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Shown a metallic square frame of edga a in a vertical plane. A uniform magnetic field B exists in the space in a directon perpendicular to the plane of the figure. Two boys pull the opposite corners of the square ot deform it intoarhombus. They start pulling the corners t t=0and sisplace the corners at a uniform speed u. (a) find the induced emf in the frame at the instant when the angles at these corners reduce to 60^@. (b) find the induced current in the frame at this instatnt if the totalresistance of the frame is R. (c) Find the total charge which flows through a side of the frame by the time the square is deformed intoa straight line. |
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Answer» Solution :(a) The PERPENDICULAR component i.e a sin theta is to be taken which is perpendicular to velocity` `So, l=a sin 30^(@) =(a/2)` `Net 'a' CHARGE =4XX(a/2)=2a` `So, induced emf= Bul=2auB` `(b) Current=(e/R)=(2auB)/(R )` |
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