What is the magnitude of the magnetic field at the center of the coil?

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SUMMARY

The discussion focuses on calculating the magnetic field and induced electromotive force (emf) in a circular current loop system. The magnetic field at the center of a circular coil with a radius of 10.5 cm and 200 turns is given by the formula B_center = 0.001197 * I, where I is the current. For a time-varying current I(t) = I0 * sin(2 * π * f * t), the magnetic field becomes B_center(t) = 0.001197 * (0.01 * sin(2 * π * f * t)). The discussion also addresses calculating magnetic flux through a smaller loop placed at the center of the larger loop and the total induced emf using Faraday's law.

PREREQUISITES
  • Understanding of magnetic fields and current loops
  • Knowledge of Faraday's law of electromagnetic induction
  • Familiarity with calculus, specifically derivatives
  • Ability to apply the formula for magnetic flux: Φ = B * A
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  • Study the derivation and application of Faraday's law in electromagnetic systems
  • Learn about magnetic flux calculations for different geometries
  • Explore the relationship between current, magnetic field, and induced emf in AC circuits
  • Review calculus concepts, particularly differentiation of trigonometric functions
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jaymode
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Alright there is a problem I was given and it has four parts, I got the first two but am having trouble with the last two.

A)Consider a circular current loop of radius 10.5 cm with 200 total turns. Assume
that the current through the coil is I. What is the magnitude of the magnetic field at the center of the coil? (Your answer should be a numerical value multiplied by the current I).

B_center= [(4*pi*10^-7)*(200)*(I)]/(2*.105) = 0.001197*I

B A time-varying current of the form I(t)=Iosin(2*pi*f*t) is passed
through the circular coil in part a) where Io= 10 mA. Using your
result from a), write down the expression for the time varying
magnetic field B(t) at the center of the coil.

B_center(t) = 0.001197*(.01*sin(2*pi*f*t))

C A small 1.5 cm radius circular current loop is placed at the center of the large current loop from part a) (as shown in the photo) oriented so that the plane of the current loop is perpendicular to the magnetic field. Assume that the magnetic field from the large current loop is constant over the small loop. What is the magnetic flux through the small current loop (use your result from part b))?

This is where I do not know what to do and for the next part I do not really know what to do either.

D What is the total induced emf in the small current loop assuming that it has 2000 turns (use your result from part c))? The frequency of the time-varying current is f=1000 Hz. (Your answer should be in the form of a numerical value times a trigonometric function of 2*pi*f*t).
 
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jaymode said:
C A small 1.5 cm radius circular current loop is placed at the center of the large current loop from part a) (as shown in the photo) oriented so that the plane of the current loop is perpendicular to the magnetic field. Assume that the magnetic field from the large current loop is constant over the small loop. What is the magnetic flux through the small current loop (use your result from part b))?

This is where I do not know what to do and for the next part I do not really know what to do either.
This requires you to know the definition of magnetic flux: Magnetic flux through a loop equals the area of the loop times the component of the magnetic field perpendicular to the area. (Your text can give you a more precise definition.)
D What is the total induced emf in the small current loop assuming that it has 2000 turns (use your result from part c))? The frequency of the time-varying current is f=1000 Hz. (Your answer should be in the form of a numerical value times a trigonometric function of 2*pi*f*t).
This one requires understanding of Faraday's law, which states that induced emf is proportional to the rate of change of the flux through the coil and the number of turns in the coil. (For details, see your text or here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c1)
 
Doc Al said:
This requires you to know the definition of magnetic flux: Magnetic flux through a loop equals the area of the loop times the component of the magnetic field perpendicular to the area. (Your text can give you a more precise definition.)

This one requires understanding of Faraday's law, which states that induced emf is proportional to the rate of change of the flux through the coil and the number of turns in the coil. (For details, see your text or here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c1)


Ok so the are of the loop. I would assume that the area of the loop is pi*r^2 with the radius being that of the smaller loop which is 1.5cm?
 
and for the last part ould it be taking the derivative of B and the derivative of the area? Then using the equation -N(d(BA)/dt) ?
 
jaymode said:
Ok so the are of the loop. I would assume that the area of the loop is pi*r^2 with the radius being that of the smaller loop which is 1.5cm?
That's right.
 
jaymode said:
and for the last part ould it be taking the derivative of B and the derivative of the area? Then using the equation -N(d(BA)/dt) ?
In that equation you are taking the derivative of BA. But A is a constant so d(BA)/dt = A \; dB/dt.
 
ok thanks. My math is a little rusty. right now I have B = 0.0000197*sin(2*pi*f*t)

if i remember correctly the derivative of this would be f'(x)*g(x) + f(x)*g'(x)?
 
jaymode said:
ok thanks. My math is a little rusty. right now I have B = 0.0000197*sin(2*pi*f*t)

if i remember correctly the derivative of this would be f'(x)*g(x) + f(x)*g'(x)?
What are you taking as f(x) and g(x)? That looks like the "product rule"; it's not clear to me how that will help here.

All I see is a constant times a sin function: a \sin (bt), where a and b are constants.
 
Yes I was using the product rule. It has been a while since I last did derivatives so I need some refreshing. I was using a as f and the sin as G

But what I get from you is that the derivative would be:
bcos(bt)?
 
  • #10
jaymode said:
But what I get from you is that the derivative would be:
bcos(bt)?
Almost. Don't forget the constant "a".
 

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