A charge Q is uniformly distributed over the surface of non-conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its center with an angular velocity \omega . As a result of this rotation a magnetic field of induction B is obtained at the center of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and very the radius of the disc then the variation of the magnetic induction at the center of the disc will be represented by the figure :

Magnetics PYQ — JEE Main 2012 · Shift 0 · Q53 with Solution

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Correct answer: A

Surface charge density on the disc

Since total charge $Q$ is uniformly distributed over a non-conducting disc of radius $R$ , the surface charge density is

$\sigma = \frac{Q}{\pi R^2}.$

Take a thin ring element

Consider a ring of radius $r$ and thickness $dr$ on the disc.

Its area is

$dA = 2\pi r\,dr.$

So the charge on this ring is

$dq = \sigma \, dA = \sigma (2\pi r\,dr).$

Current due to rotating ring

When the disc rotates with angular velocity $\omega$ , each ring behaves like a current loop.

Time period of rotation is

$T = \frac{2\pi}{\omega}.$

Hence current due to the ring is

$dI = \frac{dq}{T} = dq\cdot \frac{\omega}{2\pi}.$

Substitute $dq$ :

$dI = \sigma (2\pi r\,dr)\cdot \frac{\omega}{2\pi} = \sigma \omega r\,dr.$

Magnetic field at center due to the ring

Magnetic field at the center of a circular loop of radius $r$ carrying current $dI$ is

$dB = \frac{\mu_0 dI}{2r}.$

Substitute $dI$ :

$dB = \frac{\mu_0}{2r}(\sigma \omega r\,dr)= \frac{\mu_0 \sigma \omega}{2}\,dr.$

This is independent of $r$ .

Integrate over the whole disc

= \frac{\mu_0 \sigma \omega}{2}R.$$ Now substitute $\sigma = \dfrac{Q}{\pi R^2}$: $$B = \frac{\mu_0 \omega}{2} \cdot \frac{Q}{\pi R^2} \cdot R = \frac{\mu_0 Q\omega}{2\pi R}.$$ 6. **Dependence on radius** Thus, $$B \propto \frac{1}{R}$$ when $Q$ and $\omega$ are kept constant. So the graph of $B$ versus $R$ is a decreasing rectangular-hyperbola type curve. 7. **Option selection** Therefore the correct figure is the one representing $$B = \frac{k}{R}$$ which corresponds to **Option A**.

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View written solutionFree

Correct answer: A

Surface charge density on the disc

Since total charge $Q$ is uniformly distributed over a non-conducting disc of radius $R$ , the surface charge density is

$\sigma = \frac{Q}{\pi R^2}.$

Take a thin ring element

Consider a ring of radius $r$ and thickness $dr$ on the disc.

Its area is

$dA = 2\pi r\,dr.$

So the charge on this ring is

$dq = \sigma \, dA = \sigma (2\pi r\,dr).$

Current due to rotating ring

When the disc rotates with angular velocity $\omega$ , each ring behaves like a current loop.

Time period of rotation is

$T = \frac{2\pi}{\omega}.$

Hence current due to the ring is

$dI = \frac{dq}{T} = dq\cdot \frac{\omega}{2\pi}.$

Substitute $dq$ :

$dI = \sigma (2\pi r\,dr)\cdot \frac{\omega}{2\pi} = \sigma \omega r\,dr.$

Magnetic field at center due to the ring

Magnetic field at the center of a circular loop of radius $r$ carrying current $dI$ is

$dB = \frac{\mu_0 dI}{2r}.$

Substitute $dI$ :

$dB = \frac{\mu_0}{2r}(\sigma \omega r\,dr)= \frac{\mu_0 \sigma \omega}{2}\,dr.$

This is independent of $r$ .

Integrate over the whole disc

= \frac{\mu_0 \sigma \omega}{2}R.$$ Now substitute $\sigma = \dfrac{Q}{\pi R^2}$: $$B = \frac{\mu_0 \omega}{2} \cdot \frac{Q}{\pi R^2} \cdot R = \frac{\mu_0 Q\omega}{2\pi R}.$$ 6. **Dependence on radius** Thus, $$B \propto \frac{1}{R}$$ when $Q$ and $\omega$ are kept constant. So the graph of $B$ versus $R$ is a decreasing rectangular-hyperbola type curve. 7. **Option selection** Therefore the correct figure is the one representing $$B = \frac{k}{R}$$ which corresponds to **Option A**.

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