Assume a magnetic Field B is perpendicular to a small sectionof wire of lenght L
ID: 1726060 • Letter: A
Question
Assume a magnetic Field B is perpendicular to a small sectionof wire of lenght L, resistivity , andcross-sectional area A, The wire is free to move at speedv within the field perpendicular to it .show the inducedcurrent in the wire given by: I=BvA/ Assume a magnetic Field B is perpendicular to a small sectionof wire of lenght L, resistivity , andcross-sectional area A, The wire is free to move at speedv within the field perpendicular to it .show the inducedcurrent in the wire given by: I=BvA/Explanation / Answer
In this case the Lorentz force drives the current in the twovertical arms of the moving loop downward, so current flows fromthe top disc to the bottom disc. In the conducting rims of thediscs, the Lorentz force is perpendicular to the rim, so no EMF isgenerated in the rims, nor in the horizontal portions of the movingloop.. Current is transmitted from the bottom rim to the top rimthrough the external return loop, which is oriented so theB-field is in its plane. Thus, the Lorentz forcein the return loop is perpendicular to the loop, and no EMF isgenerated in this return loop. Traversing the current path in thedirection opposite to the current flow, work is done against theLorentz force only in the vertical arms of the moving loop,where . F = B q v . Consequently emf is : . = B v L ----------------(1) ; where 'L' is the vertical length of theloop . We have : = IR = I ( L / A ) Since R = L / A . Hence equation (1) becomes ; . I ( L / A ) = B v L . ==> I = B v A / . Hope this helps u!
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