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In the figure, the rolling axle, 1.70 m long, is pushed along horizontal rails at a constant speed v = 2.60 m/s.

a)A resistor R = 0.330 Ω is connected to the rails at points a and b, directly opposite each other. The wheels make good electrical contact with the rails, and so the axle, rails, and R form a closed-loop circuit. The only significant resistance in the circuit is R. There is a uniform magnetic field B = 8.20×10-2 T vertically downward. Calculate the induced current I in the resistor.


b)What horizontal force Fb is required to keep the axle rolling at constant speed?


c)Calculate the size of the potential difference between point a and point b.

In the figure, the rolling axle, 1.70 m long, is pushed along horizontal rails at-example-1
User Andrewrk
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2 Answers

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Correct Answers:

a) To calculate the induced current I in the resistor, we can use Ohm's Law. The induced current is equal to the potential difference across the resistor divided by its resistance. Let's calculate it!

b) To keep the axle rolling at a constant speed, the horizontal force Fb must overcome the frictional force acting on the axle. We can use Newton's second law to calculate it!

c) The potential difference between point a and point b can be calculated using the formula V = I * R, where V is the potential difference and R is the resistance. Let's calculate it!
User Rhadames
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The loop experiences an induced emf of -0.213 V, resulting in an induced current of -0.645 A. The horizontal force is -0.0901 N, opposing motion, and the potential difference (Vab) is -0.213 V, indicating a lower potential at point a.

a) The induced emf (E) is given by Faraday's law:

\[E = -N \frac{\Delta \Phi}{\Delta t}\]

where N is the number of turns in the loop, Φ is the magnetic flux, and t is the time. In this case, N = 1, Φ = BA (where A is the area of the loop), and t = L/v (where L is the length of the axle and v is the speed). Therefore:

\[E = -\frac{BLv}{L} = -Bv\]

Plugging in the given values:

\[E = -(8.20 \times 10^{-2}\ T)(2.60\ m/s) = -0.213\ V\]

The induced current (I) is then given by:

\[I = \frac{E}{R + r}\]

Assuming the internal resistance (r) of the axle is negligible:

\[I = \frac{-0.213\ V}{0.330\ Ω} = -0.645\ A\]

The negative sign indicates that the current flows in the opposite direction of the assumed loop.

b) The horizontal force (Fb) is given by:

\[F_b = IBL\]

Plugging in the values:

\[F_b = (-0.645\ A)(8.20 \times 10^{-2}\ T)(1.70\ m) = -0.0901\ N\]

The negative sign indicates that the force acts in the opposite direction of the motion.

c) The potential difference (Vab) is given by:

\[V_{ab} = E - Ir\]

Plugging in the values:

\[V_{ab} = -0.213\ V - (-0.645\ A)(0) = -0.213\ V\]

The negative sign indicates that point a is at a lower potential than point b.

User August Miller
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