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

• hidemi
In summary, the conversation discusses the calculation for the magnetic field of N loops with varying radius and current. The relevant equation is provided and it is clarified that the question is about N loops, not just one. An example calculation for N=3 is also given.

#### hidemi

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

For N=1, d, not e, gives the relevant equation you wrote.

Last edited:
hidemi
anuttarasammyak said:
For N=1, d, not e, gives the relevant equation you wrote.
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)}##

hidemi
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.