How Does the Angle of Incline Affect Current in a Sliding Rod Experiment?

[kpangrace]

The two conducting rails in the drawing are tilted upwards so they make an angle of 30.0° with respect to the ground. The vertical magnetic field has a magnitude of 0.048 T. The 0.22 kg aluminum rod (length = 1.6 m) slides without friction down the rails at a constant velocity. How much current flows through the bar?

ok so I'm really completely lost here, but let me give you guys what I'm thinking, see if you could lead me in the right direction

I'm expecting that you don't use that 30 degrees as part of F= ILB sin(theta), but rather you use it to find force with F=ma right? if that's the case, how would you find the acceleration? would it just be 9.8 for gravity, or would it be something else because you're not going straight down... but then i have a problem with velocity being constant, which means there would be no acceleration... but there has to be gravity...

(i'm sorry if i seem all over the place just confused)

so here's what I've done and it didn't work

F=ma

F= .22 * 9.8sin30
F= 1.078
F= I L B sin(theta)
1.078= I 1.6*.048 sin 90
I =14.036

thats not the right answer, but to me that seems the only logical way to go about answering this problem... any info would help thank you!

 

[kpangrace]

Nevermind i figured it out, its sin(60), not sine of 30 or 90... you can kill this thread if you wish!

 

[quicknote]

I would first clarify some of the information provided in the problem. The 30-degree angle mentioned in the problem is not relevant to the calculation of the current flowing through the bar. It is only mentioned to indicate the orientation of the conducting rails.

To find the current flowing through the bar, we can use the equation F = ILBsin(theta). In this case, the force (F) is equal to the weight of the bar, which is given by mg (mass x gravity). The angle (theta) in this case is 0 degrees, as the bar is sliding straight down the rails. The magnetic field (B) is given as 0.048 T. So the equation becomes:

mg = ILBsin(0)

Solving for I, we get:

I = mg / (LB)

Substituting the given values, we get:

I = (0.22 kg) * (9.8 m/s^2) / (1.6 m * 0.048 T) = 3.426 A

Therefore, the current flowing through the bar is 3.426 A. This current will remain constant as the bar slides down the rails at a constant velocity, as there is no force acting to change its speed.

I hope this helps clarify the problem and the approach to solving it. Let me know if you have any further questions.