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?

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

since you're a real machine, your ama is always less than your ima, because you can never totally eliminate friction.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)