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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

User Raddykrish
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2 Answers

6 votes

Final answer:

The magnitude of the magnetic field at the center of the loops can be determined using Ampere's law. It is given by µ₀NI/(2a), where µ₀ is the permeability of free space, N is the number of turns, I is the current, and a is the radius of the loop.

Step-by-step explanation:

The magnitude of the magnetic field at the center of the loops can be determined using Ampere's law. According to the law, the magnetic field inside a current-carrying wire of radius a is given by µ₀NI/(2a), where µ₀ is the permeability of free space, N is the number of turns, I is the current, and a is the radius of the loop.

In this case, the radius of the first loop is a and the current is I. Since the pattern is repeated up to the Nth loop, the radius of the Nth loop is Na and the current is NI. Therefore, the magnitude of the magnetic field at the center of the loops is µ₀NI/(2a).

User Guini
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1 vote

Answer:

a) 0

Step-by-step 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:


B=(\mu_oI)/(2r)

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:


B_(T)=B_1-B_2+B_3-...-B_(n)\\\\B_(T)=(\mu_oI)/(2a)+(\mu_o(2I))/(2(2a))-...-(\mu_o(nI))/(2(na))=(\mu_oI)/(2a)+(\mu_o(I))/(2(a))-...-(\mu_o(I))/(2(a))=0T

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

hence, the total magnetic field is 0.

User Pratik Kamani
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