A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid, perpendicular to its axis, where the field is 0.27 T. The magnetic force on the wire is:

1. Force on the wire: $F = IlB\sin\theta$. 2. The wire is perpendicular to the field, so $\theta = 90^\circ$ and $\sin\theta = 1$. 3. Substitute $I = 10$ A, $l = 0.03$ m, $B = 0.27$ T: $F = 10 \times 0.03 \times 0.27$. 4. $F = 8.1 \times 10^{-2}$ N. _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.158_