Moving bar enclosing a changing magnetic field generates a current

In summary, the conversation discussed the amplitude of ##\vec{B}## and the attempts to find the flux and current. It was noted that the current depends on velocity and a differential equation must be solved. The force acting on the bar was identified as the magnetic force and the differential equation was solved using a method involving separating variables.
  • #1
besebenomo
11
1
Homework Statement
A circuit with resistance R, a moving bar of lenght l and mass m has initial velocity along the x-axis = -v and at t=0 the bar is in position x=2l. The circuit lays in a region with depending on the x-coordinate. Find the current flowing in the circuit.
Relevant Equations
Faraday's law
The amplitude of ##\vec{B}## is given by:
$$B(x) = B_{0} - B_{0} \frac{x}{2l}$$ for ##0 \leq 0 \leq 2l##
Immagine 2022-09-03 113102.png


This was my attempts at finding the flux of B:

$$\Phi(B) = (B_{0} - B_{0} \frac{x}{2l})(2l-x(t))l = B_{0}2l^2-2B_{0}x(t)l+ B_{0}\frac{x(t)^2}{2}$$

and the current: $$ i = -\frac{d \Phi(B)}{dt} \frac{1}{R} = -\frac{B_{0} v(t) (2l -x(t))}{R}$$

Is this approach correct?
 
Physics news on Phys.org
  • #2
besebenomo said:
Is this approach correct?
The current depends on the velocity, so you will have to find that first. I would start with $$I=-\frac{1}{R}\frac{d\Phi}{dt}=-\frac{1}{R}\frac{d(BA)}{dt}$$then use the magnetic force $$\mathbf{F}_M=I\mathbf{L}\times\mathbf{B}$$in Newton's second law. The acceleration is not constant. You will have to solve a differential equation.
besebenomo said:
The circuit lays in a region with depending on the x-coordinate.
Is something missing from this sentence?
 
Last edited:
  • Like
Likes besebenomo
  • #3
kuruman said:
Is something missing from this sentence?
Yes sorry, I meant B is depending on the x-coordinate.
kuruman said:
he acceleration is not constant. You will have to solve a differential equation.
Yes, I know. The problem is that I am not used to solving differential equation when also ##x(t)## is the equation... Is there any way I can rewrite ##i## as a function of only ##v(t)##? That way would be so much easier to solve. Maybe I did something wrong when I computed the flux?
 
  • #4
You need to write Newton's second law and bring the mass in the equation. What is the net force acting on the bar?
 
  • Like
Likes besebenomo
  • #5
kuruman said:
You need to write Newton's second law and bring the mass in the equation. What is the net force acting on the bar?
The force acting on the bar is the magnetic force, in this case: $$\vec{F}_{mag} = i \vec{l}\times \vec{B} = - \frac{-B_{0} v(t) (2l-x(t)) l}{R} (B_{0} - B_{0} \frac{x(t)}{2l})$$
Then I solve the differential equation, if I didn't make mistakes along the way:
$$m \frac{dv(t)}{dt} = - \frac{-B_{0} v(t) (2l-x(t)) l}{R} (B_{0} - B_{0} \frac{x(t)}{2l}) = - \frac{B_{0}^2 v(t) (2l^2+ \frac{1}{2}x(t)^2)}{R}$$
$$\Longrightarrow \frac{dv(t)}{v(t)} = -\frac{B_{0}^2}{mR}2l^2 dt -\frac{B_{0}^2}{2mR} x(t)^2 dt$$

This is where I get stuck, if there wasn't the second term to the right I would just integrate but I don't really know what to do in this case. That's why I was thinking I did make a mistake in the conceptualization of the problem
 
  • #6
A useful method for handling this (some people call it a trick, I call it the chain rule of differentiation) is this
$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$ It allows you to separate variables with ##v(x)## on one side and ##x## on the other. Then you integrate. No variable ##t## anywhere.
 
  • Like
Likes besebenomo
  • #7
kuruman said:
A useful method for handling this (some people call it a trick, I call it the chain rule of differentiation) is this
$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$ It allows you to separate variables with ##v(x)## on one side and ##x## on the other. Then you integrate. No variable ##t## anywhere.
Thanks a lot you've been really helpful!
 
  • Like
Likes berkeman

FAQ: Moving bar enclosing a changing magnetic field generates a current

1. What is a moving bar enclosing a changing magnetic field?

A moving bar enclosing a changing magnetic field refers to a conducting bar or wire that is in motion and surrounded by a magnetic field that is changing in strength or direction.

2. How does a moving bar enclosing a changing magnetic field generate a current?

When a conducting bar or wire moves through a changing magnetic field, it experiences a force known as electromagnetic induction. This force causes the electrons in the bar to move, creating an electric current.

3. What factors affect the amount of current generated by a moving bar enclosing a changing magnetic field?

The amount of current generated by a moving bar enclosing a changing magnetic field depends on the strength of the magnetic field, the speed of the bar, and the angle between the bar's motion and the direction of the magnetic field.

4. What is the relationship between the direction of the current and the direction of the magnetic field?

The direction of the current generated by a moving bar enclosing a changing magnetic field is always perpendicular to both the direction of motion of the bar and the direction of the magnetic field.

5. What are some real-world applications of moving bar enclosing a changing magnetic field generating a current?

Some examples of real-world applications include generators in power plants, electric motors, and transformers. This phenomenon is also utilized in devices such as microphones, speakers, and induction cooktops.

Back
Top