# How to determine direction of current in this two loop system?

Let's say, there are two loops made from a single wire , loop 1 of bigger area and loop 2 having smaller area.
Now let us suppose, a magnetic field exists in space perpendicular to the plane of the loops and in the outward direction and it's magnitude is increasing linearly. (To neglect effects of self inductance)

Now , what would be the direction of current in this case?
According to my teacher, direction of current in loop 1 would be in clockwise direction which would result in direction of current in loop 2 being in anticlockwise direction, and the stated reason was that, since loop 1 has bigger area, it has more contribution in the flux .

According to lenz law , flux due to current induced should oppose the already present magnetic field , but I can't see how loop 1 flux has more opposing flux to overcome supporting effect of loop 2 , even if area of loop 1 is bigger .
Isn't it also that loop 1 magnetic field strength is less (going by formulae for magnetic field at the center of loop).

So , I am just confused about determining the direction of current in this case.

• The loops can only oppose a change in the magnetic field by generating one of their owns which is given by B=ui/2R at the center of the loop. So with this relation you should be able to figure it out. Aug 21, 2021 at 11:06
• @BlackThunder Even if we were to only consider Magnetic field , then smaller loop must have current in clockwise direction , which is not ! Aug 21, 2021 at 11:31

. . . . . there are two loops made from a single wire . . . . .

If that is the case then there is one (soap film) surface to consider as shown in the image below.

The total magnetic flux through that surface can be found noting that the directions of the normals to the two surfaces change direction.
The induced emf will depend on the rate of change of flux linked with the whole (copper) loop.
The emf will in turn determine the direction of the induced current which will try to oppose the change producing it.

What happens is that overall for the whole loop the direction of the induced current does try and negate the effect of a changing magnetic field.
Looking at the image and assuming that the magnetic field is increasing in the direction of the lower unit vector $$\hat n$$, the induced current will flow in a clockwise direction in the lower part of the loop and in an anticlockwise direction in the upper part of the loop.

That upper part of the loop does try to negate the effect of the changing magnetic field (Lenz) as it does reduce the overall emf but it is not enough to make the overall induced current travel in a clockwise direction in that part of the loop.

• Usual question. Why the downvote? Aug 22, 2021 at 7:27

Theory:Due to changing magnetic field, an emf is induced in a loop which produces a current which produces a magnetic field to oppose the change in flux. Formula for induced emf will be e=d(f)/dt where f is flux and I'm not talking about direction here, it can be determined by lenz law

Now according to the given situation, the emf induced in the left loop will be more than the emf induced in the right loop as area of the left loop is more and f=BAcosx(here x=0°)

So basically to represent emf you can draw a battery and you'll see that the emfs induced in the loops oppose each other and as the emf induced in the right loop is less than left one, the emf in left loop dominates and hence now by Lenz law you can say that direction of current will be clockwise in left loop and therefore anticlockwise in right loop

Hope this clears your confusion now

According to Lenz Law E=-$$\dfrac{d\phi}{dt}$$. As magnetic field is increasing, flux through any loop is given by BA. Where B is the magnetic field, so $$\dfrac{d\phi}{dt}$$=$$\dfrac{dB}{dt}$$A. As $$\dfrac{dB}{dt}$$ is same for both the loops but A, is different, you can find the corresponding Voltages induced in both and can easily find the current direction. Hope this helps