Question

four straight conductors form a square with a magneticfield perpendicular to the square.if all conductorsmove outward with the same velocity while contacting each other at thecorners,find V(rms) induced in the square loop at the instant whenits area is 2 m^2,velocity is 4m/s , and,B=cos(2*pi*f*t)where f=2khz.
answer is 17.8 kv....how???

Answer 1

emf = rate of change of magnetic flux passing through

as the square expands, both magnetic field and area are varying with time
\[A = L ^ 2\]
so, dA/dt = 2*L*dL/dt
dL/dt = v ( velocity)
hence dA/dt = 2*L*v

\[B = \cos wt \]
where \[w = 2 * \pi * f\]
hence dB/dt = \[- w * \sin wt\]

Now, flux = B * A
differentiating with respect to time to find the rate of change of flux,
d (flux)/dt = B*dA/dt + A*d(B)/dt
= \[(2 * L * v * \cos wt ) - ( L^2 * w * \sin wt )\]

therefore V = d(flux)/dt = \[(2 * L * v * \cos wt ) - ( L^2 * w * \sin wt )\]

we can write this as (by using cos(A + B) = cosAcosB - sinAsinB)
\[V = L * \sqrt{4*v^2 + (L^2 * w^2)} * \cos (wt + \delta)\]

Now, to find V(R.M.S) , we observe that for any function of the form
\[A = Ao* \cos B\]
the root mean square value is \[Arms = Ao / \sqrt{2}\]

we get , \[Vrms = Vo/\sqrt{2}\]

where \[Vo = L * \sqrt{4*v^2 + (L^2 * w^2)} \]

now, L = 2, v = 4 and w = 2 * pi * f

thus Vrms = 17.777 KV