Magnetic field produced by coil and solenoid

AI Thread Summary
The discussion centers on the differing magnetic field strength calculations for a coil and a solenoid using their respective formulas. The coil formula yields a significantly higher magnetic field strength than the solenoid formula, which aligns with the publisher's answer. The solenoid formula is deemed more accurate under specific conditions, such as closely packed windings and a length much greater than its diameter. The inapplicability of the coil formula arises because the stacking of loops in a solenoid alters the magnetic field distribution, making the simple addition of fields incorrect. Thus, the solenoid formula provides a more reliable approximation for the given scenario.
mikelepore
Messages
550
Reaction score
2
Can someone please tell me why the formula for B at the center of a coil and the formula for B inside a solenoid give widely different answers in this problem?

From Serway and Vuille, College Physics: "A solenoid with 500 turns, 0.10 m long, carrying a current of 4.0 A and with a radius of 10^-2 m will have what strength magnetic field at its center?" -- Answer provided by the publisher: 250 X 10^-4 T

My comparison of the two formulas ...

Method 1
Using the formula for the magnetic field at the center of a coil that has N turns:

B = N mu0 I / 2 R = (500 turns) (4 pi X 10^-7 T m/A) (4.0 A) / 2 (10^-2 m)
= 0.12566 T = 0.126 T
(About five times greater than the publisher's answer)

Method 2
Using the solenoid formula, which textbooks describe as becoming more accurate as the windings become more closely packed, and as the length of the solenoid becomes much larger than its diameter, and if those assumptions are met then the formula expresses the nearly uniform field everywhere inside the solenoid, not only at the center:

turns per unit length = n = N/L = 500 turns / 0.10 m = 5000 turns per meter
B = mu0 n I = (4 pi X 10^-7 T m/A) (5000 turns per meter) (4.0 A)
= 0.025133 T = 0.025 T
(Agrees with the publisher's answer.)

Does the "coil" formula cease to become applicable just because the additional assumptions for using the "solenoid" formula have been provided? If so, why?

Thank you.
 
Physics news on Phys.org
No, method 1 won't work because you aren't simply merging the wires together, you are stacking them. Because you are stacking the loops by the time you get to the last loop you are no longer in the center of the loop and the equation you're trying to use is very wrong at this point. They won't just add one by one (for multiplication by N) as you suggested.

You could get the actual result, method 2 is only an approximation, if you were to take the equation for a loop and integrate. I seem to remember the integral is not so easy to do by hand, and you may not even know about integration, so I'm not going to go into it. The book described the conditions for using the infinite length solenoid approximation, which are mostly met by this problem.
 
My bad, my computer froze, and I accidentally double posted.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top