Question 5 of 8: EMF of a Rotating Direction-Finding Loop Antenna
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A3, Electromagnetics — National Exam, December 2016. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the eight questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value. Aids given on the paper: ε0 = 8.85×10-12 F/m, μ0 = 4π×10-7 H/m. All eight printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — transmission lines, waveguides, plane waves, antennas; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients; Pozar, Microwave Engineering (4th ed.) — transmission-line theory and stub matching; Balanis, Antenna Theory (4th ed.) — short-dipole far field.
Question 5: EMF of a Rotating Direction-Finding Loop Antenna (equal value)
Given. Vertical circular loop, area $A=25\ \text{cm}^2$, $N=20$ turns, rotating about its own VERTICAL axis at $2500$ rpm (so the loop's normal stays horizontal and sweeps around); aircraft flies horizontally north-west at $200$ km/h; ambient field $B=1.0\times10^{-5}$ T, pointing north and $45^\circ$ below horizontal.
Given data
Quantity
Symbol
Value
Loop area
$A$
25 cm²
Turns
$N$
20
Rotation rate
—
2500 rpm
Field magnitude
$B$
$1.0\times10^{-5}$ T
Field dip angle
—
45° below horizontal, pointing N
Find. (i) induced EMF; (ii) whether the aircraft's horizontal translation contributes; (iii) the loop orientation at which the EMF is zero.
Only the horizontal component of $\mathbf B$ links flux through a loop whose normal always stays horizontal; the vertical component contributes nothing.
Approach. Only the component of $\mathbf B$ parallel to the loop's (always-horizontal) normal contributes flux, so resolve $B$ into horizontal and vertical parts and apply Faraday's law to the rotating flux; check the translational contribution with the motional-EMF line integral $\oint(\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell$.
Horizontal field component and flux linkage. With the loop normal $\hat n(t)$ always horizontal, only $B_h=B\cos45^\circ$ (pointing north) links flux:
$$\Phi(t)=NAB_h\cos(\omega t)=NAB\cos45^\circ\cos(\omega t).$$
Peak EMF. $\omega=2500\ \text{rpm}=2500\times\dfrac{2\pi}{60}=261.8\ \text{rad/s}$, so
$$\varepsilon_{pk}=-\frac{d\Phi}{dt}\bigg|_{max}=NAB\cos45^\circ\,\omega=(20)(25\times10^{-4})(10^{-5})(0.7071)(261.8)$$
$$=\boxed{92.6\ \mu\text{V}}.$$
Part (ii): does the aircraft's translation contribute? The motional EMF around any closed, rigid loop is $\oint(\mathbf v\times\mathbf B)\cdot d\boldsymbol\ell$. Pure translation gives every point on the loop the SAME velocity $\mathbf v$, so $(\mathbf v\times\mathbf B)$ is a constant vector, and $\oint d\boldsymbol\ell=0$ around any closed path — the two combine to give exactly $\boxed{\text{zero contribution}}$. (Earth's field is essentially uniform over the loop's dimensions, so no flux-gradient effect arises either.) Only the loop's OWN rotation, which gives points on the loop different velocities, produces an EMF.
Part (iii): orientation of zero EMF. $\varepsilon(t)\propto\sin(\omega t)$, which is zero exactly when $\cos(\omega t)=\pm1$, i.e. when the loop's normal is aligned with (or opposite to) the horizontal field component — when the loop's plane contains the north–south line (normal pointing due north or south). At that instant the flux is at its extremum, so its instantaneous rate of change is $\boxed{\text{zero}}$.
