The horizontal component of the earth's magnetic field at a certain place is $3.0 \times 10^{-5} \; T$ and the direction of the field is from the geographic south to the geographic north. $A$ very long straight conductor is carrying a steady current of $1 \; A$. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is $(a)$ east to west; $(b)$ south to north?

(A) The magnetic force on a current-carrying conductor is given by $\vec{F} = I(\vec{l} \times \vec{B})$.
The magnitude of the force is $F = I L B \sin \theta$.
The force per unit length is $f = F/L = I B \sin \theta$.
$(a)$ When the current flows from east to west, the angle between the current direction and the magnetic field (south to north) is $\theta = 90^{\circ}$.
Thus, $f = I B \sin 90^{\circ} = I B = (1 \; A) \times (3.0 \times 10^{-5} \; T) = 3.0 \times 10^{-5} \; N/m$.
The direction of the force is downwards (using the right-hand rule).
$(b)$ When the current flows from south to north, the angle between the current direction and the magnetic field is $\theta = 0^{\circ}$.
Thus, $f = I B \sin 0^{\circ} = 0 \; N/m$.
There is no force on the conductor in this case.