# Magnetic Force, Resistance, Faraday's Law

• Eris13
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.
Eris13

## 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.

good problem, i think. not sure why you haven't some help. Likely tomorrow am.

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.

• Introductory Physics Homework Help
Replies
42
Views
3K
• Introductory Physics Homework Help
Replies
11
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
692
• Introductory Physics Homework Help
Replies
2
Views
780
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
23
Views
1K
• Introductory Physics Homework Help
Replies
30
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
989
• Introductory Physics Homework Help
Replies
6
Views
947