transformer
I'm trying to solve this question.
For a specified input voltage and frequency, if the equivalent radius
of the core of a transformer is reduced by half, the factor by which
the number of turns in the primary should change to maintain the same
no load current I
If radius is halved, area becomes \$ 1/4\$ times. Hence to maintain the same no load current, there needs to be 4 times the number of turns we had earlier.
But the given answer is 2 times the number of turns. Why is that?
Here (from Wikipedia) is a fairly complete linear model of a transformer:
Note the magnetizing inductance Xm across the primary. If that inductance is too low, you'll get excessive current flowing even with no load on the secondary.
While a single-turn primary is certainly possible, with sensibly-sized cores it implies either a very low voltage (for example, a current transformer, which is typically toroidal) or a very high frequency (or both).
The inductance is proportional to the number of turns squared, and a small 120/240V 50/60Hz mains transformer primary might be some hundreds of turns, so you can see how far off a single turn is. At a fraction of a volt, or higher frequencies at relatively low voltage, a single-turn primary might make some sense.
I see: you are examing the transformer you bought and your calculations don't match.
Volts per turn is 0.7292 V/turn. Let's ask how many turns are required for 250V? Answer is: \$N=\dfrac{250}{0.7292} \approx 343\$. The first mystery is solved.
The real transformer has 375 turns, \$\dfrac{375}{343}\cdot 1.3= 1.42\$, you are lucky, because if you buy from ebay you can get a transformer with \$B_{max}\$ at 1.9 Tesla.
The wires are too thin: maybe you have a diffent table, or they cheated on copper. If you see that the window has a lot of space left, then they cheated. If you see that window is full, then a thicker wire couldn't be used.
Best Answer
This answer assumes that "no load current" refers to the magnetization current in the primary of the transformer and that is defined by the magnetization inductance.
If the radius is halved then the inductance falls by 4 times because area is quartered. If the core is a regular shaped core with a decent value of permeability then you can assume that inductance is proportional to the square of turns hence, to restore the inductance, you need to double the turns to increase the inductance 4 times: -
Picture from here.
For a closed magnetic core \$\ell\$ becomes the mean length around the core.