Answer: \(e = \dfrac{1}{2}B\omega l^2\); coefficient of coupling \(K \approx 0.335\).
(i) Motional e.m.f. in a rotating bar: A conducting bar OP of length \(l\) rotates with constant angular velocity \(\omega\) about an axis through its end O, in a plane perpendicular to a uniform magnetic field \(\vec{B}\).
[Figure: rod OP rotating about O in a circle of radius \(l\); \(\vec{B}\) perpendicular to the plane (into the page); a small element \(dr\) at distance \(r\) from O.]
Consider a small element of length \(dr\) at distance \(r\) from O. Its linear speed is \(v = r\omega\), perpendicular to both the rod and \(\vec B\). The motional e.m.f. across the element is
\[de = Bv\,dr = B\omega r\,dr\]
All elements are in series, so the total e.m.f. between O and P is
\[e = \int_0^l B\omega r\,dr = B\omega\frac{l^2}{2} = \frac{1}{2}B\omega l^2\]
Alternative: in time \(T = 2\pi/\omega\) the rod sweeps area \(\pi l^2\), cutting flux \(B\pi l^2\), so \(e = \dfrac{B\pi l^2}{2\pi/\omega} = \dfrac{1}{2}B\omega l^2\).
(ii) Numerical: \(M = 1.5\) H, \(L_1 = 5\) H, \(L_2 = 4\) H.
\[K = \frac{M}{\sqrt{L_1L_2}} = \frac{1.5}{\sqrt{5\times4}} = \frac{1.5}{\sqrt{20}} = \frac{1.5}{4.472} = 0.3354\]
