Calculation of charge flowing through a loop

[Antoha1]

Homework Statement

A frame (a loop) with an area of 5 cm^2 and a resistance of 2 Ω is placed in a uniform magnetic field with an induction (B) of 0.1T. A galvanometer is connected to the frame. What charge (q) will flow through the galvanometer when the frame is rotated through an angle of $\beta$=120 degrees? At the beginning of the observation, the plane of the frame is perpendicular to the lines of magnetic field.

Relevant Equations

$$\phi=BAcos\theta$$

$$\Delta\phi=\phi_{1}-\phi_{2}=BA_{1}-BA_{2}=B(A_{1}-A_{2})=B(Acos\alpha+Acos(\alpha+\beta))$$

$$\varepsilon=\frac{\Delta\phi}{\Delta t}=IR=\frac{\Delta q}{\Delta t}R$$

$\varepsilon$ - EMF

from the written formulas, charge is:

$$\Delta q=\frac{BA}{R}(cos\alpha-cos(\alpha+\beta))$$ .

It is said that in the beginning, the Area was perpendicular to magnetic field lines so the

$\alpha=0$
$\phi_{1}=BAcos0=BA$ ,

Refered to that:

$$\Delta q=\frac{BA}{R}(1-cos\beta)$$

but, thinking logically, cosine function becomes negative at 120 degrees so the subtraction of cosines becomes the sum. And I do not think that $\Delta q$ can be bigger than BAcos0

I'm thinking of this formula as final:

$$\Delta q=\frac{BA}{R}(1-\left| cos\beta \right|)$$

 

[Gordianus]

Why did you change from cos(beta) to |cos(beta)| ?

 

[Antoha1]

Why did you change from cos(beta) to |cos(beta)| ?

I think that change in magnetic flux cannot be bigger than this exact $\phi=BAcos0$ initial magnetix flux.
Well, imagining from other perspective:
If the loop would have been turned not 120 degrees as it's said, but 60 degrees (just to the other side) wouldn't the change in magnetic flux $\Delta\phi$ be the same? (as 120 degrees)

 

[Gordianus]

Keep in mind that flux carries a dot product and sign matters.

 

[Antoha1]

Keep in mind that flux carries a dot product and sign matters.

so without absolute of cosB it would be correct?

 

[nasu]

The current through the loop will change direction for the last part of the rotation. Would you add the charge going in the opposite direction to the "charge flowing through the galvanometer" or subtract it? For me, it's not clear what is expected.

 

[kuruman]

The current through the loop will change direction for the last part of the rotation. Would you add the charge going in the opposite direction to the "charge flowing through the galvanometer" or subtract it? For me, it's not clear what is expected.

I think it is fair to say that the direction of the current density vector $\mathbf J$ does not matter. Counting coulombs going past a point in a circuit is analogous to traffic flow. If you count cars going past a point on a highway over a time interval, you might say "I saw 100 cars go by" or maybe "I saw 60 cars traveling North and 40 traveling South" but you will not say "I saw 20 cars go by" or, even worse, "I saw -20 cars go by." In this example, if the angle by which the loop rotated were 360° instead of 120° would it be fair to say that because $\Delta \Phi =0$, zero charge has flown through the galvanometer?

 

[Gordianus]

Let me plot with with Geogebra (can I do this?) the cosine part of the magnetic flux:

The sign of the first derivative doesn't change in the range from zero to 120 degrees. Thus, if we assume the loop's resistance is high enough to neglect inductance effects, the induced flows in the same direction.