Math Problem Statement

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 164 cm , but its circumference is decreasing at a constant rate of 11.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EE induced in the loop after exactly time 1.00 ss has passed since the circumference of the loop started to decrease.

Express your answer numerically in volts to three significant figures.

Solution

We can solve this problem using Faraday's Law of Induction, which states that the induced electromotive force (emf) in a loop is given by:

E=dΦBdt\mathcal{E} = -\frac{d\Phi_B}{dt}

Where:

  • E\mathcal{E} is the induced emf,
  • ΦB\Phi_B is the magnetic flux, and
  • dΦBdt\frac{d\Phi_B}{dt} is the rate of change of the magnetic flux.

The magnetic flux ΦB\Phi_B through the loop is given by:

ΦB=BA\Phi_B = B \cdot A

Where:

  • BB is the magnetic field strength,
  • AA is the area of the loop.

Now, the area of the circular loop is related to its circumference. If CC is the circumference, then the radius rr of the loop is:

C=2πrC = 2\pi r

Thus, the radius of the loop is:

r=C2πr = \frac{C}{2\pi}

The area AA of the loop is then:

A=πr2=π(C2π)2=C24πA = \pi r^2 = \pi \left(\frac{C}{2\pi}\right)^2 = \frac{C^2}{4\pi}

Thus, the magnetic flux is:

ΦB=BC24π\Phi_B = B \cdot \frac{C^2}{4\pi}

To find the induced emf, we differentiate the flux with respect to time:

dΦBdt=Bddt(C24π)\frac{d\Phi_B}{dt} = B \cdot \frac{d}{dt} \left( \frac{C^2}{4\pi} \right)

Using the chain rule:

dΦBdt=B14π2CdCdt\frac{d\Phi_B}{dt} = B \cdot \frac{1}{4\pi} \cdot 2C \cdot \frac{dC}{dt}

Simplifying:

dΦBdt=BC2πdCdt\frac{d\Phi_B}{dt} = \frac{B C}{2\pi} \cdot \frac{dC}{dt}

Now, we can substitute the given values:

  • The magnetic field B=0.500TB = 0.500 \, \text{T},
  • The rate of change of the circumference dCdt=11.0cm/s=0.11m/s\frac{dC}{dt} = -11.0 \, \text{cm/s} = -0.11 \, \text{m/s},
  • The initial circumference C=164cm=1.64mC = 164 \, \text{cm} = 1.64 \, \text{m}.

Thus, the rate of change of magnetic flux becomes:

dΦBdt=0.500×1.642π×(0.11)\frac{d\Phi_B}{dt} = \frac{0.500 \times 1.64}{2\pi} \times (-0.11)

Now, calculating this expression:

dΦBdt=0.8202π×(0.11)0.0144Wb/s\frac{d\Phi_B}{dt} = \frac{0.820}{2\pi} \times (-0.11) \approx -0.0144 \, \text{Wb/s}

Finally, the induced emf is:

E=dΦBdt=0.0144V\mathcal{E} = -\frac{d\Phi_B}{dt} = 0.0144 \, \text{V}

Thus, the magnitude of the induced emf is:

0.0144V\boxed{0.0144 \, \text{V}}

Let me know if you need further details or clarifications!


Related Questions:

  1. What is the relationship between magnetic flux and the area of a loop?
  2. How does the rate of change in the magnetic flux affect the induced emf?
  3. What is the significance of the negative sign in Faraday’s Law?
  4. How can the induced emf be affected by the strength of the magnetic field?
  5. What would happen to the induced emf if the magnetic field strength were doubled?

Tip: Always ensure the units are consistent when solving electromagnetism problems—converting all measurements to SI units is crucial!

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Math Problem Analysis

Mathematical Concepts

Magnetic Flux
Faraday's Law of Induction
Differentiation

Formulas

Faraday's Law: E = -dΦ_B/dt
Magnetic Flux: Φ_B = B * A
Area of a Circle: A = π * (C/(2π))^2 = C^2 / 4π

Theorems

Faraday's Law of Induction

Suitable Grade Level

Grades 11-12