electromagnetic - Find the current flowing in the load of a transmission line?

My Try:

Now

\$\beta l=\frac{2\pi\times37.5\times10^6\times10}{3\times10^8}=\frac{5\pi}{2}\$

so its becomes

\$Z_{in}=\frac{Z_0^2}{Z_L} \$

\$Z_{in}=\frac{200\times200}{100}=400\Omega\$

So for part \$(a)\$

\$I_{i}=\frac{200}{600}=\frac{1}{3}A\$ also \$V_{i}=\frac{400\times200}{600}=\frac{400}{3}\$

Now for part \$(b)\$ from the generator we know the equation for lossless line is \$ I(z)=\frac{-j}{Z_0}sin\beta z V_i+Cos\beta z I_{i}\$

so current at the load, that means at \$\lambda/4\$ distance from the generator will be \$I=\frac{-j}{200}sin(\frac{5\pi}{2})\times\frac{400}{3}+0\times\frac{1}{3}=\frac{2}{3}\angle-90^{\circ}\$

but for part\$(b)\$ the actual answer is \$\frac{1}{3}\angle-90^{\circ}\$

What is the mistake i am doing can anyone help please?