## 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.What is the terminal speed of the bar?

Solution

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

$\begin{gathered} F_g = mg \ \sin{\phi} \end{gathered}$

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:

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

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

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

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

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

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

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

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