Induced current in a closed loop in opposing directions

Question

When a magnetic field is given to the loop as in the figure(sorry for the horrible drawing);

I'm told that there will be an anticlockwise current induced in both the 'inner' and 'outer' loops, resulting in opposing currents, and since the area of the outer loop is larger, the net current flows so that it's anticlockwise in the outer loop.

I'm confused. Isn't there just one closed loop here? Doesn't a circuit have to be closed for current to be induced in it?

Answer 1

When the B-field is applied as indicated in the picture, according to the Maxwell-Faraday law:

$$\nabla\times E = -\frac{\partial B}{\partial t}$$

In order to know the direction of the induced E-field, the curl follows the right hand rule. So, because of the minus sign, the conventional current flows counterclockwise.

The only role of the inner part of the loop is to reduce its area $S$, reducing the magnetic flow. If the same length of wire formed a circle for example, $S$ would be bigger. The integral form of the equation above is:

$$\oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S}$$

Or, renaming some terms: $$\epsilon= -(d\phi/dt)= -(d(BS)/dt) $$

Answer 2

For simplicity assume that magnetic field is uniform in a downward direction and increasing in magnitude at a constant rate. I will explain in a minute why I have made this assumption from the information that you have provided.

Without saying anything about the nature of the loop other than it is of constant area $A$, the rate of change of magnetic flux is $\frac {d(BA)}{dt} = A \frac {dB}{dt}$ and this is the induced emf in the loop (Faraday).

If the loop is conducting then the induced emf induces a current in the circuit which you have stated to be anticlockwise in the top left hand corner and the same current flow though each part of the loop although there may be ambiguity about direction if one tries to use the terms anticlockwise and clockwise for each segment of the loop.

At each part of the loop the induced current produces a magnetic field which is upwards within the loop ie in the opposite direction to the increasing external magnetic field.
This is Lenz's law in action as the induced current is trying to reduce the rate at which the magnetic field (and hence the magnetic flux) is changing within the loop.
That is happening for all segments of the conducting loop and so those bits of the loop where the current seems to go clockwise still produce a magnetic field in opposition to the changing external magnetic field.

So the shape of the loop does not matter.
What does matter is the enclosed area $A$, the rate of change of external magnetic field (and the resistance of the loop).

Answer 3

Well you are right in your understanding that there is only one loop. Anyone who has used the terms "inner" and "outer" must specify what he/she means.

There is one more thing about your question. You seem to have a bit of problem with the concept here. Let me explain.

This is what your loop looks like with the magnetic field. Now the current that is induced in this is zero. The reason for this is Faraday's law that governs electromagnetic induction.

$$\epsilon = -N\frac{d\phi}{dt}$$

where, $\epsilon$ is the induced electromotive force (or more appropriately voltage)

$\phi$ is the flux of the magnetic field through the loop

$N$ is the number of loops.

For our case, $N$ is $1$.

Now if our loop is a conductor, we will observe current. And in such a case, how do we determine the direction of our current? The magnitude is given by ohm's law.

There are two ways which you can use to arrive at the direction.

Method 1: Using Lenz's law

According to Lenz's law, the induced current will be such that the magnetic field due to it will oppose the original magnetic field. If you are a high-school student, it is highly recommended you use this law because of its simplicity.

Method 2: Using Faraday's Law

Faraday's law is self sufficient in determining the direction of the current. For this you need to have a look at one of the Maxwell's equation:

$$\nabla \times \mathbf{E} = -\frac{\partial{\mathbf{B}}}{\partial{t}}$$

The mathematical complexity should be evident by now. However, if you are well acquainted with vector calculus, you may proceed as follows:

You have the Magnetic Field, therefore you can obtain the Electric field in its vector form. After doing so(it will be a lengthy process) you can analyze the effect of the electric field on an electron in the conductor. With the direction of electron movement you can get the direction of current.

Now use these laws in the case you have mentioned, and you will see that the current is zero.

Hint: The magnetic field and the area of the loop are both constant with no time dependence.

Hope this helps you! Feel free to comment and ask any questions you have!

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EDIT 1:

Okay so in the comments the OP asked about why the current is zero. This edit is regarding that comment.

Since you want to know why the current is zero, you must take a look at the laws governing the magnitude of the current.

In Faradays law, we see the term $\phi$ which is the flux of the magnetic field through the loop. This flux is given by,

$$\phi=\mathbf{B.A}$$

Now we have three quantities that may change with time:

1. The Magnetic field $\mathbf{B}$
2. The magnitude of Area of the loop $A$
3. The angle between the normal to the loop surface(direction of the Area vector) and the magnetic field.

Now look at your case to see if $\phi$ which is given by the above expression is a function of time. You will find the answer.