Finding coil area from charge and magnetic field using Faraday's law

[AlexPilk]

Homework Statement

In a uniform magnetic field with induction of 0.1 Teslas - a coil is located perpendicular to the lines of induction (I suppose it's something like a ring of wire, meaning N=1). Resistance = 2 Ohms. What is the area of the "ring" if when the field is switched on - it will charge 50 microcoulombs?

The Attempt at a Solution

The area can be found from this equation I believe: Ф=B*S*cosa --> S=Ф/(B*cosa)
B = 0.1 teslas
cos(90 degrees) = 0, which doesn't make a lot of sense, since the flux won't be 0 if the lines of the magnetic field are going straight into the "ring", so I must be doing something wrong and it = 1.
If so: S = Ф/0.1
Now I have to find the flux given the charge and resistance.
I know that the voltage induced = change in flux/change in time. V=dФ/dt
I=dq/dt. V=I*R, so dФ/dt=dqR/dt --> dФ=dqR
If so: S=(5*10^-5*2)/0.1=100*10^-5=10^-3 m^2.

Is it correct?

 

[BOAS]

Homework Statement

In a uniform magnetic field with induction of 0.1 Teslas - a coil is located perpendicular to the lines of induction (I suppose it's something like a ring of wire, meaning N=1). Resistance = 2 Ohms. What is the area of the "ring" if when the field is switched on - it will charge 50 microcoulombs?

The Attempt at a Solution

The area can be found from this equation I believe: Ф=B*S*cosa --> S=Ф/(B*cosa)
B = 0.1 teslas
cos(90 degrees) = 0, which doesn't make a lot of sense, since the flux won't be 0 if the lines of the magnetic field are going straight into the "ring", so I must be doing something wrong and it = 1.
If so: S = Ф/0.1
Now I have to find the flux given the charge and resistance.
I know that the voltage induced = change in flux/change in time. V=dФ/dt
I=dq/dt. V=I*R, so dФ/dt=dqR/dt --> dФ=dqR
If so: S=(5*10^-5*2)/0.1=100*10^-5=10^-3 m^2.

Is it correct?

Is the axis of the coil perpendicular or parallel to the magnetic field?

It is the angle between the normal to the surface of the coil, and the magnetic field that you must consider.