Magnetic Force, Resistance, Faraday's Law

In summary, a conducting bar of length 0.218 m and mass 0.08 kg slides along conducting rails with an initial speed of 42.0 m/s. There is a resistor with a value of 20.0 Ohms at one end and a uniform magnetic field of 2.2 T perpendicular to the plane of the rod and rails. The equation for velocity is v = e^[(B^2)(L^2)t/(Rm)+ln(v0)], and to find the distance the bar slides before coming to rest, the equation for velocity can be integrated and solved for t and then used to solve for position.
  • #1
Eris13
1
0

Homework Statement


A conducting bar of length L = .218 m and mass M = .08 kg lies across a pair of conducting rails. The contact friction between the bar and the rails is negligible, but there is a resistor at one end with a value R = 20.0 Ohms. Initially the rod is given an initial speed of v0 = 42.0 meters per second. There is a uniform magnetic field perpendicular to the plane containing the rod and rails of magnitude B = 2.2 T.
What is the speed of the rod at time t = 15.303 s?
How far does the rod slide before coming to rest?

Homework Equations


F=ILB
Ohm’s law: I=V/R
Faraday’s law: V = dΦ/dt = B(dA/dt)

The Attempt at a Solution


dA = Ldx, giving V=BL(dx/dt)=BLv

F=BLv/R(LB)=(B^2)(L^2)v/R
F=ma=m(dv/dt)
(B^2)(L^2)v/R= m(dv/dt)

If I rearrange this I get:
(B^2)(L^2)/R dt= m/v dv

Taking the integrals of both sides gives:
(B^2)(L^2)t/R (from t=0 to t=t) = m*ln(v) (from v0 to v)
So…
(B^2)(L^2)t/R = m*ln(v) – m*ln(v0)

Solve for v:
v= e^[(B^2)(L^2)t/(Rm)+ln(v0)]

When I put in the numbers I’m not getting the right answer. Is what I did right? Is there something that I’m doing wrong?

Regarding the question on how far the bar goes...I'm guessing that once I get the right equation for velocity, I would set v=0 and solve for t. Then integrate v to find and equation for position and use the t to solve for it. Is this right?

Thanks in advance for any help.
 
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  • #2
good problem, i think. not sure why you haven't some help. Likely tomorrow am.
 
  • #3


Your approach to solving for the velocity is correct. However, there are a few errors in your calculations.

Firstly, in your equation for the force, F=BLv/R(LB), the L's should cancel out, leaving you with F=B^2Lv/R.

Secondly, when you take the integral of both sides, you should have (B^2)(L^2)/R * t = m*ln(v) - m*ln(v0).

Thirdly, when solving for v, you need to divide both sides by m, giving you v = v0 * e^[(B^2)(L^2)t/(Rm)].

Using these corrections, you should be able to get the correct answer for the velocity at t=15.303s.

As for the distance the bar slides, your approach is correct. Once you have the equation for velocity as a function of time, you can set v=0 and solve for t. Then, you can integrate the velocity equation to get the position equation and plug in the value of t to solve for the distance.

Hope this helps!
 

1. What is magnetic force?

Magnetic force is a force that is caused by the interaction between two magnetic objects or a magnetic field and a magnetic object. It is responsible for the attraction or repulsion between objects with magnetic properties.

2. What is resistance in the context of electromagnetism?

In electromagnetism, resistance is a measure of how much a material or object resists the flow of electric current. It is influenced by factors such as the material's composition, temperature, and physical dimensions.

3. What is Faraday's Law?

Faraday's Law is a fundamental law of electromagnetism that describes the relationship between a changing magnetic field and the induction of an electromotive force (EMF) in a conductor. It states that the induced EMF is directly proportional to the rate of change of the magnetic field.

4. How does magnetic force affect electric current?

Magnetic force can affect electric current by exerting a force on the charged particles within the current. This can cause the current to change direction or alter its path, depending on the orientation of the magnetic field.

5. What are some real-life applications of Faraday's Law?

Faraday's Law has numerous real-life applications, including the operation of electric generators, motors, and transformers. It is also used in devices such as microphones, speakers, and magnetic levitation trains. Additionally, it plays a crucial role in the production and distribution of electricity.

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