Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Magnetic field and current intensity of a loop of wire

1 vote

A loop of wire of radius a, auto-induction L and resistance R is located at XY plane. The magnetic field goes through the loop having +Z direction. The modulus of the magnetic field has the equation B(t) = At (A is constant). Calculate :
a) Total Magnetic flux on the loop
b) Current intensity along the loop in function of time

What I attempted to do:

a) Magnetic flux is defined as $\phi_{m} = \int_{S} \overrightarrow{B} \hat{n} dA$

My result is $\phi _{m}$ = $\pi a^{2}At$. I just plugged in the value of the circle's area.
I would like to be rigorous. Shouldn't I obtain the area of the loop through setting up an integral based on the equation of the circumference $(x-p)^2+(y-q)^2=r^2$? Defining the integral from 0 to $a$ and placing the circle at the origin of coordinates:

$$4\int_{0}^{a} \sqrt{a^2-x^2} dx$$

b) we know that

If the magnetic flux through the loop is not constant, a emf (which is equal to the variation of such a magnetic flux per unit of time) is induced on the surface's loop. Therefore:

$$\epsilon = - \frac{d\phi_{m}}{dt}$$

Based on the answer at a)

$$\epsilon= \pi a^{2}A$$

Once I obtained the emf (its modulus) I can get the current. Then:

$$I=\frac{\pi a^{2}A}{R}$$

I am struggling to get the answer $I(t)=C_0(1-e^{-Rt/L})$

Solving the differential equation I obtained:

$$t = \frac{-L}{R}ln(\frac{\epsilon}{R}-I)$$

By which mathematical method can I reach the answer (from the solution I got)?

asked May 14, 2018 in Physics Problems by Jorge Daniel (676 points)
edited May 18, 2018 by Jorge Daniel
What is a *spire*? My guess is that you mean a *spiral* (ie a coil of wire, like a solenoid) or a *loop* of wire. Whichever it is, I presume the axis of the loop or solenoid is in the z direction. Is that correct?
I don't quite understand the problem. The magnetic flux through the loop induces an emf (Faraday's Law) which is constant because the rate of change of flux is constant. So the current in the loop is constant. Self-induction occurs when the current is changing. But it doesn't change in this question so why is self-induction mentioned? Is it an intentional *red herring*?

Have you typed the question exactly as it is printed? What answers have you been given?
Exactly, it is a spiral (like a solenoid) and the axis of the solenoid is in the z direction that is right.
Your diagram is confusing. It does not seem to match your question (which does not mention any point P), or your comment that there is a solenoid not a loop. It seems to be for a different problem - eg calculating the magnetic field on the axis of a loop of current.

How many coils does the solenoid have?
To be honest I am struggling to understand it (description of the  provided problem poorly detailed). It comes from an exam and I translated it literally. I do not have the solutions, but I know my answer to a) is wrong since I used $4 \pi a^{2}$ as the area (it is not a sphere right?)
It is a spiral (like a solenoid). As my diagram is misleading you,  I remove it.
The problem does not specify how many coils have the solenoid. I might be misunderstanding the problem and it is a loop instead of a solenoid. It is always regarded as  ‘espira’ (spanish word), which I am not sure if it means loop or spiral (solenoid).
Assume that it means a single circular loop of wire. Can you solve the problem from the hints in my 2nd comment? You have the correct equation for flux but the wrong formula for area of the plane loop. I do not see how self-induction comes into this problem so assume it is a *red herring*.

By the way, No the differential equation does not help. You only need Faraday's Law of Induction.
Then $B= \frac{\mu I}{2} \pi a$? I supposed that the problem is asking for B at the center of the circular loop. Let me read a little bit of Faraday’s Law of induction before answering b)
No. Where did you get that from? B is given in the question. I is not given and is to be found. Use Faraday's law to find the emf in the loop, then use the emf and resistance to find current.
I updated my question again Sr
You can verify that $I(t)=C_0(1-e^{-Rt/L})$ is a solution by substituting into the equation. Any equation of the form $\frac{dI}{dt}+bI=0$ has a solution of the form $I(t)=c_1e^{-bt}$ (complementary function).  Any equation of the form $\frac{dI}{dt}+bI=B$ also has a solution of the form $bI(t)=B$ (particular integral). The general solution is the sum of the CF and PI.
Yes I know, the problem is I don’t know how to get it. But if this specific point is not suitable for this site, I can ask for it at Math SE no problem
Yes it would be better to use a different site to learn how to solve differential equations. eg https://math.stackexchange.com/questions/2211874/differential-equation-first-order

1 Answer

1 vote
Best answer

a) Your answer for $\phi_m$ is correct. The magnetic field is uniform over the area of the loop, which we assume is a circle. The magnetic flux through the loop is simply $B\times$ area.

No you do not need to derive the area of a circle from 1st principles. That is not being rigorous, it is being pedantic. You always have to make some assumptions without proof, particularly mathematical formulae (eg Pythagoras' Theorem, trigonometry relations, sums of arithmetic and geometric series) which are either elementary or given in data sheets.

b) This is also correct. The changing magnetic flux through the loop induces an emf (Faraday's Law), which drives a current round the loop.

It is puzzling why the question mentions self-inductance, because it is not needed in the calculation.

If the induced emf (and therefore also the current in the loop) were changing with time, instead of being constant as here, then the self-inductance of the loop generates a 'back emf' $L\frac{dI}{dt}$ which opposes the increase in the current. Applying Kirchhoff's Voltage Law : $$V=\frac{d\phi}{dt}-L\frac{dI}{dt}=IR$$

In the present case in which $\frac{d\phi}{dt}$ is a constant, and the magnetic field is switched on suddenly at time $t=0$ then $I(0)=0$. The solution is $$I(t)=C_0(1-e^{-Rt/L})$$ ie the current rises exponentially to its saturation value $C_0=A\pi a^2/R$. On the other hand, if the magnetic field were switched on a long time ago then $t\to \infty$ so $$I(t)=C_0$$ In a more complex situation $\frac{d\phi}{dt}$ depends on $t$.

answered May 15, 2018 by sammy gerbil (28,806 points)
edited May 17, 2018 by sammy gerbil
I will think about why it mentions self-inductance. Please apologize me for being pedantic, it was not my intention at all :)
I found out why I was not getting $\pi a^2$. The problem was I just took into account the first quadrant. I should have multiplied the integral by 4
As always great help, thank you
About the possible *red herring*. The merely explanation coming to my mind is that b) asks for an expresion depending on time. If current were time-dependent, then there would be self-induction. But how could we do so?