Electromotive force induced in a conducting rod

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The discussion centers on the direction of induced electromotive force (emf) in a conducting rod moving in a magnetic field. The induced current is counterclockwise because the magnetic flux is decreasing as the rod moves upward, which requires the induced magnetic field to oppose the change. If the area were increasing, the induced current would flow clockwise to counteract the increase in magnetic flux. Clarification is sought regarding the positioning of the rod in relation to the resistor in the solution provided. Understanding the conventions of magnetic field direction and motion is crucial for determining the correct current flow.
DottZakapa
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Homework Statement
Two long, parallel and perfectly conducting rails at a distance l = 50 cm from each other are placed in a homogeneous magnetic field with magnetic induction B = 2 mT, which is perpendicular to the rail plane. The rails are conductively connected at one point.
Electrical resistance of the connection is R = 5 Ω. A perfectly conducting rod slides on these rails with a constant speed v = 2 ms-1. Determine:
a) the electromotive force induced in the conducting rod, the value and direction of induced electric current in the closed part of the circuit,
b) the mechanical pulling force needed to keep the rod at the constant speed and mechanical powe
Relevant Equations
electromotive force
Screen Shot 2019-09-12 at 11.35.53.png

i don't understand why in the solution of this exercise, the induced electromotive force / current is counterclockwise.
Shouldn't it be clockwise?
##emf=-\frac {d \Phi_B} {dt} ##
According to the picture the rod is moving upward, the magnetic flux (entering the page) is decreasing due to a reduction of the area, consequently the induced current will flow in such way that the magnetic field produced is in the same direction of the magnetic field that induced it. It would go counterclockwise only if the area was increasing which would imply an increase of magnetic flux, hence the magnetic field produced by the induced current will oppose it.
Isn't it?
 
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I agree with you. The convention is that a cross is into the page (circle with center point is out of the page) and then ##\vec v \times \vec B ## points to the left.

Any indication in the solution (which solution?) that they have the red rod on the top of the picture, having passed the resistor ?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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