Answer Posted / 7anoter4

Since V=K*B*frq [V= primary supply voltage -approx.
equal to E-emf]; B=magnetic flux density ; frq=frequency]
then if frq=5 Hz and V=constant B[5Hz]=10*B[50 Hz].The no-
load[excitation] current will be more than 300% rated
current.

You have to reduce the voltage to 10% from rated at 50 Hz
in order to get the same B.

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