Production of Induced EMF by Changing Relative Orientation

What this figure shows. A rectangular coil of N turns and area A sits inside a uniform magnetic field $\vec B$, and is shown rotating anticlockwise with angular velocity $\omega$ about an axis perpendicular both to the field and to the plane of the page. At time t the coil's plane has rotated through angle $\theta=\omega t$ from its starting position (in which the coil's plane was perpendicular to the field, giving maximum flux linkage $\Phi_m=NBA$). The figure decomposes the field at this deflected instant into a component $B\cos\omega t$ normal to the coil's plane (the only component that contributes to flux linkage) and a component $B\sin\omega t$ lying IN the coil's plane (which contributes nothing to the flux), directly motivating the flux-linkage expressio …

What this figure shows. A sequence of six small coil diagrams, at $\omega t = 0, \pi/2, \pi, 3\pi/2, 2\pi$ and back near $0$, each paired with the corresponding value of the induced emf ($\varepsilon_m\sin\omega t$ evaluated as $0$, $\varepsilon_m$, $0$, $-\varepsilon_m$, $0$, and back towards $\varepsilon_m\sin0=0$ respectively), traces out how the emf rises to its positive peak, falls back through zero to its negative peak, and returns to zero over one complete rotation. The figure is the direct graphical companion to the derived formula $\varepsilon = \varepsilon_m\sin\omega t$, making visible that a full mechanical rotation of the coil produces exactly one full sinusoidal cycle of induced emf, …

Worked out. A rectangular coil of area 70 cm$^2$ with 600 turns rotates about an axis perpendicular to a field of 0.4 Wb/m$^2$, completing 500 revolutions per minute, and the instantaneous emf is required when the coil's plane is (i) perpendicular to the field, (ii) parallel to the field, and (iii) inclined at $60^{\circ}$ to the field. First the peak emf is found: $\varepsilon_m = NBA\omega = NBA(2\pi f)$, with $f=500/60$ rev/s, giving $\varepsilon_m\approx88$ V. (i) Plane perpendicular to field means the coil's normal is IN the plane of the field ($\omega t=0$ in the usual convention used here), so $\varepsilon=\varepsilon_m\sin0^{\circ}=0$. (ii) Plane parallel to field corresponds to $\omega t=90^{\circ}$, giving the peak value $\varepsilon=\varepsilon_m\sin90^{\circ}=88$ V. (iii) At $\omega t=90^{\circ}-60^{\circ}=30^{\circ}$, $\varepsilon=\varepsilon_m\sin30^{\circ}=88\times0.5=44$ V. The example is a direct, numeric drill on rea …