# Magnetic field and current intensity of a loop of wire

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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)?

edited May 18, 2018
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 vote

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 (28,448 points)
edited May 17, 2018
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?