Answer: 1.72*10^-7
Explanation:
Given
Radius of the ring, r = 2.68 cm = 0.0268 m
Charge on the ring, q = 6.08 µC
Angular speed of the ring, w = 4.21 rad/s
First, we know that the charge per unit area, σ = q / πr²
Also, the charge on ring of width, dr = σ⋅2πrdr
The Magnetic moment of this ring of width dr.dμ = i⋅A
If we integrate dr at R(top) and at 0(bottom), we get
∫dµ = ∫(R, 0) T⋅2πrdr.(w/2π).πr²
On finding at (R, 0), we get
μ = qwR² / 4
On substituting our values, we have
μ = (6.08*10^-6 * 4.21 * 0.0268) / 4
μ = (6.08*10^-6 * 0.113) / 4
μ = 6.87*10^-7 / 4
μ = 1.72*10^-7
The magnitude of the magnetic moment is 1.72*10^-7
