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Question:

An insulating thin rod of length l has a linear charge density ρ(x) = ρ₀x/l on it. The rod is rotated about an axis passing through the origin (x=0) and perpendicular to the rod. If the rod makes n rotations per second, then the time-averaged magnetic moment of the rod is?

nρl³

π3nρl³

π4nρl³

πnρl³

Solution:

The magnetic moment M of a current loop is given by M = IA, where I is the current and A is the area of the loop. In this case, the rod is rotating, creating a current. Let's consider a small element dx of the rod at a distance x from the origin. The charge dq on this element is given by dq = ρ(x)dx = (ρ₀x/l)dx. The time period T of one rotation is T = 1/n. The current due to this element is dI = dq/T = n(ρ₀x/l)dx. The current element creates a magnetic moment dM. The area of the loop formed by this element in one rotation is dA = πx². Thus, the magnetic moment due to this element is dM = dI * dA = n(ρ₀x/l)dx * πx² = (nπρ₀/l)x³dx. To find the total magnetic moment, we integrate over the length of the rod (from 0 to l): M = ∫₀ˡ (nπρ₀/l)x³dx = (nπρ₀/l) ∫₀ˡ x³dx = (nπρ₀/l) [x⁴/4]₀ˡ = (nπρ₀/l) (l⁴/4) = (1/4)nπρ₀l³. Therefore, the time-averaged magnetic moment of the rod is (π/4)nρ₀l³