# How to find the direction of current in this rotating loop in a magnetic field using right-hand rule for moving charges?

So how can you determine the direction of the current here using just the right-hand rule but without using Lenz's law? The textbook says I should use the right-hand rule applied to the velocity of a charge but how does that help find the current? It only finds the resultant magnetic force on the moving charge. In this case, the magnetic flux is increasing, so applying Lenz's law means that the induced current's magnetic field will oppose the the flux which makes sense here with the direction of the current shown. But how can I just use the right-hand rule to find the current direction in this case? That's what my book says. Is there such a way?

The Lorentz force is given by $$\mathbf{F} = q \mathbf{v}\times \mathbf{B}$$ and indicates in which directions the charged particles (electrons here which have negative q) are pushed. If you consider an electron in the upper part of the loop, then we know the velocity direction of this electron and also the magnetic field at the location of the electron. If you then take the cross product of these two quantities, you know the force on the electron (via right-hand rule). The direction of the force will be parallel with the direction of the movement of the electron. The conventional current (movement of positive charges) will be in the direction opposite the flow direction of the electrons. This results in the current indicated on the figure.

• The cross product given above indicates the force on a positive charge and hence the direction of positive current flow. For any cross product, start with your right hand fingers in the direction of the first named vector (in this case,v). Hold your hand so that you can bend your fingers in the direction of the second named vector, (B). The direction of your thumb indicates the direction of the product (back along the top of the square loop). Commented Mar 14, 2020 at 19:35

For the sake of simplicity, let's assume this loop is in a perfect vertical position, and is slowly moving clockwise as shown in my diagram below. Let us take two points on this loop. Point A denotes the midpoint of the upper arm while Point B denotes the midpoint of the lower arm. We would not talk about the vertical arms for this analysis.

Now, when we look at Point A, let's say a force acts on it such that in every instant it is tangential to the radius of rotation. Let us call this force 'Force F'. This force F is the cause for the rotation of the loop.

The magnetic field at A is uniform in both, magnitude and direction. The force F is also a constant in magnitude, but its direction changes at every instant. Now, let us resolve this force into its horizontal and vertical components. The horizontal component makes an angle of zero degrees with the magnetic field. We know: $$F=IlB\sin(\theta)$$ $$\therefore F_x=IlB\sin(0°)$$ $$\therefore I=\frac{F_x}{Bl\sin(0°)}$$ Thus, we know that the current produced by the horizontal component of the force cannot be mathematically defined.

Let us analyze the vertical component next. This makes an angle of 90 degrees with the magnetic field. Therefore: $$F=IlB\sin(\theta)$$ $$\therefore F_y=IlB\sin(90°)$$ $$\therefore I=\frac{F_y}{Bl\sin(90°)}=\frac{F_y}{Bl}=\frac{F\sin(\phi)}{Bl}$$

Due to the sine of 90 degrees being 1, we know that there is a well defined current produced by the vertical component of the force. Now, by using Fleming's RHR here, we can predict a direction of the current:

• Stretch your thumb downward in the direction of the vertical component of the force at A
• Stretch your index finger in the direction of the magnetic field at A (to the right)
• Extend your central finger; its direction denotes the induced current (you should find it to be out of the page)

On doing this same process at B, we can find the induced current there too; note that the induced current at B would always oppose the direction of that at A. (the induced current at B goes into the page while that at A goes out of the page)

Two important points one should note regarding this:

1. The direction of the induced current will always change after every half rotation. The diagram shown by you shows the loop with the first half of its rotation while my diagram shows the loop in the second half of its rotation. See the difference in the current directions?

2. The current in the loop is never going to be constant in magnitude. The magnitude of the current is in direct proportion with the sine of $$\phi$$. This therefore generates an Alternating EMF and an Alternating Current (AC).

3. No current flows when the loop is perfectly in a vertical position, due to a zero vertical component of the force, while maximum current flows when the loop is in a perfectly horizontal position, due to maximum vertical component of the force.

4. Your diagram uses charge and velocity instead of current and force. To prevent any confusion, the formula I used, $$F = IlB\sin(\theta)$$ can also be written as $$F=qvB\sin(\theta)$$. Therefore, either formula works. I've learnt to use the IlB version, that's why I prefer it; but if you replace the force vector with a velocity vector and the current with the charge, you should get the same result.

One final thing, your diagram is actually describing the functioning of a Simple A.C. Generator. And the loop you mention is analogous to an armature coil which turns under the effect of an external torque. I would suggest that you read up on the concept of AC Generators too to get any further insight on this concept.

I hope my answer here explained this mechanism in enough detail. If you have any doubts, ask away. Thank you!