Answer
$I=\frac{5.0mg }{\ell B}$, toward the left.
Work Step by Step
The 2 forces act uniformly along the entire length, so we may consider the two forces to act at the rod’s center of mass. To be in equilibrium, the net torque is zero.
Find an expression for the torque.
$$\Sigma \tau = 0 = Mg(\ell/4)-mg(\ell/4)-F_{mag}(\ell/4)$$
$$ F_{mag}=(M-m)g$$
Use equation 20–2 and solve for the current.
$$I=\frac{ F_{mag}}{\ell B}=\frac{(M-m)g }{\ell B}$$
$$I=\frac{(6.0m-m)g }{\ell B}$$
$$I=\frac{5.0mg }{\ell B}$$
Use the right-hand rule. The magnetic force must be downward, therefore the current flows toward the left.
