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22 January, 04:08

A uniform non-conducting ring of radius 2.68 cm and total charge 6.08 µC rotates with a constant angular speed of 4.21 rad/s around an axis perpendicular to the plane of the ring that passes through its center. What is the magnitude of the magnetic moment of the rotating ring?

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  1. 22 January, 04:16
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    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
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