Let's solve the problem step by step. --- ## **Step 1: Understanding the Problem** - **Disk of radius** \( a \) **in the \(xy\)-plane** - **Uniform surface charge density:** \( \rho_s \) (C/m\(^2\)) - **Disk rotates about \(z\)-axis** at angular velocity \( \omega \) (rad/s) - **Find:** Magnetic field intensity \( \vec{H} \) at any point on the \(z\)-axis --- ## **Step 2: Surface Current Density** When charged disk rotates, each point at radius \( r \) has velocity \[ \vec{v} = \omega r \, \hat{\phi} \] The **surface current density** is: \[ \vec{K} = \rho_s \vec{v} = \rho_s \omega r \, \hat{\phi} \] --- ## **Step 3: Magnetic Field on the \(z\)-axis** By **symmetry**, the field on the \(z\)-axis points along \( \hat{z} \). We can use the **Biot-Savart Law** for surface currents: \[ \vec{H} = \frac{1}{4\pi} \int_S \frac{\vec{K} \times \vec{R}}{R^3} dS \] where: - \(\vec{R}\) is the vector from the surface element to the field point (on z-axis at \(z\)) --- ## **Step 4: Setting Up the Integral** Let a surface element on the disk be at \( (r', \phi', ) \). Let the field point be at \( (, , z) \). \[ \vec{R} = (,,z) - (r'\cos\phi', r'\sin\phi', ) = (-r'\cos\phi', -r'\sin\phi', z) \] \[ R = \sqrt{r'^2 + z^2} \] \[ \vec{K} = \rho_s \omega r' \hat{\phi}' \] where \( \hat{\phi}' = -\sin\phi' \hat{x} + \cos\phi' \hat{y} \) --- ## **Step 5: Cross Product** \[ \vec{K} \times \vec{R} \] Let's write out: \[ \vec{K} = \rho_s \omega r' (-\sin\phi' \hat{x} + \cos\phi' \hat{y}) \] \[ \vec{R} = -r'\cos\phi' \hat{x} - r'\sin\phi' \hat{y} + z \hat{z} \] Now, cross each term: - \( (-\sin\phi' \hat{x}) \times (-r'\cos\phi' \hat{x}) = \) - \( (-\sin\phi' \hat{x}) \times (-r'\sin\phi' \hat{y}) = -\sin\phi' \hat{x} \times -r'\sin\phi' \hat{y} = r' \sin^2\phi' \hat{z} \) - \( (-\sin\phi' \hat{x}) \times (z \hat{z}) = -z\sin\phi' (\hat{x} \times \hat{z}) = -z\sin\phi' (-\hat{y}) = z\sin\phi' \hat{y} \) - \( (\cos\phi' \hat{y}) \times (-r'\cos\phi' \hat{x}) = -r'\cos\phi' \cos\phi' (\hat{y} \times \hat{x}) = -r'\cos^2\phi' (-\hat{z}) = r'\cos^2\phi' \hat{z} \) - \( (\cos\phi' \hat{y}) \times (-r'\sin\phi' \hat{y}) = \) - \( (\cos\phi' \hat{y}) \times (z \hat{z}) = z\cos\phi' (\hat{y} \times \hat{z}) = z\cos\phi' \hat{x} \) Sum all z-components: \[ (r'\sin^2\phi' + r'\cos^2\phi') \hat{z} = r'(\sin^2\phi' + \cos^2\phi') \hat{z} = r' \hat{z} \] Sum all x and y components: - x: \( z\cos\phi' \) - y: \( z\sin\phi' \) So, \[ \vec{K} \times \vec{R} = \rho_s \omega r' \left[ z\cos\phi' \hat{x} + z\sin\phi' \hat{y} + r' \hat{z} \right] \] --- ## **Step 6: Integrate Over the Disk** Only the \( \hat{z} \) component survives when integrating over \( \phi' \) from to \( 2\pi \): \[ \vec{H}(,,z) = \frac{1}{4\pi} \int_^a \int_^{2\pi} \frac{\rho_s \omega r' \cdot r' \hat{z}}{(r'^2 + z^2)^{3/2}} r' d\phi' dr' \] (Area element: \( dS = r' d\phi' dr' \)) \[ \vec{H}(,,z) = \frac{\rho_s \omega}{4\pi} \hat{z} \int_^a \int_^{2\pi} \frac{r'^2}{(r'^2 + z^2)^{3/2}} d\phi' dr' \] \[ \int_^{2\pi} d\phi' = 2\pi \] \[ \vec{H}(,,z) = \frac{\rho_s \omega}{2} \hat{z} \int_^a \frac{r'^2}{(r'^2 + z^2)^{3/2}} dr' \] --- ## **Step 7: Solve the Integral** Let’s solve: \[ I = \int_^a \frac{r'^2}{(r'^2 + z^2)^{3/2}} dr' \] Let \( u = r'^2 + z^2 \implies du = 2r' dr' \implies dr' = \frac{du}{2r'} \) When \( r' = , u = z^2 \); when \( r' = a, u = a^2 + z^2 \) But let's use substitution: Let’s use \( r' = z \tan\theta \implies dr' = z \sec^2\theta d\theta \) Then \( r'^2 + z^2 = z^2 \sec^2\theta \), \( r'^2 = z^2 \tan^2\theta \) So: \[ I = \int_{\theta=}^{\arctan(a/z)} \frac{z^2 \tan^2\theta}{(z^2 \sec^2\theta)^{3/2}} z \sec^2\theta d\theta \] \[ = \int_^{\arctan(a/z)} \frac{z^2 \tan^2\theta}{z^3 \sec^3\theta} z \sec^2\theta d\theta \] \[ = \int_^{\arctan(a/z)} \frac{z^3 \tan^2\theta \sec^2\theta}{z^3 \sec^3\theta} d\theta \] \[ = \int_^{\arctan(a/z)} \tan^2\theta \frac{1}{\sec\theta} d\theta = \int_^{\arctan(a/z)} \tan^2\theta \cos\theta d\theta \] Let’s use \( \tan^2\theta = \sec^2\theta - 1 \): \[ = \int (\sec^2\theta - 1) \cos\theta d\theta = \int \sec^2\theta \cos\theta d\theta - \int \cos\theta d\theta \] \[ \int \sec^2\theta \cos\theta d\theta = \int \frac{1}{\cos^2\theta} \cos\theta d\theta = \int \frac{1}{\cos\theta} d\theta = \int \sec\theta d\theta \] So, \[ I = \int \sec\theta d\theta - \int \cos\theta d\theta = \ln|\sec\theta + \tan\theta| - \sin\theta \] Back to the limits: When \( r' = \), \( \tan\theta = \rightarrow \theta = \) When \( r' = a \), \( \tan\theta = a/z \rightarrow \theta = \arctan(a/z) \) So, \[ I = \left[\ln|\sec\theta + \tan\theta| - \sin\theta \right]_^{\arctan(a/z)} \] Compute at \( \theta = \arctan(a/z) \): - \( \tan\theta = a/z \) - \( \sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + (a/z)^2} = \sqrt{z^2 + a^2}/z \) - \( \sin\theta = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}} = \frac{a/z}{\sqrt{1 + (a/z)^2}} = \frac{a}{\sqrt{z^2 + a^2}} \) At \( \theta = \): - \( \tan = \), \( \sec = 1 \), \( \sin = \) So, \[ I = \left[ \ln\left( \frac{\sqrt{z^2 + a^2}}{z} + \frac{a}{z} \right) - \frac{a}{\sqrt{z^2 + a^2}} \right] - [\ln(1) - ] \] \[ = \ln\left( \frac{\sqrt{z^2 + a^2} + a}{z} \right) - \frac{a}{\sqrt{z^2 + a^2}} \] --- ## **Step 8: Final Expression** \[ \boxed{ \vec{H}(,,z) = \frac{\rho_s \omega}{2} \left[ \ln\left( \frac{\sqrt{z^2 + a^2} + a}{z} \right) - \frac{a}{\sqrt{z^2 + a^2}} \right] \hat{z} } \] --- ### **Summary** - Compute surface current density: \( \vec{K} = \rho_s \omega r \hat{\phi} \) - Use Biot-Savart law for surface currents - Integrate over the disk and evaluate on the \(z\)-axis - Result above gives the magnetic field intensity at any point on the \(z\)-axis due to the rotating charged disk --- **Let me know if you need further clarification or a check for special cases (e.g., on the axis near the center or far away)!**