The figure below shows a bar of mass m that can slide without friction

The figure below shows a bar of mass m = 0.200 kg that can slide without friction on a pair of rails separated by a distance l = 1.20 m and located on an inclined plane that makes an angle ϴ = 25.0° with respect to the ground. The resistance of the resistor is R = 1.00 Ω, and a uniform magnetic field of magnitude B = 0.500 T is directed downward, perpendicular to the ground, over the entire region through which the bar moves. With what constant speed v does the bar slide along the rails?

[connorw]

you sir, have singlehandedly guided me through my physics class. God bless you

[erickrobles]

To add on to what you said about using the right-hand rule: You can point your fingers towards the direction the bar is sliding (velocity vector) and curl your fingers to the direction of the magnetic field. The thumb will point in the direction of Induced current; which flows from high to low potential (CCW)

[rodolfomagana]

Super Awesome and clear explanation, thank you, and great teaching

[classicDropwig]

the things i do for 1 point on homework in a class outside of my major

[bentpc]

Why don't you just consider sum of forces parallel to slope? Magnetic force cos theta = mg sin theta

[aonsi]

Even though the obtained final formula is correct however the procedure followed to obtain the final answer from minute 6:52 is not quite correct, where it is given in the problem that the bar slides without friction, as the surface is friction-less this implies that there is no normal force n, hence equating the x and y component of the force is not correct. It is simply enough to equate the forces along which the bar slides to have the same equations, where we know that the bar would reach a terminal velocity ``a non-accelerating velocity''. Hence applying equilibrium condition the magnetic force makes an angle theta with the direction of inclination and hence Fm cos theta = mg sin theta and from there you can have the same final value. Yours