Magnetic field in a solenoid - -

[pat666]

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
$$B=\mu_o*n*I$$
$$B=0.16T$$
I'm not sure if this is right? I didn't use the area and am not sure about my calculation for n.

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

Thanks

 

[anathema]

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.