In a turn the angular momentum is an effect of the gravity acting on the bike, rider, and lean of the bike.
If you start with Newton’s second law for a body in motion being acted on by a force we know: F = dp/dt where p is momentum (m*v).
Now if we imagine a circle the rider is turning through and since angular momentum must always have a reference point we’ll use the Center of the circle and imagine a vector r from the rider to the Center of the turning circle.
Multiple both sides of the equation by r: r*F = r * dp/dt
Of course r* dp is the change in angular momentum or dL.
And r*F is torque.
Torque = dL/dt
Or changing your angular momentum always results in torque.
Then I think the torque, ignoring friction, in this case should be something like or close to: torque = m*g*sin(theta). Where m is the mass of the bike and rider, g is gravitational force of the Earth, and theta is the angle of the plane of the bike to its vertical plane. (at zero degrees offset to the vertical plane the angular momentum is zero.) As you straighten the bike back up to reduce the torque from the turn to zero you regain your gravitational potential energy which is where the torque originally came from.
Could be wrong though.