How Does Changing Magnetic Field Influence Current in a Two-Loop Wire Circuit?

[Tekmachine]

Homework Statement

A two-loop wire circuit is 53.4069 cm wide
and 35.6046 cm high. The wire circuit in
the figure is located in a magnetic field whose
magnitude varies with time according to the
expression B = (0.001 T/s) t and its direction
is out of the page.

Assume The resistance per length of the
wire is 0.113 Ohm/m.

http://img87.imageshack.us/img87/9058/figureqm2.th.jpg http://g.imageshack.us/thpix.php

When the magnetic field is 0.2 T, find the
magnitude of the current through middle leg
PQ of the circuit. Answer in units of μA.

Homework Equations

$$\mathcal{E} = -A\frac{dB}{dt}$$

$$\mathcal{E} = I R$$

The Attempt at a Solution

$$\mathcal{E} = -(0.356046m)(0.356046m)(0.001T/s) = -1.267687541\times 10^{-4} v$$

$$\mathcal{E} = -(0.178023m)(0.356046m)(0.001T/s) = -6.338437706\times 10^{-5} v$$

$$\frac{(-1.267687541\times 10^{-4}v) - (-6.338437706\times 10^{-5}v)}{0.113}$$ = -560.92 μA

I might have messed up on the signs, and also I'm not sure how "When the magnetic field is 0.2 T" applies to my equations.

 

[anathema]

Thank you for your post. I am a scientist and I would like to help you with your question.

Firstly, I would like to clarify that the equations you have used are correct. However, there are some minor errors in your calculations. Let me guide you through the correct solution.

The first step is to calculate the induced electromotive force (EMF) in the circuit, using the formula \mathcal{E} = -A\frac{dB}{dt}. Here, A is the area of the circuit, which is given by the width multiplied by the height of the circuit. So, A = (53.4069 cm)(35.6046 cm) = 1.9005 x 10^-3 m^2. The rate of change of magnetic field, dB/dt, is given by 0.001 T/s. Therefore, the induced EMF is \mathcal{E} = -(1.9005 x 10^-3 m^2)(0.001 T/s) = -1.9005 x 10^-6 V.

Next, we need to calculate the current through the middle leg PQ of the circuit. This can be done using the formula \mathcal{E} = IR, where R is the resistance per unit length of the wire. The length of the middle leg PQ is given by 0.356046 m. Therefore, the resistance of this leg is R = (0.356046 m)(0.113 Ohm/m) = 0.040267 Ohm. Substituting these values in the equation \mathcal{E} = IR, we get I = -1.9005 x 10^-6 V/0.040267 Ohm = -47.211 μA.

Now, to find the magnitude of the current, we need to take the absolute value of the current, which is 47.211 μA. However, this is the current when the magnetic field is changing at a rate of 0.001 T/s. When the magnetic field is 0.2 T, we need to scale up the current by a factor of 0.2/0.001 = 200. Therefore, the magnitude of the current through the middle leg PQ when the magnetic field is 0.2 T is 200 x 47.211 μA = 9.442 μA.

I hope this helps you

 

[thomate1]

I would first like to commend you for your attempt at solving this problem. Your use of the equations for induced EMF and Ohm's law is a good start. However, there are a few things that need to be addressed in your solution.

Firstly, the equations you have used are correct, but it seems that you have used the wrong values for the length and width of the circuit. The given dimensions of the circuit are 53.4069 cm for the width and 35.6046 cm for the height, not 0.356046 m for both. This would change your calculations significantly.

Secondly, the given magnetic field is not constant, but rather varies with time. This means that the induced EMF will also vary with time. In order to find the magnitude of the current, you need to consider the instantaneous EMF at the time when the magnetic field is 0.2 T. This can be calculated by substituting t=0.2 T/s into the equation for induced EMF.

Finally, the units of your final answer should be in amperes (A) instead of microamperes (μA). This is because the resistance per length is given in Ohm/m, and not Ohm/cm.

With these corrections, your final answer should be in the range of 0.5-1.0 A. I would also recommend using more significant figures in your calculations to avoid rounding errors. Overall, your approach was good, but make sure to pay attention to the given values and units in the problem.