ideal for simplicity.

The transformer formula relating the number of turns and the voltage in primary and secondary windings is given by:

$$\frac{V_p}{V_s} = \frac{N_p}{N_s}$$

Where:

$$V_p$$ is the primary voltage
$$V_s$$ is the secondary voltage
$$N_p$$ is the number of turns in the primary winding
$$N_s$$ is the number of turns in the secondary winding

Given:

$$V_p = 220 \, \mathrm{V}$$
$$N_p = 200$$
$$N_s = 40000$$

We need to find $$V_s$$, the secondary voltage. Rearrange the formula to solve for $$V_s$$:

$$V_s = V_p \times \frac{N_s}{N_p}$$

Substitute the given values:

$$V_s = 220 \, \mathrm{V} \times \frac{40000}{200}$$

Calculate $$V_s$$:

$$V_s = 220 \, \mathrm{V} \times 200 = 44000 \, \mathrm{V}$$

Now, we need to find the potential difference per turn. For an ideal transformer, this is the same for both primary and secondary because the voltage per turn is consistent in an ideal transformer. We can use the primary coil to find this:

Potential difference per turn (in primary) is:

$$\frac{V_p}{N_p}$$

Substitute the values for $$V_p$$ and $$N_p$$:

$$\frac{220 \, \mathrm{V}}{200 \, \text{turns}} = 1.1 \, \mathrm{V \, per \, turn}$$

Therefore, the potential difference per turn is:

Option D: 1.1 V

Electromagnetic Induction

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