Q.(a)

(a) A changing enclosed area induces a motional emf $\varepsilon=Bvl$; for a disc of area 0.03 m² spinning at 20 rev/s in a 0.4 T field, the induced emf between axis and rim works out to about 0.24 V. (b) Brewster's law $\tan\theta_B=\mu$ gives the polarising angle; for glass of $\mu=1.65$, the plate must be tilted about 31.2° to the horizontal. Both alternatives answered below.

(a)(i) Inducing emf by changing the enclosed area

Consider a conducting rod sliding with velocity $v$ along two parallel rails, separation $l$, in a uniform field $B$ perpendicular to the plane of the rails. The flux linked, $\Phi=BA$, changes as the enclosed area $A$ changes at rate $dA/dt=lv$:

$\varepsilon = -\dfrac{d\Phi}{dt} = -B\dfrac{dA}{dt} = -Bvl$

The magnitude, $\varepsilon=Bvl$, is called the motional emf; by Lenz's law its polarity opposes the rod's motion.

(a)(ii) EMF of a rotating disc

Formula. For a conducting disc of radius $R$ rotating at angular speed $\omega$ about its centre, with a uniform field $B$ parallel to the rotation axis, the emf induced between the centre (axis) and the rim is (obtained by integrating $\varepsilon=B\omega r\,dr$ from $0$ to $R$, since each annular element at radius $r$ moves at speed $\omega r$):

$\varepsilon = \dfrac{1}{2}B\omega R^2$

Given: Area $A=0.03\ \text{m}^2$, so $R=\sqrt{A/\pi}=\sqrt{0.03/\pi}\approx0.0977$ m. $B=0.4$ T. Frequency $=20$ rev/s, so $\omega=2\pi\times20\approx125.66\ \text{rad/s}$.

$\varepsilon = \dfrac{1}{2}(0.4)(125.66)(0.0977)^2 = \dfrac{1}{2}(0.4)(125.66)(0.00955)$

$\varepsilon \approx 0.24\ \text{V}$

(The resistance of the disc, 4 Ω, is extra information not needed to find the induced emf itself -- it would be needed to find the current, $I=\varepsilon/R_{disc}\approx0.06$ A.)

(b)(i) Brewster's Law

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