The figure below shows two loops of wire having the same axis. The smaller loop has radius a andresistance R and the larger loop has radius b. The smaller loop is above the larger one, by a distance z,which is large compared to the radius b of the larger loop, (z >> b). Hence with current I through the largerloop as indicated, the consequent magnetic field is nearly constant through the plane area bounded bythe smaller loop. Suppose now that z is not constant but is changing at the positive constant rate v z = dz /dt > 0 (z increasing).(a) Determine the magnetic flux across the area bounded by the smaller loop as a function of z.
Question
The figure below shows two loops of wire having the same axis. The smaller loop has radius a andresistance R and the larger loop has radius b. The smaller loop is above the larger one, by a distance z,which is large compared to the radius b of the larger loop, (z >> b). Hence with current I through the largerloop as indicated, the consequent magnetic field is nearly constant through the plane area bounded bythe smaller loop. Suppose now that z is not constant but is changing at the positive constant rate v z = dz /dt > 0 (z increasing).(a) Determine the magnetic flux across the area bounded by the smaller loop as a function of z.
Solution
The magnetic field B due to a current I in a loop of radius b at a distance z along the axis is given by Ampere's Law:
B = μ0 * I / (2 * π * z)
where μ0 is the permeability of free space.
The magnetic flux Φ through the smaller loop of radius a is given by the magnetic field B times the area A of the smaller loop:
Φ = B * A
Substituting the expression for B from Ampere's Law and the area of a circle π * a^2 for A, we get:
Φ = μ0 * I / (2 * π * z) * π * a^2
Simplifying, we find the magnetic flux Φ as a function of z:
Φ = μ0 * I * a^2 / (2 * z)
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