Magnetic force from BIL

Fnet = m(dv/dt) = Fexternal - Fmagnetic), we can write m(dv/dt) = mg - (Vinduced*A)/Bo. Since the external force is mg, we can rewrite the equation as m(dv/dt) = mg - (mg/Vinduced)*Vinduced*A/Bo.Integrating both sides of the equation with respect to time, we get the velocity of the loop as a function of time (v(t)).Similarly, we can use the equations for induced EMF and the net force on the loop to find the displacement of the loop (y(t)) and the induced current (i(t)) as
  • #1
KleZMeR
127
1

Homework Statement



There is a rectangular loop submerged in uniform magnetic field, where all but one edge is in the field, and I am given W (length), Bo, R (resistance), L (self-inductance), m.

I am to find v(t), y(t), and i(t). when R = 0 and L =/ 0 (L is NOT =0)


Homework Equations



Fnet = m(dv/dt) = Fexternal - Fmagnetic

Fexternal = mg

Emfinduced = Vinduced= -(dflux/dt) = v*B*W

Fmagnetic = Iinduced*W*Bo

flux=L*I

Emfinduced = -L*(dI/dt)


The Attempt at a Solution




So I think the result is that two parallel sides with induced currents in opposite directions will cancel out, and only one side, length W will contribute to force, because if total loop was submerged into B field region then force would be zero.

And, so, I solved this for the conditions R=/0 and L = 0 (R is NOT =0), so I have not included any equations with R because I am not using it, so what I am trying to do is solve it in similar fashion as before, using the equation of motion.

So I think I am to use again the Fmagnetic = Iinduced*W*Bo

What I am having trouble with is finding Fmagnetic in terms of Iinduced in terms of the given variables.
I see that there is an equation relating flux to Iinduced, as well as Emfinduced to Iinduced

I'm not using the flux equation, because I don't see a differential area, because I am dealing with a wire here, or do I use the differential area covered by the whole rectangle even though there is no contribution from the two parallel wires?

My guess would be to again use:

Fmagnetic = Iinduced*W*Bo

and (dI/dt) = (vBoW/L) = (dy/dt)*(BoW/L)

and then divide out dt, because current over a change in time is not a current?
so I get:

(dI) = (dy)*(BoW/L)

and so then I = y*(BoW/L)

and then use:

m(dv/dt) = Fexternal - Fmagnetic = mg - (y*(BoW/L)*WBo)

and integrate?

but I don't know if I can do this...

any help would be graetly appreciated
 
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  • #2
!

Thank you for your post. I am a scientist and I would be happy to help you with your problem.

First, I would like to clarify a few things about the problem. You mentioned that all but one edge of the rectangular loop is in the magnetic field. Can you please specify which edge is not in the field? This will help me better understand the geometry of the problem.

Secondly, you mentioned that you are trying to solve the problem for the conditions R = 0 and L =/= 0. I believe you meant to say L = 0 and L =/= 0. Can you please confirm this?

Now, onto the solution. From the given equations, it seems like you are dealing with a Faraday's law problem. In this case, the induced EMF (voltage) is given by the equation Vinduced = -N(dflux/dt), where N is the number of turns in the loop and flux is the magnetic flux through the loop.

Since you are dealing with a rectangular loop, the flux through the loop can be calculated as flux = B*A*cos(theta), where B is the magnetic field strength, A is the area of the loop and theta is the angle between the magnetic field and the normal to the loop.

In this case, since the loop is submerged in the magnetic field, theta = 0 degrees and the flux equation becomes flux = B*A.

Now, let's consider the two sides of the loop that are parallel to the magnetic field. As you correctly stated, the induced currents in these sides will cancel out and there will be no net force on the loop due to these sides. However, the two other sides of the loop that are perpendicular to the magnetic field will have induced currents in the same direction, resulting in a net force on the loop.

Using the equation Fmagnetic = Iinduced*W*Bo, we can find the net force on the loop due to the two perpendicular sides. Since the force is in the same direction as the induced current, we can write the equation as Fmagnetic = Iinduced*Bo*A, where A is the area of one of the perpendicular sides.

Now, using the equation for induced EMF, we can write Iinduced = -Vinduced/Bo*A. Substituting this into the equation for Fmagnetic, we get Fmagnetic = -(Vinduced*A)/Bo.

Next, using
 

1. What is BIL in relation to magnetic force?

BIL stands for Biot-Savart law, which states that the magnetic field generated by a current-carrying wire is directly proportional to the current, the length of the wire, and the sine of the angle between the wire and the point where the field is being measured.

2. How does BIL affect the strength of magnetic force?

BIL is one of the factors that determines the strength of magnetic force. As the current, length of the wire, or angle between the wire and the point of measurement increases, so does the strength of the magnetic force.

3. Can BIL be used to calculate the strength of a magnetic field?

Yes, BIL can be used to calculate the strength of a magnetic field as long as all the necessary variables are known and the Biot-Savart law equation is applied correctly.

4. Is BIL a constant value?

No, BIL is not a constant value as it depends on the variables of current, length of the wire, and angle. The value of BIL can vary for different situations.

5. What are some real-life applications of BIL and magnetic force?

BIL and magnetic force are used in various technologies such as electric motors, generators, and MRI machines. They are also used in everyday objects like speakers, doorbells, and credit cards.

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