Time Dependent Current in a wire

In summary, the conversation discusses a problem involving an infinite straight wire carrying a current that varies with time. A conducting loop is also present at a fixed distance from the wire. The task is to find the induced emf in the loop at a specific time, with the definition of emf being positive for a clockwise current and negative for a counter-clockwise current. The attempt at a solution involves calculating the flux through the loop, which cannot be done using the formula B*A since the magnetic field is not uniform. Instead, an integral must be used.
  • #1
kjlchem
24
0

Homework Statement


An infinite straight wire carries a current I that varies with time as shown above. It increases from 0 at t = 0 to a maximum value I1 = 2.1 A at t = t1 = 14 s, remains constant at this value until t = t2 when it decreases linearly to a value I4 = -2.1 A at t = t4 = 24 s, passing through zero at t = t3 = 21.5 s. A conducting loop with sides W = 20 cm and L = 57 cm is fixed in the x-y plane at a distance d = 49 cm from the wire as shown.

What is ε1, the induced emf in the loop at time t = 7 s? Define the emf to be positive if the induced current in the loop is clockwise and negative if the current is counter-clockwise.

Homework Equations



B = μI/2∏r

Flux = B*A

-dflux/dt = ε


The Attempt at a Solution



I don't understand what I'm doing wrong with this problem.

This is what I have so far...

(dB*A)/dt= ε, A = L(W)

μ(dI)(L)W/(2∏rdt) = ε

μ=12.566*10^-7
dI = 2.1 A
L = .57 m
W = .2 m
dt=14 s.

On the left side of the box, r = .49 m and the current is negative, so the emf is positive.
On the right side of the box, r = 1.06 m and the current is positive, so the emf is negative.
Putting the 2 emf's together by subtracting the right side from the left side, I get an emf of -3.753*10^-9V.

What am I doing wrong?
 

Attachments

  • t.png
    t.png
    2.1 KB · Views: 532
Physics news on Phys.org
  • #2
kjlchem said:
On the left side of the box, r = .49 m and the current is negative, so the emf is positive.
On the right side of the box, r = 1.06 m and the current is positive, so the emf is negative.
Putting the 2 emf's together by subtracting the right side from the left side, I get an emf of -3.753*10^-9V.

What am I doing wrong?
I'm not sure about the method you are using. For calculating EMF for a straight section of wire, I am only familiar with doing that for the wire moving through a magnetic field.

Since the wire loop is not moving, I think you have to use ε=-dflux/dt instead. So first I would first calculate the flux through the loop -- as a function of time, during the time interval that contains 7 s.
 
  • #3
Yeah, I used ε = -dflux/dt. The flux as a function of time = μ(dI)(L)W/(2∏rdt)
 
  • #4
kjlchem said:
Yeah, I used ε = -dflux/dt. The flux as a function of time = μ(dI)(L)W/(2∏rdt)
That won't work here; for one thing, there is no r given in this problem. I think I see your problem though:
kjlchem said:
Flux = B*A
That only works if B is uniform over the whole area. It isn't; B is stronger at the side of the rectangle closest to the wire, and weaker at the far side.

Instead, you'll need to do an integral to calculate the flux:
Flux = [itex]\int B \cdot dA[/itex]​
 
  • #5




Your approach to the problem is correct, but there are a few errors in your calculations. Here is a step-by-step solution to help you understand where you went wrong:

1. First, we need to find the magnetic field at the location of the loop, which is at a distance of 49 cm from the wire. Using the equation B = μI/2∏r, we get:

B = (4∏*10^-7)(2.1)/((2∏)(0.49)) = 1.357*10^-5 T

2. Next, we need to find the flux through the loop, which is given by the equation Flux = B*A. The area of the loop is L*W = (0.57)(0.2) = 0.114 m^2. Therefore, the flux through the loop is:

Flux = (1.357*10^-5)(0.114) = 1.548*10^-6 Wb

3. Now, we can use the equation -dflux/dt = ε to find the induced emf at time t = 7 s. We know that the flux is changing linearly between t = 0 and t = 14 s, so we can use the formula for a linear change in flux:

ΔFlux/Δt = (1.548*10^-6)/14 = 1.106*10^-7 Wb/s

4. Finally, we can plug this value into the equation -dflux/dt = ε to find the induced emf at t = 7 s:

-ε = (1.106*10^-7)/(-7) = 1.58*10^-8 V

Since the current in the wire is increasing at t = 7 s, the induced current in the loop will be clockwise, making the induced emf positive. Therefore, the value of ε1 at t = 7 s is 1.58*10^-8 V.

Hope this helps clarify the solution for you!
 

1. What is time dependent current in a wire?

Time dependent current in a wire refers to the flow of electric charge through a wire that changes over time. This means that the amount of current passing through the wire varies as time passes.

2. What causes time dependent current in a wire?

Time dependent current in a wire can be caused by a variety of factors, such as changing voltage or resistance in the circuit, as well as the presence of magnetic fields or changing electric fields.

3. How is time dependent current measured?

Time dependent current can be measured using an ammeter, which is a device that measures the amount of electric current in a circuit. The ammeter can be connected in series with the wire to measure the current passing through it.

4. What is the difference between time dependent current and steady state current?

The main difference between time dependent current and steady state current is that steady state current remains constant over time, while time dependent current varies. Steady state current is typically seen in DC circuits, while time dependent current is more common in AC circuits.

5. How is time dependent current related to the concept of capacitance?

Capacitance is a measure of a material's ability to store an electric charge. In the case of a wire, the capacitance can affect the amount of time dependent current that flows through it. In circuits with capacitors, time dependent current can be affected by the charging and discharging of the capacitor.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
758
  • Introductory Physics Homework Help
Replies
5
Views
901
  • Introductory Physics Homework Help
Replies
4
Views
816
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
648
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top