Conducting rod creates a complete circuit

In summary: It looks like it's meant to be moving in the direction of the magnetic field (so the rod is moved leftward) but if it's not, the force will be in the opposite direction to the velocity, so the rod will be slowed down. That would make the problem a bit more complicated. If you assume the rod has a certain mass, you can use ∑F=ma to find the acceleration, a, then use either the SUVAT equations or find a function for v(t) by integrating the acceleration function. I'm not sure how to do this without knowing the mass though. You could try using the work-energy theorem but that involves finding the distance travelled, which is the next part of the question, making this approach circular
  • #1
scoutdjp2012
2
0

Homework Statement


A conducting rod makes contact with rails to complete a circuit. If the rails are 50 cm apart ina uniform magnetic field B = 0.38 [T] directed out of the paper. The total resistance of the circuit is R = 9 [ohms] and is constant.
1) what is the magnitude and direction of EMF induced in the rod when it is moved to the left with a speed of 5 m/s
2) what force is required to keep the rod in motion (without acceleration)
3) how long would it take the rail to come to a stop?
4) what distance would it travel in that time?

Homework Equations


e = Blv
I = e/R
F = ILB

The Attempt at a Solution


1) e = Blv = 0.38 * 0.5 * 5 = .95 V
2) I = e/R = 0.95 / 9 = 0.106
F = 0.106 * .5 * .38 = 0.2N

I need help on number 3 and 4
 
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  • #2
Thread moved from Calculus section. This problem is more of a physics problem than a mathematics problem.

The description of the problem isn't clear to me. Is there a drawing that goes with this?

In the future, please post one or at most two problems per thread, not four or five as you have done in this and another thread you started.
 
  • #3
Since you already calculated the force as a function of the rod's velocity in part (2), you can just apply Newton's second law in the form ##m\dot{v} = F##. This gives a simple differential equation for ##v(t)##. Once you have the solution, find ##T## such that ##v(T) = 0##--that's your answer to (3). (Hint: the answer may be a bit surprising.) To solve (4), just compute the integral ##\int_0^T v(t) dt##.
 
  • #4
scoutdjp2012 said:
3) how long would it take the rail to come to a stop
Rod, not rail, surely?
Is the mass of the rod given?
 

Related to Conducting rod creates a complete circuit

1. How does a conducting rod create a complete circuit?

A conducting rod creates a complete circuit by connecting two ends of a power source, such as a battery, with the rest of the components in the circuit. This allows for the flow of electric current from the power source to the components and back to the source through the conducting rod.

2. What is the purpose of a conducting rod in a circuit?

The purpose of a conducting rod in a circuit is to provide a path for the flow of electric current. It helps to connect all the components in the circuit and complete the circuit, allowing for the transfer of energy from the power source to the desired output.

3. Can any type of material be used as a conducting rod?

No, not all materials can be used as conducting rods. A material must have low resistance to the flow of electric current in order to be an effective conducting rod. Some common materials used as conducting rods include copper, aluminum, and silver.

4. What happens if the conducting rod is broken or disconnected from the circuit?

If the conducting rod is broken or disconnected from the circuit, the flow of electric current will be interrupted and the circuit will be incomplete. This can result in the circuit not functioning properly or not working at all.

5. How can the effectiveness of a conducting rod be improved?

The effectiveness of a conducting rod can be improved by using a material with lower resistance, such as copper, and by ensuring that the rod is securely connected to all components in the circuit. Additionally, reducing the length and increasing the thickness of the rod can also improve its effectiveness.

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