Did I get it right by coincidence?

[flyingpig]

Homework Statement

OKay, PF again erased my hard 30mins work of LaTeX and wouldn't let me backspace. So I apologize if I miss any crucial detail

Two parallel rails with negligible resistance are 10.0 cm apart and are connected by a 5.00 resistor. The circuit also contains two metal rods having resistances of R = 10.0Ω and 15.0 Ω sliding along the rails. The rods are pulled away from the resistor at constant speeds 4.00 m/s and 2.00m/s, respectively. A uniform magnetic field of magnitude 0.0100 T is applied perpendicular to the plane of the rails. Determine the magnitude and direction of the current in the 5.00 Ωresistor

The Attempt at a Solution

http://img219.imageshack.us/img219/7499/p3131alt.th.gif

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So I did my loop

Left Loop $$-I_1 R_1 - Blv_1 - I_1 R_3 + I_2 R_3 = 0$$

Right Loop $$-Blv_2 - I_2 R_2 - I_2 R_3 + I_1 R_3 = 0$$

Add some numbers

Left Loop $$-15I_1 + 5_2 = 4Bl$$

Right Loop $$5I_1 - 20I_2 = 2Bl$$

$$\begin{bmatrix} -15 & 5& 4\\ 5&-20 & 2 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0& \frac{-18}{55}\\ 0& 1 & \frac{-10}{55} \end{bmatrix}$$

So I₁ = 18/55A and I₂ = 10/55A

So the current down the middle resistor must be 8/55A which is 0.145A

Here is the book solution

http://img853.imageshack.us/img853/9262/81989521.th.gif

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I didn't even use the 10cm...

 

[flyingpig]

TeX fixed

 

[SammyS]

You didn't use the B field either AND your answer is 1000 times what the books solution is.

0.145A=145 mA. not μA.

Also, you didn't mention the direction of the induced EMF.

 

[flyingpig]

I didn't mention or the question? Probably me, the book is better than me lol. Oh yeah the B-field too. But how did they get 145μA?

Sammy, what is your time zone? In case I need help and I move during the Summer lol

 

[SammyS]

(Central zone. I often stay up way too late. fp, what's your Time Zone?)

B=0.010 T, ℓ (or d) = 0.10 meters, so Bℓ = 0.0010 = 1/1000, and your answer is 1000 times the correct answer.

 

[flyingpig]

I live in Canada right now, so Pacific? It's like 12:23pm right now.

Also I thought l is the length of the rod and the "l" (they have d) is the distance travelled

 

[SammyS]

They plug in 0.100m for d. d is the distance between the rails.

 

[flyingpig]

But that isn't right

$$\varepsilon = Blx$$

x is the distance traveled and l is the length of the rod. The "d" in the problem is said to be the distance between them, but in Blv, the v is dx/dt which already took care of the distance travelled, but l is still untouched

 

[SammyS]

You have $$Blv\,,\ ($$and you threw away the $$Bl\,).$$

The posted solution has $$B\,d\,v\,.$$ Their d corresponds to your ℓ.
It's very clearly the case.

BTW: Your equation, $$\varepsilon = Blx$$ is in error.

$$\mathcal{E}=-\,\frac{d\Phi}{dt}=-B\ell\frac{dx}{dt}=-B\ell v$$

 

[flyingpig]

I didn't throw it away! I just forgot to use B = 0.001T!

$$-I_1 R_1 - Blv_1 - I_1 R_3 + I_2 R_3 = 0$$

I know it corresponds to my "l", but the "l" given is wrong, the 10cm is the distance between the wires, l is supposed to be the length of the rod according to my textbook.

For $$\varepsilon = Blx$$ I thought we only put the minus sign when we want the direction as in this https://www.physicsforums.com/showthread.php?t=486781?

Sigh, I think I have to realize that I am just not cut out for physics, I don't understand why E&M is so difficult for me lol.

 

[SammyS]

The minus sign isn't the problem.

 

[flyingpig]

No, but the d and l is, I don't understand why they think the distance between the rods is the length of the rod themselves!?

 

[SammyS]

Homework Statement

...

Two parallel rails with negligible resistance are 10.0 cm apart and are connected by a 5.00 resistor. The circuit also contains two metal rods having resistances of R = 10.0Ω and 15.0 Ω sliding along the rails. The rods are pulled away from the resistor at constant speeds 4.00 m/s and 2.00m/s, respectively. A uniform magnetic field of magnitude 0.0100 T is applied perpendicular to the plane of the rails. Determine the magnitude and direction of the current in the 5.00 Ωresistor
...

No, but the d and l is, I don't understand why they think the distance between the rods is the length of the rod themselves!?

In the problem as you posted it, no variable is assigned to the length of the rods. (BTW: to get technical, it's the distance between the rails that's important. Any portion of the rod extending past the rails doesn't carry current.)

It's clear that your solution uses l as the effective length of each rod. It's equally clear (to me at least) that the book solution uses d as the effective length of each rod.

 

[flyingpig]

Two parallel rails with negligible resistance are 10.0 cm apart

Doesn't that mean the rods' distance is 10.0cm apart? Not the length of the each rod

 

[SammyS]

The two rails are stationary & are 10.0 cm apart. The rods sit on the rails, so each rod is 10.0 cm long. This is the d in the book solution.

 

[flyingpig]

The two rails are stationary & are 10.0 cm apart.

Yup

The rods sit on the rails, so each rod is 10.0 cm long.

How did you deduct that? Of course without the solution from the book?

 

[SammyS]

How did you deduct that? Of course without the solution from the book?

I suppose there should have been a figure showing the rods to be perpendicular to the rails, thus the portion of each rod that carries current is equal to the distance from one rail to the other.