To understand how the magnetic field at the center of a rotating charged disc varies with its radius, let's break down the problem step by step.

Step 1: Magnetic Field Due to a Rotating Charged Disc at Its Center
A disc of radius R has a uniform surface charge density σ. The total charge Q is constant, so σ = Q / (πR²). The disc rotates with angular velocity ω. Each point on the disc acts like a current loop. The magnetic field at the center due to a ring of radius r and width dr is given by the formula for a circular loop: dB = (μ₀ / 2) * (dI / r), where dI is the current due to the ring.

Step 2: Find the Current dI for a Ring
The charge on a ring of radius r and width dr is dq = σ * 2πr dr. As the disc rotates, this charge moves, creating a current. The current dI = (charge per unit time) = (dq) * (frequency) = dq * (ω / 2π) = (σ * 2πr dr) * (ω / 2π) = σ ω r dr.

Step 3: Magnetic Field Contribution from the Ring
Substituting dI into dB: dB = (μ₀ / 2) * (dI / r) = (μ₀ / 2) * (σ ω r dr / r) = (μ₀ / 2) σ ω dr.

Step 4: Integrate to Find Total Magnetic Field B
Integrate dB from r = 0 to r = R: $B={\int }_0^R\frac{{\mu }_0}{2}\sigma \omega dr=\frac{{\mu }_0\sigma \omega }{2}{\int }_0^{R}dr=\frac{{\mu }_0\sigma \omega R}{2}$

Step 5: Substitute σ in Terms of Q and R
Since σ = Q / (πR²), substitute: $B=\frac{{\mu }_0\omega R}{2}·\frac{Q}{\pi R^2}=\frac{{\mu }_{0}Q\omega }{2\pi R}$

Step 6: Analyze Variation with R
From B = (μ₀ Q ω) / (2π R), we see that B is inversely proportional to R, i.e., B ∝ 1/R. Since Q and ω are constant, as R increases, B decreases. The graph of B vs. R should be a rectangular hyperbola in the first quadrant, starting from a high value at small R and approaching zero as R becomes large.

Final Answer: The correct figure is the one where B decreases hyperbolically with increasing R.

Related Topics

Magnetic Field Due to Current Loops: Understanding how current-carrying loops produce magnetic fields is fundamental. For a circular loop, B at center is μ₀I/(2R).

Surface Charge Density: Uniform distribution on surfaces and how it relates to total charge and area.

Integration in Physics: Summing infinitesimal contributions, like here for rings on a disc, is a common technique.

Formulae

Magnetic field at center of a circular loop: $B=\frac{{\mu }_{0}I}{2R}$

Surface charge density: $\sigma =\frac{Q}{A}=\frac{Q}{\pi R^2}$

Current due to rotating charge element: dI = (dq) * (ω / 2π)

Magnetic field at center of rotating disc: $B=\frac{{\mu }_{0}Q\omega }{2\pi R}$