Question

Guys anyone :/

Answer 1

Below is the diagram http://screencast.com/t/3kmKno7na

Question 1 --- http://screencast.com/t/y63LYrmyR


@Vincent-Lyon.Fr

Answer 2

@AravindG @robtobey

Answer 3

hall voltage Vh=[Rh*(I*B)]/t
so vh is inversely prop to time so i think graph should be a rectangular hyperbola

Answer 4

well that's on the qustn

Answer 5

so it's a step function but how?

Answer 6

hmmm it cn be either then

Answer 7

cuz it say EMF Vh means it shuld be constant step up function

Answer 8

@Vincent-Lyon.Fr

Answer 9

This is what I would draw, taking into account the effect when entering and exiting the magnetic zone where B is uniform.

Answer 10

yes thats what i got too

Answer 11

yes thats what i got to because VH is constant due to uniform magnetic flux?

Answer 12

would u mind helping me with another one, i dont get it :/

Answer 13

http://screencast.com/t/lL8JdtiPTi

Answer 14

could u help me on how to approach that question

Answer 15

i know that EMF induced = Rate of change in magnetic flux, so due to the motion of the coil, the flux is changing, at a constant rate ?

Answer 16

?

Answer 17

@AravindG

Answer 18

@ganeshie8 @theEric

Answer 19

@waterineyes @radar

Answer 20

\(e=-\dfrac{d\Phi}{dt}\)

now, you are told that the orientation (and the shape/area) of the coil does not vary.

So, what is the varying factor in the magnetic flux \(\Phi\) ?
Once you have answered that, going from your first curve to the second one is not difficult.

Answer 21

hmmm im confused :/

Answer 22

@theEric

Answer 23

Since \(\Phi = NBS\cos \theta\), then B is the quantity responsible for flux variation.

Answer 24

okay so, since b is constant, there wouldnt be any change in magnetic flux, except at the start and the end?