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Question: A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant a...

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40π rad s⁻¹ about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8×10⁻⁹ T, then the charge carried by the ring is close to (µ₀ = 4π×10⁻⁷ N/A²).

Answer

3.0×10⁻⁵ C

Explanation

Solution

Solution Explanation:

  1. A rotating charged ring is equivalent to a current loop with current
      I=QT=Qω2π.I=\frac{Q}{T}=\frac{Q\omega}{2\pi}.

  2. The magnetic field at the centre of a current loop is given by
      B=μ0I2R=μ0Qω4πR.B=\frac{\mu_0I}{2R}=\frac{\mu_0Q\omega}{4\pi R}.

  3. Solving for Q,
      Q=4πRBμ0ω.Q=\frac{4\pi R B}{\mu_0 \omega}.

  4. Substitute the given values:
      - R=0.1 mR=0.1\,m
      - B=3.8×10−9 TB=3.8\times10^{-9}\,T
      - μ0=4π×10−7 N/A2\mu_0=4\pi\times10^{-7}\,N/A^2
      - ω=40π rad/s\omega=40\pi\,rad/s

    Thus,

    Q=4π(0.1)(3.8×10−9)(4π×10−7)(40π)=1.52π×10−9160π2×10−7=1.52160×1π×10−2.Q=\frac{4\pi (0.1)(3.8\times10^{-9})}{(4\pi\times10^{-7})(40\pi)}=\frac{1.52\pi\times10^{-9}}{160\pi^2\times10^{-7}} =\frac{1.52}{160}\times\frac{1}{\pi}\times10^{-2}.

    Calculating,
    1.52160≈0.0095\frac{1.52}{160}\approx 0.0095 and so

    Q≈0.0095×10−2π≈9.5×10−53.14≈3.0×10−5 C.Q\approx\frac{0.0095\times10^{-2}}{\pi}\approx\frac{9.5\times10^{-5}}{3.14}\approx3.0\times10^{-5}\,C.

Answer:
3.0×10−5 C\boxed{3.0\times10^{-5}\,C} (Closest value)