Basic Coil Turn question please see

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In summary, the power factor of an electrical appliance is significantly lower when operated on a 60Hz supply. This is because the 60Hz supply creates a slower electrical waves, which causes the power factor to decrease. It is important to keep in mind that the power factor is not always a good indicator of the quality of an electrical supply. For example, if an electrical appliance is rated for 2 kW, it will still operate at full power on a 60Hz supply, as long as the supply voltage is equal to or greater than the appliance's rated voltage.
  • #1
helpphysics
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Homework Statement


200-turn,coil of 8.0cm length is made of 0.50mm-diameter copper wire.
Magnetic field is perpendicular to coil. At what rate B should increase to induce a 2.0 A current in the coil?

Homework Equations





The Attempt at a Solution


i am not getting anthing
 
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  • #2
helpphysics said:
i am not getting anthing

Hello helpphysics,

Why don't start by stating any relevant equations, and then describe where and how you are getting stuck (showing your work as you go).
 
  • #3
i am sorry i am not getting it
i just know that emf = A*∂B/∂t & thus
2R = A*∂B/∂t
 
  • #4
Hello helpphysics,

I've been thinking about this problem a lot, and a couple of things are troubling me.

Is this a homework problem?

Is there anything missing in the problem statement, such as the radius, diameter, or cross sectional area of the coil? I see that the length is given. The length is useful if the coil is producing a magnetic field, but typically when discussing induced emf, something relating to the cross sectional area is given.

Are there any other coils or solenoids involved in this problem?

The problem statement says, "Magnetic field is perpendicular to coil." Initially I took this to mean that the cross sectional plane of the coil was perpendicular to the magnetic field (meaning the normal vector of the cross sectional area is parallel to the field). But thinking about it more, I think most people might interpret the problem statement as saying the major axis of the coil is perpendicular to the magnetic field. But if that was the case, there would be no induced emf at all! All I'm saying is that "Magnetic field is perpendicular to coil" is kind of a weird way to state the problem.

The more I look at this problem, the more it seems that maybe something is missing. :confused:
 
Last edited:
  • #5
Sorry mate. No more info has been given.
Don't worry you are not the only one going nuts over this. Physics is a hard nut! :tongue2:
 
  • #6
Anyway can you please help me out with the following?

240V , 50 Hz electrical appliance is rated at 2 kW. It has a lagging power factor of 0.7
(a) What is appliances power factor when it is used on a 60Hz supply.
(b) What is supply voltage required to maintain appliance at its rated power when operated off a 60 Hz supply?

Please look into this.
Thanks
 

1. What is a basic coil turn?

A basic coil turn is a fundamental technique used in the construction of electronic coils. It involves winding a wire around a cylindrical or rectangular core in a continuous manner, creating multiple turns or loops.

2. Why is the number of turns in a coil important?

The number of turns in a coil determines the strength and efficiency of the magnetic field it produces. More turns result in a stronger magnetic field, while fewer turns result in a weaker field.

3. How do you calculate the number of turns in a coil?

The number of turns in a coil can be calculated by dividing the length of the coil by the wire diameter and then multiplying it by the desired coil diameter. This is known as the fill factor formula.

4. What factors affect the number of turns in a coil?

The number of turns in a coil is influenced by factors such as the type and gauge of wire used, the size and shape of the core, and the desired magnetic field strength.

5. How can the number of turns in a coil be optimized?

The number of turns in a coil can be optimized by experimenting with different wire types and gauges, as well as core materials and sizes. Additionally, the use of computer simulations can help determine the most efficient number of turns for a specific application.

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