Is My Answer Correct for Calculating Average Current on a Ring?

[songoku]

Homework Statement

What is the average current on the circumference of an insulator ring if it is rotated with frequency f? (Please see the figure below)
a) 4Qf
b) 8Qf/r
c) 4Qf/r
d) 8πQf

Relevant Equations

I = q/t

$$I=\frac{q}{t}$$
$$=\frac{4Q}{T}$$
$$=4Qf$$

My answer is (a). Is that correct?

Thanks

 

[jbriggs444]

My answer is (a). Is that correct?

I agree. Total charge of 4Q rotating past a given point at f repetitions per unit time.

One can rule out b) and c) immediately because radius does not enter in.

One can rule out d) because even if they were talking about angular frequency instead of regular frequency, the $\pi$ is in the numerator, not the denominator where it would belong.

 

[Orodruin]

I agree. Total charge of 4Q rotating past a given point at f repetitions per unit time.

One can rule out b) and c) immediately because radius does not enter in.

One can rule out d) because even if they were talking about angular frequency instead of regular frequency, the $\pi$ is in the numerator, not the denominator where it would belong.

Most of all, all except (a) can be ruled out on the basis of having the wrong physical dimension.

 

[kuruman]

One can rule out d) because even if they were talking about angular frequency instead of regular frequency, the π is in the numerator, not the denominator where it would belong.

Why? If somehow one thinks that $~I=q\dfrac{d\phi}{dt}=\omega q,~$ what does one get after substituting $q=4Q$ and $\omega=2\pi f$?

 

[jbriggs444]

Why? If somehow one thinks that $~I=q\dfrac{d\phi}{dt}=\omega q,~$ what does one get after substituting $q=4Q$ and $\omega=2\pi f$?

The wrong answer.

If one is going to assert that $~I=q\dfrac{d\phi}{dt}=\omega q,~$ then one must be interpreting $q$ as charge per radian. But the given charge of $4Q$ is spread across $2 \pi$ radians. So the actual charge density is lower. By a factor of $2\pi$ radians per cycle.

If one expected that $f$ is an angular frequency in units of radians per unit time then the substitution $\omega = 2 \pi f$ is also wrong-headed. A more reasonable substitution would be $\omega = f$.

With the corrected substitutions in mind, a correct response would have been: $~I = \frac{4Qf}{2\pi}$ (for $f$ interpreted as an angular frequency).

As previously pointed out, answer d) has the $\pi$ in the numerator when it would belong in the denominator.

 

[kuruman]

The wrong answer.

I agree. I will stop here because I came to my senses and realized that it is unprofitable to hypothesize what could be going through one's mind when one puts down a specific incorrect answer.

 

[DaveE]

I agree. I will stop here because I came to my senses and realized that it is unprofitable to hypothesize what could be going through one's mind when one puts down a specific incorrect answer.

Except that that's what instructors do every time they make a good multiple choice exam question. The right answer is the easy part.

 

[Orodruin]

Except that that's what instructors do every time they make a good multiple choice exam question. The right answer is the easy part.

In particular, I’d expect at least one false answer with the correct physical dimension. This one was just silly. Even if one did believe that radius could play a role - the physical dimension of those answers rule them out immediately.

 

[songoku]

Thank you very much jbriggs444, Orodruin, kuruman, DaveE