Why Integrate Over Entire Conducting Rail Instead of Bar Area?

[baby_1]

Hello
here is a problem that we want to find induced voltage in two different questions.

1-here is part (a) solution:

question 1:why we get integral in all space of conducting rail instead of bar space?(Area of bar)for example if we enlarge the space and set bar fix again induced voltage increase?
2-here is part (b) solution

1-why we get integral from (.06) to (0) instead of y changes?(0 to .08) because velocity is in y direction?and why form .06 to 0 not 0 to .06?

any help appreciated

 

[collinsmark]

It looks to me as though your textbook's/courswork's style is to first write out everything in general form, even if it can be easily simplified. And only then simplify it.

In other words, even if the integral is a trivial integral, it still writes it out first in the solution, before evaluating the integral.

Hello
here is a problem that we want to find induced voltage in two different questions.

1-here is part (a) solution:

question 1:why we get integral in all space of conducting rail instead of bar space?(Area of bar)for example if we enlarge the space and set bar fix again induced voltage increase?

The integral represents the area of the loop, not the area of the bar itself.

Imagine connecting a voltmeter at the ends of the two rails, where y=0.

Or if it helps, imagine that the bar has some resistance to it, and the rails have zero resistance.

Or imagine that nothing is connected to the ends of the rails, and the bar and the rails all have zero resistance. What is the potential at the end of the rails (from one rail to the other, where y=0)?

I agree, your textbook/coursework could have been more clear where this potential is to be measured. It's a little ambiguous the way it is.

2-here is part (b) solution

1-why we get integral from (.06) to (0) instead of y changes?(0 to .08) because velocity is in y direction?and why form .06 to 0 not 0 to .06?

any help appreciated

The integral from \ell to 0 is just a fancy way of saying that the length of the rod is of length \ell in this case (and it also gives you indication of the polarity of the emf). It's a trivial integral in this case. It seems to me like it's just your textbook's style to do that.

Keeping it integral form can come in handy though, if the bar wasn't straight. Suppose it was some sort of curvy shape. Or suppose the rails were not parallel, and length changed with time. Then you might actually need to integrate something more complicated. Setting things up in integral form is something I suspect your textbook/coursework is doing to prepare you for more complicated problems in the future.