A long straight conductor carrying 1 A lies on a horizontal table. The horizontal component of the earth's field there is 3.0 × 10⁻⁵ T pointing south to north. When the current flows east to west, the force per unit length on the conductor is:

1. Force per unit length: $f = IB\sin\theta$. 2. The current (east–west) is perpendicular to the field (south–north), so $\theta = 90^\circ$, $\sin\theta = 1$. 3. Substitute $I = 1$ A, $B = 3.0\times10^{-5}$ T: $f = 1\times3.0\times10^{-5}$. 4. $f = 3\times10^{-5}$ N m⁻¹ (it would be zero only if current were along the field). _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.156_