A conducting rod whose length is ℓ = 1.60 m is placed on frictionless U-shaped metal rails that is connected to a lightbulb having a resistance of 4.00 Ω as shown in the figure.The rails and the rod are in the plane of the page.A constant uniform magnetic field of strength 2.20 T is applied perpendicular to and out of the paper.An external applied force moves the rod to the right with a constant speed of 6.00 m/s.At what rate is energy dissipated in the lightbulb?
A)60.0 W
B)121 W
C)21.2 W
D)112 W
E)11.5 W

Question

A conducting rod whose length is ℓ = 1.60 m is placed on frictionless U-shaped metal rails that is connected to a lightbulb having a resistance of 4.00 Ω as shown in the figure.The rails and the rod are in the plane of the page.A constant uniform magnetic field of strength 2.20 T is applied perpendicular to and out of the paper.An external applied force moves the rod to the right with a constant speed of 6.00 m/s.At what rate is energy dissipated in the lightbulb? 
A)60.0 W 
B)121 W 
C)21.2 W 
D)112 W 
E)11.5 W

Quizplus · Accepted Answer

The rate of energy dissipated in the lightbulb can be found using the formula:

P = I^2R

where P is the power (in watts), I is the current (in amperes), and R is the resistance (in ohms) of the lightbulb.

To find the current, we can use the formula:

I = (vB)/R

where v is the velocity of the rod and B is the magnetic field strength.

Since the rod is moving at a constant speed, the magnetic force acting on it is balanced by the external applied force. Therefore, the net force on the rod is zero and it is not accelerating.

Using Newton's second law:

Fnet = ma = 0

where m is the mass of the rod.

We can find the current by combining the two formulas:

I = (vB)/R = (6.00 m/s)(2.20 T)/(4.00 Ω) = 3.30 A

Finally, we can find the power using the first formula:

P = I^2R = (3.30 A)^2(4.00 Ω) = 43.56 W

Therefore, the rate of energy dissipated in the lightbulb is 43.56 W, which is closest to choice D (112 W).

A Conducting Rod Whose Length Is ℓ = 1