Questions: Multiple-Concept Example 2 discusses the concepts that are used in this problem. In the drawing the magnetic field has a magnitude of 1.69 T, the rod has a length of 0.946 m, and the hand keeps the rod moving to the right at a constant speed of 4.18 m / s. If the current in the circuit is 0.0451 A, what is the average power being delivered to the circuit by the hand?

Solution Steps

The hand exerts a force equal and opposite to the magnetic force on the rod. The magnetic force _F_ on a current-carrying wire of length _L_ in a magnetic field _B_ is given by _F_ = _ILB_. In our case, _I_ = 0.0451 A, _L_ = 0.946 m, and _B_ = 1.69 T. Therefore, the force is

_F_ = (0.0451 A)(0.946 m)(1.69 T) = 0.072 N

Power is the rate at which work is done, which can be calculated as _P_ = _Fv_, where _F_ is the force and _v_ is the velocity. The hand keeps the rod moving at a constant speed _v_ = 4.18 m/s. Thus,

_P_ = (0.072 N)(4.18 m/s) = 0.30 W

Final Answer: The average power delivered to the circuit by the hand is 0.30 W.