A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic field is turned on perpendicular to the plane of the coil. This field changes at a constant rate from 20 to 80 mT in a time of 20 ms. What is the magnitude of the induced emf in the coil at the instant the magnetic field has a magnitude of 50 mT?

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Answer:

Approximately 18 volts when the magnetic field strength increases from [tex]\rm 20\; mT[/tex] to [tex]\rm 80\;mT[/tex] at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF [tex]\epsilon[/tex] that a changing magnetic flux induces in a coil is:

[tex]\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}[/tex],

where

  • [tex]N[/tex] is the number of turns in the coil, and
  • [tex]\displaystyle \frac{d\phi}{dt}[/tex] is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux [tex]\phi[/tex] is equal to

[tex]\phi = B \cdot A\cdot \cos{\theta}[/tex],

where

  • [tex]B[/tex] is the magnetic field strength at the coil, and
  • [tex]A\cdot \cos{\theta}[/tex] is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so [tex]\theta = 0[/tex] and [tex]A\cdot \cos{\theta} = A[/tex]. The area of this circular coil is equal to [tex]\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}[/tex].

[tex]A\cdot \cos{\theta} = A[/tex] doesn't change, so the rate of change in the magnetic flux [tex]\phi[/tex] through the coil depends only on the rate of change in the magnetic field strength [tex]B[/tex]. The size of the magnetic field at the instant that [tex]B = \rm 50\; mT[/tex] will not matter as long as the rate of change in [tex]B[/tex] is constant.

[tex]\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}[/tex].

As a result,

[tex]\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V[/tex].

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