Solve EMF & Ohm's Law for Coil w/Magnetic Field Angle

[takumi_91]

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 don't know how to get 'dt' for my equation. i know there is another way to do this problem but i can't really come up with anything at the moment

 

[waht]

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)


\frac{dB}{dt}

we know that

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

so what is?

\Phi_B

 

[overt26]

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)


\frac{dB}{dt}

we know that

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

so what is?

\Phi_B

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.

 

[Triangulum]

\Phi_B = B * A (dot product)

 

[Triangulum]

Magnetic Flux = B dot A

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

so emf = N d\Phi/dt

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

Therefore,

iR = dB/dt * N * A cos20

dB/dt = (iR)/(NA cos20)