Question

A circular uniform disc of radius $r=2m$, mass $M=2kg$ and uniform charge distribution $\sigma=2C/m^2$ is initially at rest. A uniform magnetic field of initial value $B_0=0.2T$, perpendicular to the plane of the disc, exists in a circular region of radius $a=1m$, concentric with the disc. The magnetic field is then switched off and uniformly decreases from $B_0$ to zero in time $\tau$. Assuming that the disc is free to rotate about its center, find the angular velocity of the disc.

Answer 1

0.175π rad/s

Answer 2

The problem asks us to find the final angular velocity of a charged disc that starts rotating due to an induced electric field when a perpendicular magnetic field decreases.

1. Induced Electric Field (E):

A changing magnetic flux induces an electric field. According to Faraday's Law: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$

The magnetic field $B$ is uniform within a radius $a$ and zero outside. It decreases uniformly from $B_0$ to $0$ in time $\tau$. So, $\frac{dB}{dt} = -\frac{B_0}{\tau}$.

Consider a circular path of radius $x$ concentric with the disc. Due to symmetry, the induced electric field $\vec{E}$ will be tangential and uniform in magnitude along this path.

• For $x \le a$ (inside the magnetic field region): The magnetic flux is $\Phi_B = B \cdot (\pi x^2)$. $E(2\pi x) = -\frac{d}{dt}(B\pi x^2) = -\pi x^2 \frac{dB}{dt}$ $E_1 = -\frac{x}{2} \frac{dB}{dt} = -\frac{x}{2} \left(-\frac{B_0}{\tau}\right) = \frac{B_0 x}{2\tau}$

• For $x > a$ (outside the magnetic field region, but within the disc): The magnetic flux is $\Phi_B = B \cdot (\pi a^2)$ (as the field only extends to radius $a$). $E(2\pi x) = -\frac{d}{dt}(B\pi a^2) = -\pi a^2 \frac{dB}{dt}$ $E_2 = -\frac{a^2}{2x} \frac{dB}{dt} = -\frac{a^2}{2x} \left(-\frac{B_0}{\tau}\right) = \frac{B_0 a^2}{2x\tau}$

2. Torque on the Disc (τ):

The disc has a uniform charge distribution $\sigma$. Consider an elemental ring of radius $x$ and thickness $dx$. The charge on this ring is $dq = \sigma (2\pi x dx)$. The force on this elemental charge is $dF = E dq = E (\sigma 2\pi x dx)$. The torque due to this force about the center is $d\tau = x dF = x E (\sigma 2\pi x dx)$.

We integrate this torque from $x=0$ to $x=r$. Since the expression for $E$ changes at $x=a$, we integrate in two parts:

• For $0 \le x \le a$: $d\tau_1 = x \left(\frac{B_0 x}{2\tau}\right) (\sigma 2\pi x dx) = \frac{\pi \sigma B_0}{\tau} x^3 dx$ $\tau_1 = \int_0^a \frac{\pi \sigma B_0}{\tau} x^3 dx = \frac{\pi \sigma B_0}{\tau} \left[\frac{x^4}{4}\right]_0^a = \frac{\pi \sigma B_0 a^4}{4\tau}$

• For $a < x \le r$: $d\tau_2 = x \left(\frac{B_0 a^2}{2x\tau}\right) (\sigma 2\pi x dx) = \frac{\pi \sigma B_0 a^2}{\tau} x dx$ $\tau_2 = \int_a^r \frac{\pi \sigma B_0 a^2}{\tau} x dx = \frac{\pi \sigma B_0 a^2}{\tau} \left[\frac{x^2}{2}\right]_a^r = \frac{\pi \sigma B_0 a^2}{2\tau} (r^2 - a^2)$

The total torque acting on the disc is $\tau_{total} = \tau_1 + \tau_2$: $\tau_{total} = \frac{\pi \sigma B_0 a^4}{4\tau} + \frac{\pi \sigma B_0 a^2}{2\tau} (r^2 - a^2)$ $\tau_{total} = \frac{\pi \sigma B_0 a^2}{4\tau} [a^2 + 2(r^2 - a^2)]$ $\tau_{total} = \frac{\pi \sigma B_0 a^2}{4\tau} [a^2 + 2r^2 - 2a^2]$ $\tau_{total} = \frac{\pi \sigma B_0 a^2}{4\tau} (2r^2 - a^2)$

3. Angular Impulse-Momentum Theorem:

The change in angular momentum of the disc is equal to the angular impulse applied by the torque. Since the torque is constant over the time $\tau$ (because $\frac{dB}{dt}$ is constant), the angular impulse $J_\theta = \tau_{total} \cdot \tau$. $J_\theta = \left(\frac{\pi \sigma B_0 a^2}{4\tau} (2r^2 - a^2)\right) \cdot \tau = \frac{\pi \sigma B_0 a^2}{4} (2r^2 - a^2)$

The initial angular momentum of the disc is $L_i = 0$ (since it's at rest). The final angular momentum is $L_f = I\omega$, where $I$ is the moment of inertia of the disc and $\omega$ is its final angular velocity. For a uniform disc, $I = \frac{1}{2} M r^2$.

According to the angular impulse-momentum theorem, $L_f - L_i = J_\theta$: $I\omega - 0 = J_\theta$ $\frac{1}{2} M r^2 \omega = \frac{\pi \sigma B_0 a^2}{4} (2r^2 - a^2)$

Solving for $\omega$: $\omega = \frac{\pi \sigma B_0 a^2 (2r^2 - a^2)}{2 M r^2}$

4. Substitute the given values: $r = 2m$ $M = 2kg$ $\sigma = 2C/m^2$ $B_0 = 0.2T$ $a = 1m$

$\omega = \frac{\pi (2) (0.2) (1)^2 (2(2)^2 - (1)^2)}{2 (2) (2)^2}$ $\omega = \frac{\pi (0.4) (1) (2(4) - 1)}{4 (4)}$ $\omega = \frac{\pi (0.4) (8 - 1)}{16}$ $\omega = \frac{\pi (0.4) (7)}{16}$ $\omega = \frac{2.8\pi}{16}$ $\omega = \frac{7\pi}{40}$ rad/s $\omega = 0.175\pi$ rad/s

The direction of rotation will be clockwise, as the decreasing magnetic field (into the page) induces a clockwise electric field, which exerts a clockwise force on the positive charges.