chaseashley24
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A 100-turn coil with an average radius of 0.04 m is placed in a uniform magnetic field so that ϕ = 60°. The strength of the field increases at a rate of 0.250 T/s. What is the magnitude of the resulting emf?

A 100turn coil with an average radius of 004 m is placed in a uniform magnetic field so that ϕ 60 The strength of the field increases at a rate of 0250 Ts What class=

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jcherry99

Five

This is exactly like your question 2. You need only substitute the given numbers here.

Givens

  • N = 100 turns
  • delta B = 0.25
  • r = 0.04
  • Area = pi * r^2 = 3.14 * 0.04^2 = 5.02 * 10^-3
  • Cos 60 = 1/2 = 0.5

Formula

e = N * delta B  * Area * cos phi

Solution

e = - 100 * 0.25 *  5.02 * 10^-3 * 0.5

e = 0.0628 = 0.063

Answer

A

Six

This is a notes question. The result is a sine curve. I don't know how they are counting the change in direction. Is a rising curve different from a falling curve? For sure going from plus to minus is a change in direction, but I still don't know how to count it. It will depend on what you  have been told.

I would answer 4 but this is by no means certain. The only other possible answer is 2

sebasacevedop

Answer:

[tex]\epsilon=0.063 V[/tex]

Explanation:

Accord to Faraday's law, the magnitude of the electromotive force is given by:

[tex]\epsilon=\frac{d\Phi_B}{dt}(1)[/tex]

Magnetic flux is defined as:

[tex]\Phi_B=NBAcos(\phi)[/tex]

Here, N are the coil turns, B is the magnitud of the magnetic field, A is the area and [tex]\phi[/tex] is  the angle between the magnetic field lines and the normal to A.

Then:

[tex]\frac{d\Phi_B}{dt}=N\frac{dB}{dt}Acos(\phi)(2)[/tex]

Replacing (2) in (1) we have:

[tex]\epsilon=N\frac{dB}{dt}Acos(\phi)\\\epsilon=100*0.25\frac{T}{s}*pi*(0.04m)^2cos(60^\circ )\\\epsilon=0.063 V[/tex]

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