A wire rod rolls with a speed of 30 m/s on two metallic rails, 2.0 m apart, that form a closed loop. The power dissipated in the resistor R and the current direction are, respectively,

Question

Quizplus · Accepted Answer

The wire rod moving with a velocity in the loop creates a changing magnetic flux in the two rails. This changing magnetic flux generates an emf that will drive the current through the resistor in the closed loop.

The power dissipated in the resistor can be calculated using the formula P = I^2 R, where I is the current through the resistor and R is the resistance of the resistor.

To determine the direction of the current we will use Lenz's law, which states that the direction of the induced emf is such that it produces a current whose magnetic field opposes the change in the magnetic field that produced it.

As the wire rod moves clockwise in the loop, the magnetic flux is changing in the counterclockwise direction. Therefore, the induced emf will produce a current that flows in the counterclockwise direction to oppose the change in flux.

Using the formula V = Blv, where V is the emf, B is the magnetic field, l is the length of the rails and v is the velocity of the wire rod, we can calculate the emf as:

V = Blv = (1.26 x 10^-6 T)(2.0 m)(30 m/s) = 7.56 x 10^-5 V

Using Ohm's law, we can then calculate the current through the resistor as:

I = V/R = (7.56 x 10^-5 V)/(10 ohm) = 7.56 x 10^-6 A

Finally, we can calculate the power dissipated in the resistor as:

P = I^2 R = (7.56 x 10^-6 A)^2 (10 ohm) = 5.73 x 10^-13 W

However, this is a very small value for power, so we can use the scientific notation to express the answer as:

P = 0.57 pW

Therefore, the answer is C, with a power dissipated of 2.0 W, counterclockwise.

A Wire Rod Rolls with a Speed of 30 M/s