Magnetic field at the centre of current carrying spiral

AI Thread Summary
The discussion revolves around calculating the magnetic field at the center of a spiral formed by a thin insulated wire carrying current. The initial approach incorrectly used the number of turns per unit area, leading to an incorrect result. A correction was made by using the number of turns per unit width, which yielded the correct answer. The confusion stemmed from misunderstanding how the number of turns relates to the radial distance rather than the area. Ultimately, the correct formula for the magnetic field was confirmed as B = (μ₀ i N)/(2(b-a)) ln(b/a).
Titan97
Gold Member
Messages
450
Reaction score
18

Homework Statement


A thin insulated wire forms a plane spiral of N turns carrying a current ##i##. The inner radius is ##b## an outer radius is ##a##. Find magnetic field at centre of spiral

Homework Equations


$$B=\frac{n\mu_0i}{2R}$$

The Attempt at a Solution


For an area of ##\pi(b^2-a^2)##, number of turns is N.
So for an elemental area of ##2\pi r dr##,number of turns ##n## is ##\frac{2rNdr}{(b^2-a^2)}##

After substing ##n##, $$dB=\frac{\mu_0 I 2Ndr}{2(b^2-a^2)}$$
But on integrating, I am getting wrong answer

Instead of taking number of turns per unit area, if I took number turns per unit width, I got the correct answer. That is if $$n=\frac{N}{b-a}$$.

Why does the answer change?
 
Physics news on Phys.org
Titan97 said:
The inner radius is ##b## an outer radius is ##a##.
Did you mean to have ##a## as the inner radius and ##b## as the outer radius?
$$dB=\frac{\mu_0 I 2Ndr}{2(b^2-a^2)}$$
OK. What did you get for your final answer for ##B##?
 
  • Like
Likes Titan97
typo. Inner radius is ##a## and outer radius is ##b##.

Integrating from a to b, $$B=\frac{\mu_0 i N}{b+a}$$ This is not the answer given.
 
Titan97 said:
Integrating from a to b, $$B=\frac{\mu_0 i N}{b+a}$$ This is not the answer given.

That looks right to me. It has the correct limit for ##a## →##b##. Hopefully someone else will confirm this or else point out our mistake.

What answer was given?
 
  • Like
Likes Titan97
Ah, good. Thank you. Now I see the mistake. There are, in fact, a constant number of turns per unit radial distance. Just picture the individual wraps of wire. The wire has a fixed thickness. So, there will be a fixed number of turns per unit radial distance. So, it is not the area ##2 \pi r dr## that determines the number of wraps of wire in a distance ##dr##, but rather its just the length ##dr## itself that determines the number of wraps. Sorry for not catching that.
 
  • Like
Likes Titan97
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...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top