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A metal bar with length LL, mass mm, and resistance RR is placed on frictionless metal rails that are inclined at an angle ϕ\phi above the horizontal. The rails have negligible resistance. A uniform magnetic field of magnitude BB is directed downward as shown in Fig. we saw earlier. The bar is released from rest and slides down the rails.What is the terminal speed of the bar?

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(b) The bar starts to slide under its weight component along the rails which is given by:

Fg=mg sinϕ\begin{gathered} F_g = mg \ \sin{\phi} \end{gathered}

Now, the bar is current-carrying conductor moving in a magnetic field, so there is a magnetic force on the bar (equation 3).
Using the right-hand rule for the cross product, the direction of the magnetic force is horizontally backward.
And from equation (3), its magnitude is:

FB=ILB\begin{gathered} \textbf{F}_B = ILB \end{gathered}

From Ohm’s law, I=ε/RI = \varepsilon/R ,thus

FB=LBεR\begin{gathered} \text{F}_B = \dfrac{LB\varepsilon}{R} \end{gathered}

Substituting for ε\varepsilon from equation (*), we get:

FB=L2B2vR cosϕ\begin{gathered} \text{F}_B = \dfrac{L^2B^2v}{R}\ \cos{\phi} \end{gathered}

The component of this force in a direction parallel to the rails is:

FB=L2B2vR cos2ϕ\begin{gathered} F_B = \dfrac{L^2B^2v}{R}\ \cos^2{\phi} \end{gathered}

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