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.
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3 Answers
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}}
- Shaswata Roy Sorry!I had mistaken sigma to be the net charge on the sphere.Multiply by 4pi R^2 to get the required result.Upvote·0· Reply ·2013-08-20 11:15:05
- Soumyadeep Basu Thanks.
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}
- Soumyadeep Basu Thanks.