Consider a very long straight solenoid having n turns per meter, radius Rbig, an
ID: 1699582 • Letter: C
Question
Consider a very long straight solenoid having n turns per meter, radius Rbig, and carrying a current I(t). Embedded inside this solenoid is a smaller solenoid whose axis is the same as the axis of the big solenoid. The small solenoid contains N turns total of wire and has a radius Rsmall.a) If the current I(t) is constant in time, explain why the potential difference measured across the two ends of the small inner solenoid is zero.
b) If the current varies with time as I(t) = I0 sin(?t), derive an expression for the potential difference measured across the small solenoid as a function of time.
Explanation / Answer
a). it I(t) is constant then the magnetic field inside the solenoid is constant. So the magnetix flux is unchanged. so that U=0. b). Magnetic field inside the big solenoid. I*n*L*muy0=B*L so that B=I*n*muy0 so that Magnetix flux through the inner solenoid. N*I*n*muy0*pi*r^2 so M=I*(N*n*muy0*pi*r^2) so E=I0*?*cos(?t)*(N*n*muy0*pi*r^2)