Question

A right circular solid cone of radius 10 cm, made of with non-conducting material has its axis aligned along z-axis as shown. The cone has a mass $m$ and uniform surface charge $q=2C$. At $t=0$, a time varying magnetic field $\vec{B}=(2t\hat{k})T$ is switched on. The torque on the cone is $\alpha \times 10^{-3}$ Joule. Find $\alpha$

Answer 1

10

Answer 2

The problem asks for the torque on a non-conducting cone with a uniform surface charge in a time-varying magnetic field.

1. Induced Electric Field: The magnetic field is given by $\vec{B} = (2t\hat{k}) \text{ T}$. This is a time-varying magnetic field, which induces an electric field according to Faraday's Law of Induction: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$ Consider a circular loop of radius $r$ in the xy-plane. The magnetic flux through this loop is: $\Phi_B = \vec{B} \cdot \vec{A} = (2t\hat{k}) \cdot (\pi r^2 \hat{k}) = 2t \pi r^2$ The rate of change of magnetic flux is: $\frac{d\Phi_B}{dt} = \frac{d}{dt}(2t \pi r^2) = 2\pi r^2$ The induced electric field $\vec{E}$ will be azimuthal (tangential to the circular loop). Let its magnitude be $E$. The line integral around the loop is $E(2\pi r)$. So, $E(2\pi r) = -2\pi r^2$. This gives $E = -r$. The negative sign indicates the direction. By Lenz's law, if $\vec{B}$ is increasing in the $+\hat{k}$ direction, the induced electric field will drive a current that produces a magnetic field in the $-\hat{k}$ direction. This corresponds to a clockwise direction when viewed from above (positive z-axis). In cylindrical coordinates, the azimuthal unit vector $\hat{\phi}$ points counter-clockwise. Therefore, the induced electric field is: $\vec{E} = -r \hat{\phi}$

2. Torque on a Charge Element: The cone has a uniform surface charge $q$. Let $dq$ be an infinitesimal charge element on the surface of the cone. The force on this charge element is $d\vec{F} = dq \vec{E} = dq (-r \hat{\phi})$. The position vector of the charge element from the origin (center of the base) can be written as $\vec{r}_{pos} = r\hat{r} + z\hat{k}$, where $r$ is the radial distance from the z-axis and $z$ is the height from the base. The torque on this charge element is $d\vec{\tau} = \vec{r}_{pos} \times d\vec{F}$: $d\vec{\tau} = (r\hat{r} + z\hat{k}) \times (-r dq \hat{\phi})$ Using the cross product identities $\hat{r} \times \hat{\phi} = \hat{k}$ and $\hat{k} \times \hat{\phi} = -\hat{r}$: $d\vec{\tau} = -r^2 dq (\hat{r} \times \hat{\phi}) - zr dq (\hat{k} \times \hat{\phi})$ $d\vec{\tau} = -r^2 dq \hat{k} - zr dq (-\hat{r})$ $d\vec{\tau} = -r^2 dq \hat{k} + zr dq \hat{r}$

3. Total Torque on the Cone: The total torque is the integral of $d\vec{\tau}$ over the entire surface of the cone: $\vec{\tau} = \int d\vec{\tau} = \int (-r^2 dq \hat{k} + zr dq \hat{r})$ $\vec{\tau} = \left( -\int r^2 dq \right) \hat{k} + \int zr dq \hat{r}$ Due to the symmetry of the cone and the uniform charge distribution, the integral $\int zr dq \hat{r}$ will cancel out to zero. For every charge element $dq$ at $(x,y,z)$, there are corresponding elements at $(-x,y,z)$, $(x,-y,z)$, and $(-x,-y,z)$ that will cause the $\hat{i}$ and $\hat{j}$ components of the torque to sum to zero. Thus, the net torque is purely in the $\hat{k}$ direction: $\vec{\tau} = -\left( \int r^2 dq \right) \hat{k}$

4. Calculating $\int r^2 dq$: The total charge $q$ is uniformly distributed over the surface of the cone. The surface consists of the circular base and the curved lateral surface. Let $\sigma$ be the uniform surface charge density. $dq = \sigma dA$ So, we need to calculate $\int r^2 \sigma dA = \sigma \int r^2 dA$. Let $A_{total}$ be the total surface area of the cone. $\sigma = q/A_{total}$. The integral $\int r^2 dA$ can be split into two parts: one for the base and one for the curved surface.

• For the base (a disk of radius $R$ at $z=0$): $dA_{base} = r' dr' d\phi'$, where $r'$ is the radial coordinate. $\int_{base} r^2 dA = \int_0^{2\pi} \int_0^R (r')^2 r' dr' d\phi' = 2\pi \int_0^R (r')^3 dr' = 2\pi \left[ \frac{(r')^4}{4} \right]_0^R = \frac{\pi R^4}{2}$

• For the curved surface: Let $s$ be the distance along the slant height from the vertex. The radius of the cone at this position is $r' = \frac{R}{L} s$, where $L$ is the slant height of the cone ($L = \sqrt{R^2+H^2}$). An infinitesimal area element on the curved surface is $dA_{curved} = 2\pi r' ds$. $\int_{curved} r^2 dA = \int_0^L (r')^2 (2\pi r' ds) = \int_0^L \left(\frac{R}{L}s\right)^2 2\pi \left(\frac{R}{L}s\right) ds$ $= 2\pi \frac{R^3}{L^3} \int_0^L s^3 ds = 2\pi \frac{R^3}{L^3} \left[ \frac{s^4}{4} \right]_0^L = 2\pi \frac{R^3}{L^3} \frac{L^4}{4} = \frac{\pi R^3 L}{2}$

Now, sum these contributions for $\int r^2 dq$: $\int r^2 dq = \sigma \left( \frac{\pi R^4}{2} + \frac{\pi R^3 L}{2} \right)$ The total surface area $A_{total} = \pi R^2 + \pi R L = \pi R (R+L)$. Substitute $\sigma = \frac{q}{\pi R (R+L)}$: $\int r^2 dq = \frac{q}{\pi R (R+L)} \left( \frac{\pi R^4}{2} + \frac{\pi R^3 L}{2} \right)$ $= \frac{q}{\pi R (R+L)} \frac{\pi R^3 (R+L)}{2}$ $= \frac{q R^2}{2}$ This result is independent of the cone's height $H$ (which is implicitly in $L$).

5. Calculate the Torque Magnitude: The magnitude of the torque is $\tau = \frac{1}{2} q R^2$. Given values: $q = 2 \text{ C}$ $R = 10 \text{ cm} = 0.1 \text{ m}$ $\tau = \frac{1}{2} (2 \text{ C}) (0.1 \text{ m})^2$ $\tau = 1 \times (0.01) \text{ Joule}$ $\tau = 0.01 \text{ Joule}$ To express this in the form $\alpha \times 10^{-3}$ Joule: $\tau = 10 \times 10^{-3} \text{ Joule}$ Comparing this with $\alpha \times 10^{-3}$ Joule, we find $\alpha = 10$.