Solution
Force Calculation:
The current flows between points P1 and P2. The magnetic force on the current distribution in the disc is equivalent to the force on a straight wire connecting the entry and exit points (effective length). The chord length connecting P1 and P2 is:
$$ L_{eff} = 2r \sin(\theta/2) $$The magnetic force is horizontal: $F_m = I L_{eff} B = 2IrB \sin(\theta/2)$.
Sliding Condition:
The disc begins to slide when the magnetic force overcomes the maximum static friction.
$$ F_m \ge f_{friction} $$ $$ 2IrB \sin(\theta/2) = \mu N = \mu mg $$Solving for current $I$:
$$ I = \frac{\mu mg}{2Br \sin(\theta/2)} $$