The magnitude of the magnetic field at the center of the loops is....

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The discussion centers on calculating the magnetic field at the center of multiple loops of wire. One participant presents a formula for the magnetic field based on the number of loops (N), asserting that the relevant equation should account for N loops rather than just one. They clarify that for each loop, the radius and current increase, affecting the overall magnetic field calculation. Misunderstandings about the question's focus on multiple loops are addressed, leading to a clearer understanding of the problem. The conversation concludes with an acknowledgment of the clarification provided.
hidemi
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Homework Statement
We find that N current loops are coplanar and coaxial. The first has radius a and current I. The second has radius 2a and current 2I, and the pattern is repeated up to the Nth, which has radius Na and current NI. The current in each loop is counterclockwise as seen from above. The magnitude of the magnetic field at the center of the loops is:

a. µoI/2Na

b. µoI/Na

c. µoI/2a

d. µoNI/2a

e. µoNI/a

Ans: D
Relevant Equations
B = µoI/2a
I calculate like this and I think the answer is E not D.

N( µoI/2a + µo2I/2*2a)
= 2µoNI/2a
= µoNI/a
 
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For N=1, d, not e, gives the relevant equation you wrote.
 
Last edited:
anuttarasammyak said:
For N=1, d, not e, gives the relevant equation you wrote.
Thanks for reply.
Here's my calculation:
N( µoI/2a + µo2I/2*2a)
= 2µoNI/2a
= µoNI/a

The question is asking about the two loops combine. The relevant equation does not include the two loops.
 
hidemi said:
The question is asking about the two loops combine. The relevant equation does not include the two loops.
You have misread the question. The question is about N loops. N could be any integer, 1, 2, 3, 4, 5 ...

For example if N=3.
First loop has radius = a, current = I, its field is ##B = \frac {\mu_0 I}{2a}##
Next loop has radius = 2a, current = 2I, its field is ##B = \frac {\mu_0 (2I)}{2(2a)}##
Next loop has radius = 3a, current = 3I, its field is ##B = \frac {\mu_0 (3I)}{2(3a)}##
 
Steve4Physics said:
You have misread the question. The question is about N loops. N could be any integer, 1, 2, 3, 4, 5 ...

For example if N=3.
First loop has radius = a, current = I, its field is ##B = \frac {\mu_0 I}{2a}##
Next loop has radius = 2a, current = 2I, its field is ##B = \frac {\mu_0 (2I)}{2(2a)}##
Next loop has radius = 3a, current = 3I, its field is ##B = \frac {\mu_0 (3I)}{2(3a)}##
Ok, I see where you are getting.
Thank you so much.
 
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