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The arm PQ of the rectangular conductor is moved from x=0, outwards. The uniform magnetic field is perpendicular to the plane and extends from x=0 to x=b and is zero for xgtb. Only the arm. PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x=0 to x=2b, and is then moved back to x=0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance. |
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Answer» Solution :Let us first consider the forward motion from `x=0` to `x=2b`. The flux `phi_(B)` linked with the circuit SPQR is, `phi_(B)=B//x""0lexltb` `=B//b""blexlt2b` The induced emf is, `epsilon=-(dphi_(B))/(dt)` `=-B//upsilon""0lexltb` `=0""blexlt2b` When the induced emf is non-zero, the current I is (in magnitude) `I=(B//v)/(r )`  The force required to keep the arm PQ in constant motion is I/B. Its DIRECTION is to the left. In magnitude `F=(B^(2)l^(2)v)/(r)""0lexltb` `=0""blexlt2b` (constant) The Joule heating LOSS is `P_(J)=I^(2)r=(B^(2)l^(2)v^(2))/(r)0lexltb` `=0""blexlt2b` One obtains similar EXPRESSIONS for the INWARD motion from `x=2b` to `x=0`. The sketch of various quantities is displayed in the above figure. |
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