Magnetic field in a solenoid - -

In summary, the magnetic field in the solenoid is 0.16 T and the energy density in the magnetic field is 639999.99999 J/m^3.
  • #1
pat666
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Homework Statement


6. A solenoid 25.0 cm long with a cross-sectional area of 0.500 cm2 has 400 turns of wire and carries a current of 80.0 A. Calculate:
a) the magnetic field in the solenoid;
b) the energy density in the magnetic field if the solenoid is filled with air. (2 marks)



Homework Equations





The Attempt at a Solution


n=400/.25=6000turns per metre
[tex] B=\mu_o*n*I[/tex]
[tex]B=0.16T [/tex]
I'm not sure if this is right? I didn't use the area and am not sure about my calculation for n.

[tex] u_B= 1/2*{B^2}/{\mu_o} [/tex]
so [tex] u_B=63999.99999 [/tex]
I need someone to check my procedure please and also what are the usints for energy density, J?

Thanks
 
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  • #2
for your question! Your calculation for the magnetic field in the solenoid seems to be correct. However, there are a few things to note:

1. When calculating the number of turns per meter, you should divide by the length of the solenoid in meters (0.25 m), not in centimeters. So the correct calculation would be n = 400/0.25 = 1600 turns per meter.

2. The units for the magnetic field should be Tesla (T), not milli-Tesla (mT). So the correct answer for part (a) would be B = 0.16 T.

3. For part (b), the energy density in the magnetic field can be calculated using the equation u_B = 1/2 * B^2 / mu_0, where B is the magnetic field and mu_0 is the permeability of free space (4 pi x 10^-7 Tm/A). The units for energy density are Joules per cubic meter (J/m^3).

So the final answer for part (b) would be u_B = 1/2 * (0.16 T)^2 / (4 pi x 10^-7 Tm/A) = 639999.99999 J/m^3.

I hope this helps clarify things! Let me know if you have any other questions.
 

Related to Magnetic field in a solenoid - -

1. What is a solenoid?

A solenoid is a coil of wire that creates a magnetic field when an electric current passes through it. It is often used in electronic devices, such as motors and speakers, to convert electrical energy into mechanical energy.

2. How does a solenoid create a magnetic field?

When an electric current flows through a wire, it creates a circular magnetic field around the wire. In a solenoid, the magnetic fields from each turn of the coil add together to create a strong, uniform magnetic field along the axis of the coil.

3. What factors affect the strength of the magnetic field in a solenoid?

The strength of the magnetic field in a solenoid is affected by the number of turns in the coil, the current flowing through the coil, and the material of the core inside the coil. Increasing any of these factors will result in a stronger magnetic field.

4. How is the direction of the magnetic field in a solenoid determined?

The direction of the magnetic field in a solenoid can be determined using the right-hand rule. If you point your thumb in the direction of the current flowing through the coil, your fingers will curl in the direction of the magnetic field.

5. What are some practical applications of a solenoid's magnetic field?

Solenoids are used in a variety of applications, such as in electric door locks, MRI machines, and particle accelerators. They can also be used to create electromagnets and in the production of electricity from renewable energy sources, such as wind turbines.

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