NESA Physics Electromagnetic Induction

Increasing the number of turns on the primary coil would decrease the output voltage, as the secondary voltage is proportional to the ratio of secondary to primary turns ($V_s = V_p \times \frac{N_s}{N_p}$).

Decreasing the number of turns on the primary coil increases the turns ratio ($\frac{N_s}{N_p}$). Halving the primary turns will double the output voltage from 6 V to 12 V.

Increasing the resistance connected to the secondary coil affects the current drawn by the load, but it does not change the output voltage of an ideal transformer.

Decreasing the resistance connected to the secondary coil would increase the secondary current, but the output voltage is determined solely by the input voltage and the turns ratio.

Decreasing wire thickness increases electrical resistance, which would reduce the current $I$ (for a fixed voltage) and consequently decrease the magnetic field strength.

The magnetic field strength $B = \mu_0 \frac{N}{L} I$ is inversely proportional to the length $L$; increasing the length while keeping the number of turns constant decreases the turn density, weakening the field.

The magnetic field strength inside a solenoid is directly proportional to the number of turns ($N$); increasing the number of turns increases the turn density $n$, which increases the field strength according to $B = \mu_0 n I$.

Alternating current produces a magnetic field that fluctuates in magnitude and direction, and inductive reactance often reduces the current compared to DC, which would not increase the field strength.