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QUESTION: Hey,

I have a physics exam tomorrow covering circuits, and here is a problem from last year's exam:

I know what the answer is, but I'd like help working through it. I'm particularly unsure as to how to deal with the hanging wire given it does not make its way back to the source of Emf.

Regards,

Frank

ANSWER: Hello Frank,

I can't work that problem without knowing where nodes A and B are. I suspect allexperts will not let you post a follow-up until tomorrow. So I will tell you that no current will flow through the 5 Ohm resistor, so it will not affect the voltage across and the current thru the 10 and the 8 Ohm resistors. As far as they are concerned, the 5 Ohm resistor could be cut off and thrown away.

In case nodes A or B are at the left end of the 5 Ohm resistor, I can tell you that the voltage at the left end of the 5 Ohm resistor will equal the voltage at the right end of the 5 Ohm resistor - and will equal the voltage at the node between the 8 and 10 Ohm resistors.

I hope this helps,

Steve

[an error occurred while processing this directive]---------- FOLLOW-UP ----------

QUESTION: Thank you Steve; I got the answer; I was just supposed to add the batteries to get 30 V, then find the voltage across the 10 Ohm resistor, which is 16.7 V.

Do you think you can help me work out these problems; I am reviewing them now, but once I feel that I have the answer I want to compare my answer to your solutions:

Note: I already know the answers are 2 uT and clockwise, respectively. I am asking you so I can compare my answer to yours once I watch the videos for those topics and attempt the problems.

I will be sure to rate and nominate you again.

Regards,

Frank

Hello Frank,

5. Ampere's Law says that the amplitude of the magnetic field a distance r away from either wire is given by

B = (mu0/(2*pi))*I/r

(where mu0 is the best I can do to show what should be the lower case Greek letter mu with a subscript of zero). The value of the constant mu0 is 4*pi*10^-7 T.m/A. (I can't do the Greek letter pi either. At least not easily. I know there actually is a way. I should learn the way, but I'm lazy.)

So if you plug the data for the current I1 into the formula you find the strength of the field at P due to I1. For the direction, if you apply the right-hand-rule, you find that it goes down into the page (or x-y plane or screen). Repeat for the current I2 to find the magnitude and direction of the field at P due to I2. So the answer is the vector sum of those results.

6. In the initial state, according to the right-hand-rule, the current in the outer ring produces a magnetic field coming up out of the page inside the outer loop (including in the inner loop). It says there initially is no current in the inner loop. When the outer current increases, the field it generates increases. Faraday's Law kicks in and induces a voltage that will produce a current that will produce a field that would tend to oppose the increase in magnetic field in the inner loop. So the field produced by the inner loop has to be down into the page. The right-hand-rule says the current in the inner loop must be clockwise.

I hope this helps,

Steve

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I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.**Education/Credentials**

BS Physics, North Dakota State University

MS Electrical Engineering, North Dakota State University