Question

N loops of current carrying wire are coiled together to form a magnetic dipole. The energy required to "twist" the dipole from its lowest energy state (θ = 0^o) to its highest energy state ( θ = 180^o) (where θ is the angle between the dipole and the external B-field) is approximately U = 27 ×10⁻³ Joules. If the current in the wire is 1.0 Amps, the area of one loop is 1.0cm x 1.0cm, and the external B-field is 1.0 T through which the coil rotates, how many loops (N) make up the coil? Answer to the nearest integer.

Answer 1