Average current around a magnetic loop that changes its shape

[Alefan_]

Homework Statement

The picture shows a loop formed by two semicircles linked together by the straight sections of a wire. The loop and the dotted line lie on the same plane.
The smaller semicircle, which has a radius of 0.20m, starts rotating with an angular speed of ω=1.5 rad/s around the dotted line until it gets into a position that's perpendicular to the bigger semicircle (B).
A constant magnetic field of B= 0.35 T is directed upwards, perpendicular to the plane where the loop lies.
The loops resistance is R=0.025 Ω
Find out the induced loop current I while it changes its form, hoing from picture A to B.

Relevant Equations

V=R×I
fem=ω×A×B
Semicircle Area=½×π×r^2

To find out what the induced loop current was i used the formula:
V=R×I

To find out what the value of V was i used the formula that links electromotive force (fem) to angular speed:
Fem=ω×B×A

The only thing that's missing is the loops area but considering that it's a semicircle and that the radius value was given I used this formula:
A=½×π×r^2

Substituting all of the values the answer comes out as incorrect, I'll leave a photo of my procedure here:

The result comes out as 1.31 A while it should be 0.84 A, what's wrong with my procedure?

Thank you to anyone who will help.

 

[TSny]

The formula $\varepsilon = \omega AB$ is not going to give you the correct result. This formula is for finding the maximum emf induced in a loop of area $A$ rotating in a uniform magnetic field. You need to find the average emf induced when the smaller semicircle rotates through a quarter turn.

I recommend that you go back to basics. According to Faraday's law, the average induced emf is given by

$\overline {\varepsilon} = \frac{\Delta \Phi}{\Delta t}$. Find an expression for $\Delta \Phi$ in terms of $B$ and $r$. Find an expression for $\Delta t$ in terms of $\omega$.

 

[kuruman]

A different approach would be to find the total charge $\Delta Q$ that flows through the semicircle and then divide that by the time $T$ is takes the loop to rotate by a quarter revolution. Note that $$\Delta Q=\int_0^T I~dt=\frac{1}{R}\int_0^T\frac{d \Phi}{dt}dt=~?$$

 

[rude man]

I have nothing to add to TSny's post except that my answer was about 5% higher than the given one.
EDIT: On recomputing I also got exactly 0.84A. My bad.

 

[kuruman]

I have nothing to add to TSny's post except that my answer was about 5% higher than the given one.

I got exactly the given answer, no rounding.

 

[collinsmark]

I got exactly the given answer, no rounding.

I concur. I also got the given answer, 0.84 A, exactly (no rounding necessary).

@kuruman's method is the correct approach here. The tricky bit is that the problem statement is asking for the "average" current (not the "peak" current). That, and the current is not a constant value -- it's time varying. To complicate things further, the current does not vary linearly with time, it varies sinusoidally with time [wink, wink (i.e., that's a hint)].

So, to solve for the "average" current, the steps are

• Find an expression for the instantaneous current as a function of time.
• Integrate over time from $t = t_0 = 0$ to $t = t_1$ to find the total charge $Q$, such that $\theta_1 = \omega t_1 = \frac{\pi}{2}$.
• Calculate $i_{ave}$ by dividing the charge by $t_1$ (i.e., $i_{ave} = \frac{Q}{t_1}$).