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Solinoid EMF question

  • Thread starter takumi_91
  • Start date
  • #1
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heres the question:

A 110 turn coil has a radius of 4.1 cm and a resistance of 25 Ohm. The coil is in a uniform magnetic field that is perpendicular to the plane of the coil. What rate of change of the magnetic field strength will induce a current of 4.3 A in the coil? What rate of change of the magnetic field strength is required if the magnetic field makes an angle of 20° with the normal to the plane of the coil?




Homework Equations



i know that i need to use faradays equation:E = -n(dtheta/dt), and Ohm's law:V =IR

i dont know how to get 'dt' for my equation. i know there is another way to do this problem but i cant really come up with anything at the moment
 
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Answers and Replies

  • #2
1,482
3
you are asked to find the rate of change of the magnetic field strength (I assume B not H since it is hard to figure this out from the context)


[tex] \frac{dB}{dt} [/tex]

we know that

[tex] \mathcal{E} = -N \frac{d\Phi_B}{dt} [/tex]

so what is?

[tex] \Phi_B[/tex]
 
  • #3
22
0
you are asked to find the rate of change of the magnetic field strength (I assume B not H since it is hard to figure this out from the context)


[tex] \frac{dB}{dt} [/tex]

we know that

[tex] \mathcal{E} = -N \frac{d\Phi_B}{dt} [/tex]

so what is?

[tex] \Phi_B[/tex]
Given the question refers to magnetic field strength rather than flux density, then
I guess you could (for the purpose of obtaining a solution) "safely" assume it's an air cored coil and free of the influence of any nearby magnetic material.
 
  • #4
12
0
[tex]\PhiB[/tex] = B * A (dot product)
 
  • #5
12
0
Magnetic Flux = B dot A

Because the Area is constant
d[tex]\Phi[/tex]/dt = dB/dt * A cos 20

so emf = N d[tex]\Phi[/tex]/dt

We know emf because of ohm's law V=iR

Therefore,

iR = dB/dt * N * A cos20

dB/dt = (iR)/(NA cos20)
 
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