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Question

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 induced current in the bar when the terminal speed has been reached?

Solution

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Answered 11 months ago
Answered 11 months ago
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(c) From equation (*), the induced emf when the terminal speed is reached is:

ε=BLvt cosϕ\begin{gathered} \varepsilon = BLv_t\ \cos{\phi} \end{gathered}

From Ohm’s law, the current in the circuit is given by:

I=εR=BLvt cosϕR\begin{gathered} I = \dfrac{\varepsilon}{R} = \dfrac{BLv_t\ \cos{\phi}}{R} \end{gathered}

Substituting for vtv_t from part (b), we get:

I=BL cosϕR mgR sinϕL2B2cos2ϕI=mg tanϕLB\begin{gathered} I = \dfrac{BL\ \cos{\phi}}{R}\ \dfrac{mgR\ \sin{\phi}}{L^2 B^2 \cos^2{\phi}}\\\\ \therefore \quad \large \boxed{I = \dfrac{mg\ \tan{\phi}}{LB}} \end{gathered}

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