0 a log ""(2 eta + 1)/(2 eta -1) = 0.1 44 mu J` |
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