Determining Velocity from a Freely Falling Bar in a Magnetic Field

[guitarman]

Homework Statement

A metal bar of length 1.3 m and mass 0.04 kg slides with negligible friction but with good electrical contact down between two vertical metal posts. After speeding up initially, the bar falls at a constant speed (zero acceleration). The falling bar and the vertical metal posts have posts have negligible electrical resistance, but the bottom rod is a resistor with resistance 26 ohms. Throughout the entire region there is a uniform magnetic field of magnitude 1.5 tesla coming straight out of the page.


Find the constant speed at which the bar falls.
v = ____ m/s

Homework Equations

Fgrav=mg
Fmag=I*(deltaL X B)
deltaV=I*R

The Attempt at a Solution

I know that Fgrav = 0.04* 9.8 = 0.392, and this must be equal and opposite to Fmag. The logic behind this is that the bar eventually falls at a constant speed, meaning there is no net force acting upon it. Therefore, I do
0.392 = I *(1.3 * 1.5)
I = 0.2
Using the current, I can find the change in voltage by 0.2 * 26 = 5.2 V.
This is where I am getting confused, is there any way to readily convert from volts to the velocity? I feel that I have all the necessary information, but could anyone let me know if I am wrong? Thanks!

 

[anathema]

Your approach is correct so far. To find the constant speed at which the bar falls, you need to use the equation V = I*R, where V is the voltage, I is the current, and R is the resistance. In this case, you have already calculated the current to be 0.2 amps and the resistance to be 26 ohms. Therefore, the voltage will be 5.2 volts.

To find the velocity, you can use the equation Fmag = B* I * L, where Fmag is the magnetic force, B is the magnetic field, I is the current, and L is the length of the bar. In this case, you know all the values except for the velocity, so you can rearrange the equation to solve for v.

Fmag = B* I * L
v = Fmag / (B*I)
v = (0.392) / (1.5 * 0.2)
v = 1.307 m/s

Therefore, the constant speed at which the bar falls is 1.307 m/s. I hope this helps and let me know if you have any further questions. Good luck with your studies!

 

[kerimek]

Your approach is correct so far. To find the velocity, you can use Ohm's law which states that V=IR, where V is voltage, I is current, and R is resistance. In this case, the voltage (deltaV) is equal to the change in voltage (5.2 V) and the resistance is given as 26 ohms. So, we can rearrange the equation to solve for the current (I) which is equal to 0.2 A.

Now, to find the velocity, we can use the equation Fmag=I*(deltaL X B) where Fmag is the magnetic force, I is the current, deltaL is the length of the bar (1.3 m), and B is the magnetic field (1.5 T). We can rearrange this equation to solve for the velocity (v) which is equal to Fmag/(I*B).

Plugging in our values, we get v= 0.392/(0.2*1.5) = 1.307 m/s. Therefore, the constant speed at which the bar falls is 1.307 m/s.