The loop experiences an induced emf of -0.213 V, resulting in an induced current of -0.645 A. The horizontal force is -0.0901 N, opposing motion, and the potential difference (Vab) is -0.213 V, indicating a lower potential at point a.
a) The induced emf (E) is given by Faraday's law:
\[E = -N \frac{\Delta \Phi}{\Delta t}\]
where N is the number of turns in the loop, Φ is the magnetic flux, and t is the time. In this case, N = 1, Φ = BA (where A is the area of the loop), and t = L/v (where L is the length of the axle and v is the speed). Therefore:
\[E = -\frac{BLv}{L} = -Bv\]
Plugging in the given values:
\[E = -(8.20 \times 10^{-2}\ T)(2.60\ m/s) = -0.213\ V\]
The induced current (I) is then given by:
\[I = \frac{E}{R + r}\]
Assuming the internal resistance (r) of the axle is negligible:
\[I = \frac{-0.213\ V}{0.330\ Ω} = -0.645\ A\]
The negative sign indicates that the current flows in the opposite direction of the assumed loop.
b) The horizontal force (Fb) is given by:
\[F_b = IBL\]
Plugging in the values:
\[F_b = (-0.645\ A)(8.20 \times 10^{-2}\ T)(1.70\ m) = -0.0901\ N\]
The negative sign indicates that the force acts in the opposite direction of the motion.
c) The potential difference (Vab) is given by:
\[V_{ab} = E - Ir\]
Plugging in the values:
\[V_{ab} = -0.213\ V - (-0.645\ A)(0) = -0.213\ V\]
The negative sign indicates that point a is at a lower potential than point b.