A rod (length = 10 cm) moves on two horizontal frictionless conducting rails, as shown. The magnetic field in the region is directed perpendicularly to the plane of the rails and is uniform and constant. If a constant force of 0.60 N moves the bar at a constant velocity of 2.0 m/s, what is the current through the 12-Ω load resistor?

Question

Quizplus · Accepted Answer

Firstly, we can use the equation for force on a current-carrying conductor in a magnetic field:
$$F = BIL$$
where $F$ is the force, $B$ is the magnetic field strength, $I$ is the current, and $L$ is the length of the conductor in the magnetic field. Since the rod moves at a constant velocity, the force is balanced by the frictional force acting in the opposite direction. Therefore,
$$F_{friction} = 0.60	ext{ N}$$
Then, since the velocity is constant and there is no resistance in the rails, the net force must be zero:
$$F_{friction} = BIL \quad \implies \quad I = \frac{F_{friction}}{BL}$$
We know $F_{friction}$, $B$, and $L$ (assuming the length of the rod in contact with the rails is negligible compared to 10 cm), so we just need to find $I$. Using the given values, we get:
$$I = \frac{0.60	ext{ N}}{(1.2	ext{ T})(0.1	ext{ m})} = 0.5	ext{ A}$$
However, this is the current flowing through the entire loop (including the resistance of the rod itself), not just the load resistor. To find the current through the load resistor only, we can use the fact that the total potential difference across the loop is equal to the emf induced in the rod:
$$V_{loop} = \frac{d\Phi_B}{dt} = BLv$$
where $v$ is the velocity of the rod. Therefore,
$$I_{load} = \frac{V_{loop}}{R_{load}} = \frac{BLv}{R_{load}}$$
Again, substituting the given values, we get:
$$I_{load} = \frac{(1.2	ext{ T})(0.1	ext{ m})(2.0	ext{ m/s})}{12\Omega} = 0.2	ext{ A}$$
Therefore, the answer is:
Answer: A

A Rod (Length = 10 Cm) Moves on Two Horizontal