JEE Challenger
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PhyMedAdv 2025 • Paper 2

Ratio of Magnetic Dipole Moment to Angular Momentum for Rotating Sphere

A conducting solid sphere of radius RR and mass MM carries a charge QQ. The sphere is rotating about an axis passing through its center with a uniform angular speed ω\omega. The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as αQ2M\alpha \frac{Q}{2M}. The value of α\alpha is ____

Official Numerical Answer1.65 to 1.67

Topics & Concepts

Moving Charges and MagnetismAngular Momentum

Step-by-Step Solution

To find the value of α\alpha, we need to calculate the magnetic dipole moment (μ\mu) and the angular momentum (LL) of the rotating conducting solid sphere.

1. Calculation of Magnetic Dipole Moment (μ\mu)

Since the sphere is conducting, the charge QQ resides entirely on its outer spherical surface with a surface charge density: σ=Q4πR2\sigma = \frac{Q}{4\pi R^2}

Consider a thin circular ring on the surface of the sphere at an angle θ\theta relative to the axis of rotation, subtending an angle dθd\theta at the center.

  • Radius of the ring: r=Rsin⁡θr = R \sin\theta
  • Area of the ring strip: dA=2πRsin⁡θ⋅Rdθ=2πR2sin⁡θdθdA = 2\pi R \sin\theta \cdot R d\theta = 2\pi R^2 \sin\theta d\theta
  • Charge on the ring strip: dq=σdA=(Q4πR2)(2πR2sin⁡θdθ)=Q2sin⁡θdθdq = \sigma dA = \left(\frac{Q}{4\pi R^2}\right) (2\pi R^2 \sin\theta d\theta) = \frac{Q}{2} \sin\theta d\theta

As the sphere rotates with uniform angular speed ω\omega, the rotating ring produces a current dIdI: dI=dqT=ω2πdq=Qω4πsin⁡θdθdI = \frac{dq}{T} = \frac{\omega}{2\pi} dq = \frac{Q \omega}{4\pi} \sin\theta d\theta

The magnetic dipole moment dμd\mu of this elemental ring is: dμ=dI⋅(πr2)=(Qω4πsin⁡θdθ)π(Rsin⁡θ)2=14QωR2sin⁡3θdθd\mu = dI \cdot (\pi r^2) = \left(\frac{Q \omega}{4\pi} \sin\theta d\theta\right) \pi (R \sin\theta)^2 = \frac{1}{4} Q \omega R^2 \sin^3\theta d\theta

Integrating over the entire surface from θ=0\theta = 0 to θ=π\theta = \pi: μ=∫0π14QωR2sin⁡3θdθ=14QωR2∫0πsin⁡3θdθ\mu = \int_0^{\pi} \frac{1}{4} Q \omega R^2 \sin^3\theta d\theta = \frac{1}{4} Q \omega R^2 \int_0^{\pi} \sin^3\theta d\theta

Using the integral identity ∫0πsin⁡3θdθ=43\int_0^{\pi} \sin^3\theta d\theta = \frac{4}{3}: μ=14QωR2⋅43=13QωR2\mu = \frac{1}{4} Q \omega R^2 \cdot \frac{4}{3} = \frac{1}{3} Q \omega R^2

2. Calculation of Angular Momentum (LL)

For a solid sphere of uniform mass MM and radius RR, the moment of inertia about its central axis is: I=25MR2I = \frac{2}{5} M R^2

The total angular momentum LL about the axis of rotation is: L=Iω=25MR2ωL = I \omega = \frac{2}{5} M R^2 \omega

3. Ratio of Magnetic Dipole Moment to Angular Momentum

Taking the ratio of μ\mu to LL: μL=13QωR225MR2ω=56QM\frac{\mu}{L} = \frac{\frac{1}{3} Q \omega R^2}{\frac{2}{5} M R^2 \omega} = \frac{5}{6} \frac{Q}{M}

We can rewrite this in terms of Q2M\frac{Q}{2M}: μL=53(Q2M)\frac{\mu}{L} = \frac{5}{3} \left(\frac{Q}{2M}\right)

Comparing this with the given expression αQ2M\alpha \frac{Q}{2M}, we get: α=53≈1.67\alpha = \frac{5}{3} \approx 1.67

Ratio of Magnetic Dipole Moment to Angular Momentum for Rotating Sphere | Physics PYQ Solution - JEE Challenger