In the arrangement shown,a conducting bar of negligible resistance slides along horizontal,parallel,frictionless conducting rails connected as shown to a 2.0$\Omega$ resistor.A uniform 1.5-T magnetic field is perpendicular to the plane of the paper.If L = 60 cm,at what rate is thermal energy being generated in the resistor at the instant the speed of the bar is equal to 4.2 m/s?

Question

Quizplus · Accepted Answer

The electrical power being generated can be found using the equation:

P = I2R

where I is the current passing through the resistor and R is the resistance of the resistor.

The current can be found using:

I = V/R

where V is the voltage drop across the resistor. The voltage drop can be found using:

V = BLv

where B is the magnetic field strength, L is the length of the resistor, and v is the velocity of the conducting bar.

Putting it all together:

I = (BLv)/R 
P = I2R = (B2L2v2)/R

Plugging in the given values:

B = 1.5 T 
L = 0.6 m 
v = 4.2 m/s 
R = 10 Ω

P = (1.5)2(0.6)2(4.2)2/10 = 7.1 W

Therefore, the answer is C.

In the Arrangement Shown,a Conducting Bar of Negligible Resistance Slides Ω\OmegaΩ

In the Arrangement Shown,a Conducting Bar of Negligible Resistance Slides $\Omega$