Induced Voltage of a Magnetic Field passing through a loop

AI Thread Summary
The discussion focuses on calculating the induced voltage in a circular loop due to a time-varying magnetic field. The magnetic field is defined by the equation B = (1.25t² - 0.500t + 4.00)T, and the loop has a radius of 14 cm, with the magnetic field making a 65° angle with the plane of the loop. The participants clarify that the area and angle are constants, allowing them to treat A*cos(Θ) as constant in the flux equation. After correcting the derivative of the magnetic field and considering the angle properly, the induced voltage is calculated to be 251 mV, with the induced current expected to flow clockwise due to a decrease in inward flux. The discussion emphasizes the importance of correctly interpreting the angle and the rate of change of the magnetic field.
abstrakt!
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Homework Statement



A magnetic field passes through a circular loop of radius 14 cm and makes an angle of 65° with respect to the plane of the loop. The magnitude of the field is given by the equation:

B = (1.25t2 - .500t + 4.00)T.

a) Determine the voltage induced in the loop when t = 2.00 s.
b) What is the direction of the current induced in the loop?

Homework Equations


ξ = -\frac{d∅}{dt} where ξ is the induced voltage and ∅ is the magnetic flux.

Acircle = πr2

The Attempt at a Solution



My thinking is that area of the circle is not changing with respect to time, however, the magnetic flux is as it is a function of time. I have attempted various avenues (including simply plugging and chugging into BAcosΘ) but have not figured out how to solve this.

I wanted to use the rate of change relationship and attempted to derive an equation that I could use to solve this, however, I am stuck.

I possibly made an error in my derivative but I also need confidence whether I am on the right track or not.

-\frac{d∅}{dt} = -\frac{d}{dt}[BA cosΘ]

Using product rule, I get:

ξ = -[(\frac{dB}{dt}A cosΘ) + B(\frac{dA}{dt}cosΘ-AsinΘ)]

\frac{dB}{dt} = (2.5t +.500)T
\frac{dA}{dt} = 2πr

Plugging in:

-[(2.5t+.500(π(.14)2cos 65°) + (1.25t2 +.500t +4.00) (2π(.14) cos 65° - π(.14)2 sin 65°)] when t = 2.00 s

This looks incorrect and since the area is not changing I am not sure that I should be taking the derivative of the area.

The answer is supposedly 251 mV but I cannot get close to this and I am honestly quite stuck at this point.

I am either over-complicating this, making some careless mistakes or simply not understanding the nature of the problem.

Any help would be appreciated! Thank you for your time and assistance!
 
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As you noted, the area ##A## and the angle ##\Theta## are constants. So, you can treat ##A\cos \Theta## as a constant.

Did you get a sign wrong when calculating ##dB/dt##?
 
Also note that the angle given is the angle measured relative to the plane of the loop.
 
So it is correct to factor those out as constants and take the derivative of the magnetic field B? I think I made a typo when entering in latex but I will check my paper to see if I made the same error when I did it by hand. I will try this and see what I get. Thank you very much for such a quick reply!
 
Also note that the angle given is the angle measured relative to the plane of the loop.
Are you saying I need to find the angle measured relative to the plane of the loop as in 90-65 = 25?
 
TSny said:
As you noted, the area ##A## and the angle ##\Theta## are constants. So, you can treat ##A\cos \Theta## as a constant.

Did you get a sign wrong when calculating ##dB/dt##?

I see what you mean. Yes I typed it incorrectly. \frac{dB}{dt} = (2.5t - .500) I had typed a + sign.
 
abstrakt! said:
Are you saying I need to find the angle measured relative to the plane of the loop as in 90-65 = 25?

The angle measured relative to the plane is given to be 65o. You should think about whether or not you should use 65o for ##\Theta## in your flux formula.
 
TSny said:
The angle measured relative to the plane is given to be 65o. You should think about whether or not you should use 65o for ##\Theta## in your flux formula.
Thank you! I will reference my notes and textbook and ponder about this a bit. I considered whether that was the appropriate angle to use but I suppose I haven't thought about it enough. I will report back after I redo this problem with the new information you have provided. Thank so much!
 
OK. See if it works out.
 
  • #10
I am pretty sure I did this correctly (I got the correct answer at least). I just want to ensure that my process was correct.

I took the derivative of the magnetic field: dB/dt = (2.5t-.500)T when t = 2.00 and kept the area and the cosine of the angle constant since their rate of change is 0. Then, I measured the angle off of the area vector that is normal to the surface based on the angle between the magnetic field vector and the plane in which the circular loop lies (90-65 = 25).

The final equation is: ξ = (2.5*2-.500)(π(.14)2cos 25°) = .251 or 251 mV

Is this correct?

Thank you for your suggestions, they gave me a better perspective of the problem (assuming I am thinking about this correctly).

b) What is the direction of the current induced in the loop?

Since the inward flux is decreasing, the induced flux will be into the page so the induced current will move clockwise in order to "oppose" the induced flux? I am still fuzzy on this concept.
 
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  • #11
Your work for part (a) looks correct.

For part (b) I don't have access to the figure. I'm guessing that the loop is in the plane of the page and the B field is pointing into the page at 25o to the normal of the loop. Is that right?

Why do you say the flux is decreasing?
 
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