How is Current Determined in a Single Phase AC Circuit with a Moving Conductor?

[ric115]

Hi there

Im new to this site and i have noticed that you give advice to electronic questions. Presently I am studing electronics and on some of my course work i am having some difficulty in understanding some of the questions. such as,

Q1
A conductor 300mm long moves at a uniform speed of 4m/s at right-angles to a uniform magnetic field of flux density 1.25T. Determine the current flowing in the conductor when:

a) its end are open-circuited

b) its ends are connected to a load of 20 ohm resistance

and,

Q2
Calculate the e.m.f induced in a coil of inductance 12 H by a current changing at the rate of 4 A/s

Any advice or helpfull sites to visit would be much appreciated,

Thanking you in advance

Rich

 

[Redbelly98]

Welcome to Physics Forums.

Since you didn't post your questions in the homework section, you're unaware of our policy on this type of question (not your fault).

Can you provide:

A. any equations from your class lectures or textbook that appear to be relevant to solving the problems, and

B. provide your thoughts or attempt at solving the problems (doesn't have to lead to a final solution)

 

[Mene]

Hello Rich,

Thank you for reaching out and for your interest in electronics. I would be happy to help you with your questions.

Firstly, a single-phase AC circuit is a circuit that has a single alternating current, meaning the current changes direction periodically. This type of circuit is commonly used in household appliances and lighting.

Now, for your first question, we can use Faraday's law of induction to solve for the current in the conductor. This law states that the magnitude of the induced EMF is equal to the rate of change of magnetic flux through a closed loop. In this case, we can calculate the EMF induced in the conductor by multiplying the length of the conductor (300mm) by the magnetic field strength (1.25T) and the speed of the conductor (4m/s). This gives us an EMF of 1500V.

a) When the ends of the conductor are open-circuited, no current can flow. This means that the induced EMF is balanced by the opposing force of the electric field created by the open circuit.

b) When the ends are connected to a load of 20 ohms, we can use Ohm's law (V=IR) to calculate the current. So, the current flowing in the conductor would be 1500V/20 ohms = 75A.

For your second question, we can use the equation e.m.f = -L (dI/dt), where L is the inductance and dI/dt is the rate of change of current. Plugging in the values, we get an induced EMF of 48V.

I hope this helps and provides a better understanding of these concepts. As for helpful sites, I recommend checking out Khan Academy and All About Circuits for more in-depth explanations and practice problems. Keep up the good work in your studies!