contestada

We find that 2N 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 the loops alternates in direction from loop to loop as seen from above. Thus the current in the first loop is counterclockwise, in the next clockwise, up to the last loop where it is again clockwise. The magnitude of the magnetic field at the center of the loops is:_________


a. 0

b. µoNI/2a

c. µoI/Na

d. µoNI/a

e. µI/2Na

Respuesta :

mavila18

Answer:

a) 0

Explanation:

To find the magnitude of the magnetic field you sum the different contributions to the field by each loop.

The magnetic field for a loop is given by:

[tex]B=\frac{\mu_oI}{2r}[/tex]

I: current

r: radius of the loop

mu_o: permeability of vacuum = 4*pi*10^{-7} Tm/A

you take into account the direction of B in each loop, that is:

[tex]B_{T}=B_1-B_2+B_3-...-B_{n}\\\\B_{T}=\frac{\mu_oI}{2a}+\frac{\mu_o(2I)}{2(2a)}-...-\frac{\mu_o(nI)}{2(na)}=\frac{\mu_oI}{2a}+\frac{\mu_o(I)}{2(a)}-...-\frac{\mu_o(I)}{2(a)}=0T[/tex]

each current loop has the same magnitude of B but in opposite directions.

hence, the total magnetic field is 0.