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Magnetic flux through a square loop (rotating loop WITH non-constant magnetic field)

  • #1

Homework Statement


A square loop with sides l is centered on the origin and fixed in the center so it is free to rotate around the x-axis. A magnetic field is changing with time B=B_0(1-exp(-a*t)). I need to find a differential equation to describe the motion of the rotating loop


Homework Equations



Faraday's law: emf=-d/dt(integral(B*dA))

The Attempt at a Solution


I'm fairly certain that I need to use this to find it's motion, but when I do the math, I get EMF=-dB/dt*dA/dt, and differentiating the magnetic field gives EMF=-dA/dt*a*exp(-a*t). But I don't see how this describes the motion?
Am I doing something wrong?

Thanks in advance!
 

Answers and Replies

  • #2


Or maybe I did the flux integral wrong?
 
  • #3


Is this the entire problem? In which direction is the magnetic field supposed to be if not? I initially assumed in the y-direction, but just to be sure...
 
  • #4


Oh! oops I totally forgot, the magnetic field is in the z-hat direction, and the loop rotates around the x-axis, so when theta=pi/2, the loop is parallel to the field, making its area vector perpendicular, and vice versa for pi
 
  • #5
1
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does the area vector consistently change due to intermittent B-field? Sorry just inquiring.
 
  • #6


Yes, the flux through the loop creates a torque on the two ends of the loop making it rotate, thus making the Area vector change
 
  • #7


Alright, so try writing the angle between the area vector and B-field in terms of the variables of the problem.

Next, try expressing the equation for emf with this equation for theta. Remember that you are trying to find the time derivative of the magnetic flux, and that means you want to rewrite the flux so that it shows an explicit time dependence. You've already got the time dependence of the B-field, but what about the area component of the flux? How does the area grow or decrease according to how fast the loop is revolving about the x-axis?
 
  • #8


Okay, so doing that I got flux=B*A*cos(theta)=B*A*cos((omega)t), and EMF is -d/dt(flux) so I get EMF=(dB/dt)*A*(omega)*sin((omega)t), then differentiating B gives EMF=a*omega*B_0*l^2*sin((omega)t)*exp(-a*t). But that wouldn't make sense, would it? Because I just used l^2 as A, which should be dA/dt?
 
  • #9


Hey Khrekek1992, I'm actually a little skeptical of where this problem is going... Are you sure you've provided all the details of the problem? I will tell you what I mean when we get there, but for now I'll focus on the flux.

Okay, so doing that I got flux=B*A*cos(theta)=B*A*cos((omega)t), and EMF is -d/dt(flux) so I get EMF=(dB/dt)*A*(omega)*sin((omega)t), then differentiating B gives EMF=a*omega*B_0*l^2*sin((omega)t)*exp(-a*t). But that wouldn't make sense, would it? Because I just used l^2 as A, which should be dA/dt?
The flux is [itex]\Phi=BA[/itex]. Here, the area is [itex]A=L^{2}cos(\omega t)[/itex]. [itex]L^{2}[/itex] is the amplitude of the area corresponding to [itex]cos(\omega t)=1[/itex] (integer values of ∏). So in taking the time derivative of the area, you get [itex]\dot{A}=-L^{2}\omega sin(\omega t)[/itex] just as you showed.

and EMF is -d/dt(flux) so I get EMF=(dB/dt)*A*(omega)*sin((omega)t)
You're almost there, but there is a problem with your time derivative of the flux. Recall the product rule of differentiation. You can't just multiply the time derivatives of B and A.
 
  • #10


Ohhh okay I understand what I was doing wrong, and another thing I'm doing wrong I think, is that the Area isn't actually changing, is it? So dA/dt=0, isnt it the angle that is changing? d(theta)/dt? So (with product rule this time) -d/dt(BA)=-B*cos(theta)dA/dt-Acos(theta)dB/dt+BAsin(theta)(dtheta/dt). With the first term canceling out because the actual area is not changing. So now I have dB/dt and d(theta)/dt. dB/dt can be easily solved for and I got EMF=-L^2*B_0*a*cos(theta)*exp(-a*t)+L^2*B_0*sin(theta)(1-exp(-a*t)*d(theta)/dt. Where dtheta/dt is the angular velocity of the loop? Is this more like what I should be getting?
 
  • #11


the Area isn't actually changing, is it? So dA/dt=0, isnt it the angle that is changing?
The area of the loop is not changing, you are correct about that. What is changing, is the angle between the B-field and the A vector. That change is given by the cosine term, which comes from the dot product of B and A in the flux equation.

d(theta)/dt? So (with product rule this time) -d/dt(BA)=-B*cos(theta)dA/dt-Acos(theta)dB/dt+BAsin(theta)(dtheta/dt). With the first term canceling out because the actual area is not changing.
Not quite.

[itex]\dot{\Phi}=\frac{d}{dt}B\cdot A=\frac{d}{dt}AB cos(ωt)=L^{2}\frac{d}{dt}B cos(ωt)[/itex]

Since A here is constant, and simply equal to[itex]L^{2}[/itex], we take it outside the derivative. So now you're just taking the time derivative of B and the cosine term, on which you need to use the product rule.
 
  • #12


Oh! I get it, so

[itex]\dot{\Phi}=\frac{d}{dt}B\cdot A=\frac{d}{dt}AB cos(ωt)=L^{2}\frac{d}{dt}B cos(ωt)[/itex]
and using the product rule - I get it to equal L^2*B_0[a*exp(-a*t)*cos(w*t)-(a-exp(-a*t)*w*sin(w*t)], right? So in this case, w (omega) is the angular velocity of the system?
 
  • #13


[itex]\dot{\Phi}=\frac{d}{dt}B\cdot A=\frac{d}{dt}AB cos(ωt)=L^{2}\frac{d}{dt}B cos(ωt)[/itex]

[itex]L^{2}\frac{d}{dt}B cos(ωt)=L^{2}B_{0}\frac{d}{dt}(1-e^{-at}) cos(ωt)[/itex]

and using the product rule - I get it to equal L^2*B_0[a*exp(-a*t)*cos(w*t)-(a-exp(-a*t)*w*sin(w*t)]
You should get 3 terms from doing this derivative.

-(a-exp(-a*t)*w*sin(w*t)
I'm not sure what you intended to write in this part of the equation; there's an open parenthesis.
 
  • #14


Ah!! I got it now :) Thank you so much for your help!!! I appreciate it soo much!! :)
 

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