In Figure P31.57, the rolling axle, 2.50 m long, is pushed along horizontal rails at a constant...

Question:

In Figure P31.57, the rolling axle, 2.50 m long, is pushed along horizontal rails at a constant speed v = 9.00 m/s. A resistor R = 0.4000 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 = 0.0500 T vertically downward.

Faraday's Law

It is observed that both the electric and magnetic properties of matter are fundamentally due to the charge possessed by the matter. According to electromagnetism, electric and magnetic fields in a region can influence one another. Faraday's law captures how an electric field is produced when magnetic field (or more generally flux) in a region varies with time. According to Faraday's law, a varying magnetic flux through a closed loop induces an emf in the circuit whose magnitude is given by: {eq}E \ = \ \dfrac{d \phi}{dt} {/eq}

Where:

{eq}\phi \ = \ \int \ \vec{B} \ \cdot \ \vec{dA} {/eq} - is the magnetic flux through the loop.

Answer and Explanation:

Given:

  • Length of the axle: {eq}L \ = \ 2.50 \ m {/eq}
  • Speed at which the axle is moving: {eq}v \ = \ 9.00 \ ms^{-1} {/eq}
  • Resistance of the circuit: {eq}R \ = \ 0.4000 \ \Omega {/eq}
  • Magnetic field in the region: {eq}B \ = \ 0.0500 \ T {/eq}

Let {eq}w {/eq} be the width of the rectangular loop at any given instant. Then, magnetic fluc through the loop is: {eq}\phi \ = \ B \times A \ = \ B \times L \times w \ {/eq}

Rate at which magnetic flux through the loop is changing is: {eq}\dfrac{d \phi}{dt} \ = \ B \times L \times \dfrac{d w}{dt} \ = \ 0.0500 \times 2.50 \times 9.00 \ = \ 1.13 \ T m^2 {/eq}

From Faraday's law, magnitude of induced emf in the loop is: {eq}E \ = \ \dfrac{d \phi}{dt} \ = \ 1.13 \ V {/eq}

Therefore, the induced current in the circuit is: {eq}I \ = \ \dfrac{E}{R} \ = \ \dfrac{1.13}{0.4000} \ = \ 2.83 \ A {/eq}


Learn more about this topic:

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Faraday's Law of Electromagnetic Induction: Equation and Application

from High School Physics: Help and Review

Chapter 13 / Lesson 10
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