Let's solve this **step by step** carefully. --- ### **Given:** * Disk of radius ( a ) lies in the **x–y plane**, centered on the z-axis. * Surface charge density: ( \rho_s ) (C/m²). * Disk rotates about the **z-axis** with **angular velocity** ( \omega ) (rad/s). * We need **magnetic field intensity** ( \vec{H} ) at a point **on the z-axis**. --- ### **Step 1: Express surface current density** When a surface charge moves, it produces a **surface current density**: [ \vec{K} = \rho_s \vec{v} ] At a distance ( r ) from the axis, the **linear velocity** of rotation is: [ \vec{v} = \omega r , \hat{\phi} ] Hence, [ \boxed{\vec{K} = \rho_s \omega r , \hat{\phi}} ] --- ### **Step 2: Divide the disk into differential current loops** Consider a **ring element** of radius ( r ) and thickness ( dr ). * Circumference: ( 2\pi r ) * Surface current: ( K = \rho_s \omega r ) * The equivalent current in the ring: [ I = K \times \text{width} = \rho_s \omega r , dr ] --- ### **Step 3: Magnetic field on the z-axis due to a circular loop** For a current loop of radius ( r ) carrying current ( I ), the **magnetic field** along the axis (z-axis) is: [ dB_z = \frac{\mu_0 I r^2}{2(r^2 + z^2)^{3/2}} ] Substitute ( I = \rho_s \omega r , dr ): [ dB_z = \frac{\mu_0 \rho_s \omega r^3}{2(r^2 + z^2)^{3/2}} , dr ] --- ### **Step 4: Integrate over the entire disk** [ B_z = \int_0^a dB_z ] [ B_z = \frac{\mu_0 \rho_s \omega}{2} \int_0^a \frac{r^3}{(r^2 + z^2)^{3/2}} , dr ] --- ### **Step 5: Solve the integral** Let ( u = r^2 + z^2 \Rightarrow du = 2r , dr \Rightarrow r^3 dr = \frac{1}{2}(u - z^2) du ) [ \Rightarrow B_z = \frac{\mu_0 \rho_s \omega}{4} \int_{z^2}^{a^2 + z^2} \frac{u - z^2}{u^{3/2}} , du ] Simplify the integrand: [ \frac{u - z^2}{u^{3/2}} = u^{-1/2} - z^2 u^{-3/2} ] Now integrate: [ \int u^{-1/2} du = 2u^{1/2}, \quad \int u^{-3/2} du = -2u^{-1/2} ] So, [ B_z = \frac{\mu_0 \rho_s \omega}{4} \left[ 2u^{1/2} + 2z^2 u^{-1/2} \right]_{z^2}^{a^2 + z^2} ] Simplify: [ B_z = \frac{\mu_0 \rho_s \omega}{2} \left[ \sqrt{a^2 + z^2} + \frac{z^2}{\sqrt{a^2 + z^2}} - 2z \right] ] [ \boxed{B_z = \frac{\mu_0 \rho_s \omega}{2} \left( \frac{a^2 + 2z^2}{\sqrt{a^2 + z^2}} - 2z \right)} ] --- ### **Step 6: Magnetic field intensity ( \vec{H} )** Since ( \vec{B} = \mu_0 \vec{H} ): [ \boxed{H_z = \frac{B_z}{\mu_0} = \frac{\rho_s \omega}{2} \left( \frac{a^2 + 2z^2}{\sqrt{a^2 + z^2}} - 2z \right)} ] and the direction is **along ( +\hat{z} )** (if rotation is counterclockwise when viewed from above). --- ### ✅ **Final Answers** [ \boxed{ \begin{aligned} \vec{K} &= \rho_s \omega r , \hat{\phi} \ \vec{B} &= \frac{\mu_0 \rho_s \omega}{2} \left( \frac{a^2 + 2z^2}{\sqrt{a^2 + z^2}} - 2z \right) \hat{z} \ \vec{H} &= \frac{\rho_s \omega}{2} \left( \frac{a^2 + 2z^2}{\sqrt{a^2 + z^2}} - 2z \right) \hat{z} \end{aligned} } ] --- ### **Step 7: Check special cases** 1. **At center (z = 0):** [ B_z = \frac{\mu_0 \rho_s \omega a}{2}, \quad H_z = \frac{\rho_s \omega a}{2} ] 2. **At far field (z ≫ a):** [ B_z \approx \frac{\mu_0 \rho_s \omega a^4}{4z^3} ] (decays like a magnetic dipole field) --- Please give me this content same as it is in render equations without any erros