Figure shows the top view of a rod that can slide without friction. The resistor is $6.0 Ω$ and a $2.5 T$ magnetic field is directed perpendicularly downward into the paper. Let l= 1.20 m.
(a) Calculate the force F required to move the rod to the right at a constant speed of 2.0 m/s.
(b) At what rate is energy delivered to the resistor?

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3 N, 8 W

3 N, 5 W

3 N, 6 W

3 N, 4 W

answer is C.

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Detailed Solution

The motional emf in the rod, $e = B v l$ or $e = (2.5) (2.0) (1.2) V = 6.0 V$

The current in the circuit, $i = \frac{e}{R} = \frac{6.0}{6.0} = 1.0 A$

(a) The magnitude of force F required will be equal to the magnetic force acting on the rod, which opposes the motion.

$F = F_{m} = i l B$ or $F = (1.0) (1.2) (2.5) N = 3 N$

(b) Rate by which energy is delivered to the resistor is

$P_{1} = i^{2} R = {(1)}^{2} (6.0) = 6 W$

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