The output voltage of the ideal transformer with polarities and dots shown in the figure is given by

A

$$\left(\frac{1}{N}\right)V_1 \sin(\omega t)$$

B

$$\left(-\frac{1}{N}\right)V_1 \sin(\omega t)$$

C

$$-N_1 V_1 \sin(\omega t)$$

D

$$-N_1 V_1 \cos(\omega t)$$