Magnetic Flux of a Wire Sliding on L-Shaped Rail

[assaftolko]

A conducting wire slides with constant speed of 1 m/s on a bent L shaped metalic rail. The angke between the wire and the rail is 45 deg and it remains constant during the whole motion of the wire. Perpendicular to the plane of the rail acts a magnetic field B=0.2 T in the X direction (Into the paper). At the beginning of the wire's motion - the point A, which is at the center of the wire, is 1m away from point C.

What's the magnetic flux as function of time?

Why can't I say that the wire moves in the positive x direction (to the right) according to:
x(t)=x0+vo,x*t, and the wire moves in the positive y direction (upwards) according to:
y(t)=y0+vo,y*t

where x0=y0= AC/sin45 and vo,x=vcos45, vo,y=vsin45?

 

[Saitama]

A conducting wire slides with constant speed of 1 m/s on a bent L shaped metalic rail. The angke between the wire and the rail is 45 deg and it remains constant during the whole motion of the wire. Perpendicular to the plane of the rail acts a magnetic field B=0.2 T in the X direction (Into the paper). At the beginning of the wire's motion - the point A, which is at the center of the wire, is 1m away from point C.

What's the magnetic flux as function of time?

Why can't I say that the wire moves in the positive x direction (to the right) according to:
x(t)=x0+vo,x*t, and the wire moves in the positive y direction (upwards) according to:
y(t)=y0+vo,y*t

where x0=y0= AC/sin45 and vo,x=vcos45, vo,y=vsin45?

If you apply the equations you state in your post to, let's say, the centre of sliding wire, you would be calculating the coordinates or the distance moved in the two directions, not the "actual" distance moved.

Try to apply it on the centre of wire, it will become obvious.

 

[assaftolko]

I don't get why this isn't the actual distance.. And how then can i find this actual distance moved in the two direction?

 

[Saitama]

I don't get why this isn't the actual distance.. And how then can i find this actual distance moved in the two direction?

Did you apply your equations to the centre of wire?

By actual distance, I meant the distance traveled by centre along the given direction of velocity.

 

[assaftolko]

Did you apply your equations to the centre of wire?

By actual distance, I meant the distance traveled by centre along the given direction of velocity.

I don't understand how what you are saying explains why I can't do the analysis I've done. Maybe I didn't explain well enough what I tried to do here:
I said that the two points on the wire that, for every moment, lay on the x and y axis, and so - make up the two legs of the right triangle (one leg lays on the x-axis from the "origin" C to the right lower "end" of the wire at each moment, and the other leg lays on the y-axis from C to the left upward "end" of the wire at each moment. I write "end" because these 2 points on the wire aren't actually at its two ends, but they are "ends" in the sense that they lie on the 2 axis) move according to constant velocity kinematics, where x0 and y0 are the positions of these two points at t=0, which means that with respect to the origin C we get x0=y0=AC/sin45 or AC/cos 45 it doesn't matter of course... AC is of course 1 m since we deal with t=0.

Hence, I get that X(t), which represents the position of the right lower point of the wire on the x-axis at each moment (and hence - the length of this leg of the triangle at every moment), is: x(t) = AC/sin45 + vcos45*t, where vcos45 is the velocity in the x direction.
y(t), which represents the position of the left upper point of the wire on the y-axis at each moment (and hence - the length of this leg of the traingle at every moment), is:
y(t) = AC/sin45 +vsin45*t, where vsin45 is the velocity in the y direction.

I know that I'm mistaken, I just don't get why - how could I know up front this analysis is wrong?

 

[Saitama]

I said that the two points on the wire that, for every moment, lay on the x and y axis, and so - make up the two legs of the right triangle (one leg lays on the x-axis from the "origin" C to the right lower "end" of the wire at each moment, and the other leg lays on the y-axis from C to the left upward "end" of the wire at each moment. I write "end" because these 2 points on the wire aren't actually at its two ends, but they are "ends" in the sense that they lie on the 2 axis) move according to constant velocity kinematics, where x0 and y0 are the positions of these two points at t=0, which means that with respect to the origin C we get x0=y0=AC/sin45 or AC/cos 45 it doesn't matter of course... AC is of course 1 m since we deal with t=0.

Those two points do not always remain on the x and y axes. The point on the x-axis at t=0 does not remain on x-axis after some time. It now moves in the xy plane, not on the x-axis.

 

[assaftolko]

Those two points do not always remain on the x and y axes. The point on the x-axis at t=0 does not remain on x-axis after some time. It now moves in the xy plane, not on the x-axis.

Yep you're right... thanks man!