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Find the mutual inductance between the straight wire and the square loop of figure. |
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Answer» Solution :We know `(d PHI)/(dt)=E=Lxx(di)/(dt)` From the QUESTION (di)/(dt)=(d)/(dt) (i_(0) sin(OMEGA) t=i_(0) (omega)cos (omega)t)` (d phi)/(dt)=E=(mu_(0)ai_(0)(omega)cos(omega)t)/(2 pi)+In (1+(a)/(b))` `Now, E =M.(d phi)/(dt)` `Now, E=M.(dphi)/(dt)` `or (mu_(0)ai_(0)(omega)cos(omega)t)/(2 pi) In (1+(a)/(b))` `Mxxi_(0) (omega)cos (omega)t` `impliesM=(mu_(0)a)/(2 pi) In (1+(a)/(b))`. |
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