FIGURE 23-12
-A conducting rod with a length of 27.0 cm is placed on a U-shaped metal wire that includes a light bulb with a resistance of 60.0 Ω as shown in Figure 23-12. A constant magnetic field with a strength of 0.600 T is applied perpendicular into the paper. An applied force moves the rod to the right with a constant speed of 6.00 m/s. What is the power consumed by the light bulb?

Question

FIGURE 23-12 
-A conducting rod with a length of 27.0 cm is placed on a U-shaped metal wire that includes a light bulb with a resistance of 60.0 Ω as shown in Figure 23-12. A constant magnetic field with a strength of 0.600 T is applied perpendicular into the paper. An applied force moves the rod to the right with a constant speed of 6.00 m/s. What is the power consumed by the light bulb?

Quizplus · Accepted Answer

The power consumed by the light bulb is given by the equation:$$P = I^2R$$where I is the current flowing through the light bulb and R is the resistance of the light bulb.The current flowing through the light bulb is given by the equation:$$I = \frac{emf}{R_{total}}$$where emf is the electromotive force (emf) induced in the rod and Rtotal is the total resistance of the circuit.The emf induced in the rod is given by the equation:$$emf = Blv$$where B is the strength of the magnetic field, l is the length of the rod, and v is the speed of the rod.The total resistance of the circuit is given by the equation:$$R_{total} = R_{bulb} + R_{rod}$$where Rbulb is the resistance of the light bulb and Rrod is the resistance of the rod.Substituting these equations into the equation for power, we get:$$P = \left(\frac{Blv}{R_{bulb} + R_{rod}}\right)^2R_{bulb}$$Substituting the given values into this equation, we get:$$P = \left(\frac{(0.600 T)(0.270 m)(6.00 m/s)}{60.0 \Omega + 0.001 \Omega}\right)^2(60.0 \Omega) = 15.7 mW$$Therefore, the power consumed by the light bulb is 15.7 mW.

FIGURE 23-12 -A Conducting Rod with a Length of 27.0 Cm Is