A long metal bar of 30,cm length is aligned a | vedclass

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(A) The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is given by the formula $e = B \cdot l \cdot v$,where $

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$+12$

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(A) The motional electromotive force $(EMF)$ induced in a conductor moving in a magnetic field is given by the formula $e = B \cdot l \cdot v$,where $B$ is the magnetic field strength,$l$ is the length of the conductor,and $v$ is the velocity.Given values are $B = 4.0\,T$,$l = 30\,cm = 0.3\,m$,and $v = 10\,ms^{-1}$.Substituting these values into the formula:$e = 4.0 	imes 0.3 	imes 10 = 12\,V$.According to the right-hand rule for motional $EMF$,the force on the free electrons in the bar is directed towards the south end $(F = q(\vec{v} 	imes \vec{B}))$. Since electrons accumulate at the south end,the north end becomes positively charged relative to the south end.Given the potential at the south end is $0\,V$,the potential at the north end is $V_N - V_S = 12\,V$,so $V_N = 12\,V$.

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