Induced current in expanding metal ring

In summary, a circular metal ring with a constant magnetic field perpendicular to it expands with time, causing its radius to grow and its resistance per unit length to change. To find the current induced in the ring, use the equation e=dФ/dt, where Ф is the flux and can be expressed as Ф=BA. The area can be determined as a function of time using the given relationship between radius and time. Once the induced emf is found, divide it by the instantaneous resistance (also a function of time) to obtain the instantaneous current. The total volume of the ring is constant, so it becomes thinner as it expands. When plugging in values, be sure to use the correct units.
  • #1
arl146
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Homework Statement


A circular metal ring, as shown on the diagram below, is constructed so as to expand or contract freely. In a region with a constant magnetic field Bo oriented perpendicular to it, the ring expands, with its radius growing with time as r=r0(1+[itex]\alpha[/itex]t2). As the ring expands and grows thinner, its resistance per unit length changes according to R=Ro(1+[itex]\beta[/itex]t2). Find the current induced in the ring as a function of time. To check your answer, suppose that B0 = 7.30 mT, r0 = 11.0 cm, R0 = 3.00 m, [itex]\alpha[/itex]= 0.245 × 10(-4) s(-2), and β = 0.500 × 10(-2) s(-2). What is the value of the induced current at t = 86.0 s? (Note: Give the direction of the current where when viewed from above a positive current will move counterclockwise.)


Homework Equations


I don't even know where to start ! Can you start with flux?
Flux=BA


The Attempt at a Solution


Again, no clue. help is definitely needed, so lost!

The picture doesn't show much, just the B0 points up ( guess you can just call it +y direction ) and the radius which of a circle is obvious. Help is greatly appreciated!
 
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  • #2
the induced emf may be given by e= dФ/dt, where Ф is the flux. now Ф can also can be given by Ф/area=B0. as the relation betwn r and t is given u can always find area as a function of time, hence also u can find dФ/dt and thus the expression for e ( B0 is const. e will be depending on rate change of area with respect to time). once u find e, u divide it by the instantenuous resistance (also a function of t) and get the instantenuous current. to check the answer put the given values and verify. while doing this always remember the total volume of the ring is constant, that means it becomes thinner with increasing length.
 
  • #3
So, I have:

[itex]\frac{d\Phi}{dt}[/itex]= B0*pi*(r0)2*(1+4[itex]\alpha[/itex]t+4[itex]\alpha[/itex]2t3)

Is that correct?

For my final answer I got 3.01292E-6 A. I only have one try left on my homework and I'm afraid to try it. I'm not sure if the (1+..) should still be there?

Oh, and R0 is actually Ohm/m sorry, I didn't notice that until now. But for that, when I had to use the R equation, I just multiplied it by 2*pi*r (the total length)
 
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  • #4
it was -2.969845533E-08 A, thanks!
 
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  • #5


I would first start by identifying the key variables and equations involved in this scenario. The key variables are the magnetic field strength (B0), initial radius of the ring (r0), initial resistance per unit length (R0), and the time-dependent parameters for expansion (α) and change in resistance (β). The key equation involved is Faraday's law of induction, which states that the induced electromotive force (EMF) in a closed loop is equal to the rate of change of magnetic flux through the loop. In this case, the loop is the expanding metal ring and the magnetic flux is given by B0A, where A is the area of the ring.

To find the current induced in the ring as a function of time, we can use Ohm's law (V=IR) and combine it with Faraday's law to get the following equation:

I = -(1/R)(dΦ/dt)

Where I is the induced current, R is the resistance per unit length, and dΦ/dt is the rate of change of magnetic flux through the ring. Since the resistance per unit length changes with time according to R=Ro(1+βt^2), we can rewrite the equation as:

I = -(1/Ro(1+βt^2))(d/dt)(B0A)

To find the value of the induced current at t=86.0 s, we can plug in the given values for B0, r0, R0, α, and β into the equation and solve for I. The direction of the current can be determined by using the right-hand rule, where the direction of the current is in the direction of the curl of the fingers when the thumb points in the direction of the magnetic field. In this case, the induced current would be counterclockwise when viewed from above.

In summary, as a scientist, I would approach this problem by first identifying the key variables and equations involved, and then using those equations to solve for the induced current at a specific time.
 

Related to Induced current in expanding metal ring

1. What is induced current in an expanding metal ring?

Induced current in an expanding metal ring refers to the flow of electric charge that is generated in the ring when it is subjected to a changing magnetic field. This phenomenon is known as electromagnetic induction.

2. How does the current in an expanding metal ring get induced?

The current in an expanding metal ring is induced through the process of electromagnetic induction. This occurs when the ring is placed in a changing magnetic field, causing the electrons in the metal to move and create an electric current.

3. What factors affect the magnitude of the induced current in an expanding metal ring?

The magnitude of the induced current in an expanding metal ring is affected by several factors, including the strength and rate of change of the magnetic field, the material and size of the ring, and the resistance of the ring's material.

4. Can the direction of the induced current be changed in an expanding metal ring?

Yes, the direction of the induced current in an expanding metal ring can be changed by changing the direction of the magnetic field or by changing the direction in which the ring is expanding.

5. What are some practical applications of induced current in expanding metal rings?

Induced current in expanding metal rings has various practical applications, such as in power generation through electromagnetic generators, induction heating systems, and electromagnetic braking systems used in trains and roller coasters.

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