Flux Linkage with a coil in a solenoid

In summary, the question asks for the current needed to produce a magnetic flux density of 3*10^-3 Tesla in a solenoid with a cross-sectional area of 2*10^-4 m^2, length of 0.5m, and 500 turns. The first part was correctly solved to find a current of 2.39A. The second part requires finding the flux linkage, which is found by multiplying the magnetic flux density by the area, resulting in an incorrect answer of 6*10^-7 Webers. However, the correct answer is 3*10^-4 Webers for the entire solenoid.
  • #1
BadlyAddicted
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Homework Statement


A solenoid of cross-sectional area 2*10^-4 m^2 and length 0.5m having 500 turns is to be used to produce a uniform magnetic field some distance close to the center of the solenoid. Find the current that must flow in the solenoid to produce a magnetic flux density of 3*10^-3 Tesla. What is the flux linage with the coil?

2. The attempt at a solution

I worked out the first part. I found the current and it is 2.39A (correct according to the answer given by the book)

Now the second part, i used the equation flux linkage = magnetic flux density * Area
= (3 * 10 ^-3) * (2*10^-4)
= 6*10^-7 Webers

Apparently my second answer is incorrect. According to the book it's supposed to be 3*10^-4 Wb. Can someone please help me find my mistake? Thanks a lot
 
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  • #2
Hi BadlyAddicted,

BadlyAddicted said:

Homework Statement


A solenoid of cross-sectional area 2*10^-4 m^2 and length 0.5m having 500 turns is to be used to produce a uniform magnetic field some distance close to the center of the solenoid. Find the current that must flow in the solenoid to produce a magnetic flux density of 3*10^-3 Tesla. What is the flux linage with the coil?

2. The attempt at a solution

I worked out the first part. I found the current and it is 2.39A (correct according to the answer given by the book)

Now the second part, i used the equation flux linkage = magnetic flux density * Area
= (3 * 10 ^-3) * (2*10^-4)
= 6*10^-7 Webers

That's right for the flux through one loop of the solenoid. So how would you find the flux linkage for the entire solenoid?
 
  • #3


There may be a mistake in your calculation of the flux linkage. The correct equation for flux linkage is:

Flux linkage = number of turns * magnetic flux

In this case, the number of turns is given as 500, and the magnetic flux is given by the product of magnetic flux density and area. So the correct calculation would be:

Flux linkage = 500 * (3*10^-3) * (2*10^-4) = 3*10^-1 Wb

This is equivalent to 3*10^-4 Wb, as given in the book. Double check your calculation to make sure you are using the correct equation and plugging in the correct values.
 

1. What is flux linkage?

Flux linkage refers to the amount of magnetic flux that passes through a particular area or surface. It is a measure of the strength of the magnetic field.

2. How is flux linkage calculated?

Flux linkage is calculated by multiplying the number of turns in a coil by the magnetic flux passing through each turn. It is represented by the symbol Φ.

3. What is a coil in a solenoid?

A coil in a solenoid is a wire wound into a series of loops that creates a magnetic field when an electric current is passed through it. It is commonly used to generate and control electromagnetic fields.

4. How does flux linkage relate to the strength of a solenoid?

The strength of a solenoid is directly proportional to the amount of flux linkage. This means that the more turns in the coil and the stronger the magnetic field, the greater the flux linkage and the stronger the solenoid.

5. How can flux linkage be changed in a solenoid?

Flux linkage can be changed in a solenoid by adjusting the number of turns in the coil, the strength of the magnetic field, or the size and shape of the solenoid. It can also be changed by varying the amount of electric current passing through the coil.

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