Question

a thin dielectric disc uniformly distributed with charge q has radius r and is rotated n times per second about an Axis perpendicular to the disc and passing through the centre find the magnetic induction at the centre of the disc​

Answer 1

Answer:

the magnetic induction at the centre of the disc​

$B = \frac{\mu_0 q n}{r}$

Explanation:

As we know that the disc is uniformly charged with charge q

it is rotated by frequency "n"

Now we will consider a small ring of radius "x" of thickness "dx"

charge on this small ring is given as

$dq = \frac{q}{\pi r^2}(2\pi x dx)$

$dq = \frac{2qxdx}{r^2}$

now the electric current due to rotation of this small ring

$di = q f$

$di = \frac{2qxdx}{r^2} n$

now the magnetic field due to this ring at its center is given as

$dB = \frac{\mu_0 di}{2x}$

$dB = \frac{\mu_0 2qxdx n}{2x r^2}$

now total magnetic field due to whole disc is given as

$B = \int \frac{\mu_0 q n dx}{r^2}$

$B = \frac{\mu_0 q n}{r}$

#Learn

Topic : Magnetic field due to circular ring

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