Calculating Induced EMF for a Changing Magnetic Field Loop

[nickmanc86]

1. A rectangular wire loop (length 60 cm, width 40 cm) lies completely within a
perpendicular and uniform magnetic field of magnitude of 0.5 T. If the length of the loop
starts increasing at a rate of 20 mm/s at time t = 0, while the width is decreasing at the
same rate, what is the magnitude of the induced emf at time t = 4.0 s?

Multiple Choice
a) 6.8
b) 5.2
c) 3.6
d) 8.4
e) 10mv

Answer: C




2. \varepsilon = -N \Delta \Phi /\Delta t and \Phi = B*A



3. I took the simple approach using the given time I took 20mm/s * 4.0 s =.08m which I then add to the length and subtract from the width. A(f)=.218, A(i)=.24, t(f)=4 (i)=0 take the \Delta A/\Delta t and multiply by B=.5 and get an answer of 2.8mV

I have some other ideas I tried but all of them also failed to produce the correct answer. Any help would be appreciated. Thanks guys (I apologize if I messed up something it's my first time!)

 

[berkeman]

1. A rectangular wire loop (length 60 cm, width 40 cm) lies completely within a
perpendicular and uniform magnetic field of magnitude of 0.5 T. If the length of the loop
starts increasing at a rate of 20 mm/s at time t = 0, while the width is decreasing at the
same rate, what is the magnitude of the induced emf at time t = 4.0 s?

Multiple Choice
a) 6.8
b) 5.2
c) 3.6
d) 8.4
e) 10mv

Answer: C




2. \varepsilon = -N \Delta \Phi /\Delta t and \Phi = B*A



3. I took the simple approach using the given time I took 20mm/s * 4.0 s =.08m which I then add to the length and subtract from the width. A(f)=.218, A(i)=.24, t(f)=4 (i)=0 take the \Delta A/\Delta t and multiply by B=.5 and get an answer of 2.8mV

I have some other ideas I tried but all of them also failed to produce the correct answer. Any help would be appreciated. Thanks guys (I apologize if I messed up something it's my first time!)

I don't think you can simplify it like that. Just write out the equation for the area A(t), then differentiate it for A'(t). Use that in the equations you show to find the induced EMF at t=4s.

 

[nickmanc86]

Thank you so much that totally led me in the correct direction. I took A'(t)=L'*W+W'*L which I substituted in the equation for Induced EMF then plugged in a time of 4 and came to the correct answer of 3.6mV.

 

[berkeman]

Good job!

Welcome to the PF, BTW.

 

[berkeman]

Also, click "Quote" on your original post to see how I fixed the formating of your Latex...

 

[nickmanc86]

Cool, Thank You

 

[iriver4]

?

How did you get A'(t) ? Isn't the derivative of a constant zero ?

 

[berkeman]

How did you get A'(t) ? Isn't the derivative of a constant zero ?

Welcome to the PF.

Please note that this thread has been dormant for almost a year. The OP has not posted since this thread, so most likely they are not around to answer your question...

 

[iriver4]

Welcome to the PF.

Please note that this thread has been dormant for almost a year. The OP has not posted since this thread, so most likely they are not around to answer your question...

Thank you

And would you be able to further explain how he got the problem ? I understand everything except how A'(t) is not equal to zero.

 

[berkeman]

Thank you

And would you be able to further explain how he got the problem ? I understand everything except how A'(t) is not equal to zero.

The area is not constant. The length and width are different, so there is a change in area, even though 2 sides are expanding while 2 sides are contracting. What do you get as an equation for the area as a function of time?

 

[iriver4]

The area is not constant. The length and width are different, so there is a change in area, even though 2 sides are expanding while 2 sides are contracting. What do you get as an equation for the area as a function of time?

He posted that A'(t) = W'L + L'W
L = 60cm
W = 40cm

A'(t) = 0*60 + 0*40 = 0

Because the derivative of a constant is zero right ?

 

[berkeman]

He posted that A'(t) = W'L + L'W
L = 60cm
W = 40cm

A'(t) = 0*60 + 0*40 = 0

Because the derivative of a constant is zero right ?

I'm not sure about what he posted, but the area is A(t) = L(t)*W(t) = (L+L'*t)(W-W'*t).

If you plug in the numbers from the problem statement, that area should be changing...

 

[iriver4]

I'm not sure about what he posted, but the area is A(t) = L(t)*W(t) = (L+L'*t)(W-W'*t).

If you plug in the numbers from the problem statement, that area should be changing...

What are the values for L' and W' ? I am really sorry but I tried infinite ways and I can't get the correct answer

 

[berkeman]

What are the values for L' and W' ? I am really sorry but I tried infinite ways and I can't get the correct answer

The problem statement says "rate of 20 mm/s"