Energy density of magnetic field in a solenoid

AI Thread Summary
To calculate the energy density of the magnetic field in a solenoid, the formula B = μo*n*I is used, where n is the number of turns per unit length. The energy density is then determined using the equation energy density = B^2/(2*μo). In the attempted solution, the user incorrectly applied the permeability constant μo and miscalculated the area conversion from cm² to m². The correct calculations are essential for accurate results in both energy density and total energy stored in the magnetic field. Attention to detail in unit conversions and constants is crucial in physics problems.
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Homework Statement



A solenoid that is 117 cm long has a cross-sectional area of 20.0 cm2. There are 1010 turns of wire carrying a current of 9.56 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).

Homework Equations



B = μo*n*I, where n=N/l
energy density = B^2/(2*μo)
total energy = answer to part a times the volume

The Attempt at a Solution



Plugging the formulas into each other, I got

(((1010/1.17)*9.56*4*10^-7)^2)/(8*10^-7) = 13.6 J/m^3 for part (a) and
13.6*.2*1.17 = 3.19 for part (b)

Both are incorrect.
 
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for part a, you have used μo to be 4*10^-7 but this is not the right definition. And for part b, 20cm^2 is not equal to 0.2m^2 (remember that 100cm^2 is not 1m^2). Apart from that, your working looks good.
 
Oh, thank you! I hate when I miss stupid little things like that.
 
hehe. yeah it is annoying when that happens.
 

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