Question

A hollow sphere of mass m, radius R is rotating
about its diameter with angular velocity o. If charge
Q is uniformly distributed over the surface, then
magnetic moment of sphere is ​

Answer 1

Answer:

Explanation:

the answer is QR^2W(omega) / 3

Answer 2

It has given that, a hollow sphere of mass m radius R is rotating about its diameter with angular velocity ω. charge Q is uniformly distributed over the surface.

To find : The magnetic moment of sphere.

solution : cut an element of thickness dx at x distance from the centre of hollow sphere as shown in figure.

so, radius of disc, r = Rsinθ

area of disc dA = 2πRsinθ dx

surface charge disc, σ = q/4πR²

so charge on disc, dq = σ dA

= q/4πR² × 2πRsinθ dx

= qsinθ/2R dx

now magnetic moment, M = A × i

= π(Rsinθ)² × dq/dt

= πR²sin²θ × dq/(2π/ω)

= ωRsin²θ × qsinθ/4 dx

= ωqR/4 sin³θ dx

but dx = Rdθ [from figure ]

= ωqR²/4 ∫sin³dθ [ from 0 to π ]

= ωqR²/4 × 4/3

= ωqR²/3

Therefore the magnetic moment of hollow sphere would be ωqR²/3

method 2 : using formula, Q/2m = M/L

where M is magnetic moment and L is angular momentum.

but angular momentum, L = Iω

= (2/3 mR²) × ω [ moment of inertia of hollow sphere about its axis is I = 2/3 mR² ]

so, q/2m = M/(2/3 mR² ω)

⇒M = qωR²/3

Therefore magnetic moment of hollow sphere is qωR²/3