Magnetic induction due to a rotating sphere

A sphere of radius R and surface charge density σ is rotating about an axis passing through its centre with angular velocity ω. Find the magnetic induction at the centre of the sphere.

3 Answers

Shaswata Roy ·

We consider a small ring making a polar angle θ with the equatorial plane.

\mathbf{di = \frac{\omega}{2\pi}\frac{\sigma (2\pi r^{2})\cos \theta d\theta}{4\pi r^{2}}=\frac{\omega \sigma}{4\pi}\cos \theta d\theta}

\mathbf{dB = \frac{\mu_{0}i\cos^{2}\theta}{2R}}

\mathbf{dB = \frac{\mu_{0}\omega\sigma \cos^{3}\theta d\theta}{8\pi R}}

B = \int^{\pi/2}_{-\pi/2}\mathbf{\frac{\mu_{0}\omega\sigma \cos^{3}\theta d\theta}{8\pi R}}

\mathbf{B = \frac{\mu_{0}\sigma \omega}{6\pi R}}

Soumyadeep Basu ·

The answer is given as 23μoσωR.

Sourish Ghosh ·

di = Rd\theta .2\pi R\cos{\theta}.\sigma.\frac{\omega }{2\pi} = \omega R^{2}\sigma \cos{\theta}d\theta

dB = \frac{\mu _{0}.\omega R^{2}\sigma \cos{\theta}d\theta.R^{2}\cos^{\2}\theta}{2R^{3}}

B = \frac{\mu _{0}\omega R\sigma }{2}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos^{3}\theta d\theta

B = \frac{2\mu _{0}\omega R\sigma }{3}

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