## Related questions with answers

A metal bar with length $L$, mass $m$, and resistance $R$ is placed on frictionless metal rails that are inclined at an angle $\phi$ above the horizontal. The rails have negligible resistance. A uniform magnetic field of magnitude $B$ is directed downward as shown in Fig. we saw earlier. The bar is released from rest and slides down the rails. Is the direction of the current induced in the bar from $a$ to $b$ or from $b$ to $a$ ?

Solution

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From equation 29.2, the magnetic flux through a finite area $\vec{\textbf{\textit{A}}}$ in a magnetic field $\vec{\textbf{\textit{B}}}$ is given by:

$\begin{gathered} \Phi = \vec{\textbf{\textit{B}}} \cdot \vec{\textbf{\textit{A}}}\tag{1} \end{gathered}$

From equation 29.3, Faraday’s law states that the induced emf in a closed loop equals the negative of the time rate of change of magnitc flux through the loop:

$\begin{gathered} \varepsilon = -\ \dfrac{d\Phi}{dt}\tag{2} \end{gathered}$

From equation 27.19, the magnetic force on a current-carrying conductor with a vector $\vec{\textbf{\textit{l}}}$ along the wire, current $I$ and placed in a magnetic field $\vec{\textbf{\textit{B}}}$ is given by:

$\begin{gathered} \vec{\textbf{\textit{F}}}_B = I\ \vec{\textbf{\textit{l}}} \times \vec{\textbf{\textit{B}}}\tag{3} \end{gathered}$

From equation 25.18, the power delivered to a resistor of resistance $R$ with flowing current $I$ is gvien by:

$\begin{gathered} P = I^2 R\tag{4} \end{gathered}$

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