## Step 1: Find the **Current Density** Due to the Rotating Surface Charge The cylinder of radius \(a\) has a **surface charge density** \(\rho_s\) and rotates with **angular velocity** \(\omega\). The linear velocity at radius \(a\) is \(v = \omega a\). This moving surface charge forms a **surface current density** (\(K\)): \[ K = \rho_s v = \rho_s \omega a \] **Explanation:** The surface charge is moving in a circle at the surface, so we treat it as a current per unit length (surface current). The total current per unit length around the cylinder is \(I = K \cdot 2\pi a\). --- ## Step 2: Use **Ampère's Law** to Find the Magnetic Field Ampère's law for a long cylinder (by symmetry, \( \vec{B} \) is along the axis, i.e., azimuthal symmetry): \[ \oint \vec{B} \cdot d\vec{l} = \mu_ I_{\text{enc}} \] - **Inside the cylinder (\(r < a\)):** No current is enclosed (\(I_{\text{enc}} = \)), so \[ B_{\text{inside}} = \] - **Outside the cylinder (\(r > a\)):** The total current is surface current times the circumference: \[ I_{\text{enc}} = K \cdot 2\pi a = \rho_s \omega a \cdot 2\pi a = 2\pi \rho_s \omega a^2 \] The path integral is \(B \cdot 2\pi r\), so: \[ B_{\text{outside}} \cdot 2\pi r = \mu_ \cdot 2\pi \rho_s \omega a^2 \] \[ B_{\text{outside}} = \mu_ \rho_s \omega \frac{a^2}{r} \] **Explanation:** We use Ampère's law to relate the magnetic field to the current enclosed by a circular loop of radius \(r\). Inside the cylinder, there's no enclosed current, so the field is zero. Outside, the surface current produces a field that decreases with \(1/r\). --- ## **Final Answer** \[ \boxed{ B = \begin{cases} , & r < a \\ \mu_ \rho_s \omega \dfrac{a^2}{r}, & r > a \end{cases} } \] Where: - \(\mu_\) is the permeability of free space, - \(\rho_s\) is the surface charge density, - \(\omega\) is the angular velocity, - \(a\) is the cylinder radius, - \(r\) is the distance from the axis.

# Solution to the Problem ## Step 1: Find the **Current Density** Due to the Rotating Surface Charge The cylinder is uniformly charged with a surface charge density \(\rho_s\) and rotates with an angular velocity \(\omega\). The linear velocity of a point on the surface of the cylinder is given by: \[ v = \omega a \] The surface current density \(K\) can be calculated using the formula: \[ K = \rho_s v = \rho_s \omega a \] **Explanation:** The surface charge on the cylinder moves in a circular path due to rotation, creating an effective surface current. This current density \(K\) represents the amount of charge passing through a unit length of the cylinder per unit time. --- ## Step 2: Use **Ampère's Law** to Find the Magnetic Field ### Inside the Cylinder (\(r < a\)) For points inside the cylinder, there is no enclosed current (\(I_{\text{enc}} = 0\)). By Ampère's Law: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} = 0 \] Thus, the magnetic field inside the cylinder is: \[ B_{\text{inside}} = 0 \] ### Outside the Cylinder (\(r > a\)) For points outside the cylinder, we consider the total current enclosed by a circular loop of radius \(r\): \[ I_{\text{enc}} = K \cdot 2\pi a = \rho_s \omega a \cdot 2\pi a = 2\pi \rho_s \omega a^2 \] Using Ampère's law: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \implies B_{\text{outside}} \cdot 2\pi r = \mu_0 \cdot 2\pi \rho_s \omega a^2 \] Solving for \(B_{\text{outside}}\): \[ B_{\text{outside}} = \frac{\mu_0 \rho_s \omega a^2}{r} \] **Explanation:** The magnetic field outside the cylinder is derived from the total current produced by the rotating surface charge. The field decreases with the distance \(r\) from the axis of the cylinder, indicating that the influence of the rotating charge diminishes further away. --- ## **Final Answer** \[ \boxed{ B = \begin{cases} 0, & r < a \\ \mu_0 \rho_s \omega \frac{a^2}{r}, & r > a \end{cases} } \] Where: - \(\mu_0\) is the permeability of free space, - \(\rho_s\) is the surface charge density, - \(\omega\) is the angular velocity, - \(a\) is the cylinder radius, - \(r\) is the distance from the axis of the cylinder.