- #1

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## Homework Statement

Question and diagram below

## Homework Equations

F=BIL

F=QVB

Vback=(epsilon)-Ir

## The Attempt at a Solution

F=QvB

ma=QvB

QvB/m=a ?

I really don't know how to start this question.

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- Thread starter mrcheeses
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- #1

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Question and diagram below

F=BIL

F=QVB

Vback=(epsilon)-Ir

F=QvB

ma=QvB

QvB/m=a ?

I really don't know how to start this question.

- #2

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- #3

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I don't think gravity effects it, but I found out that the diagram is of a rail gun.

- #4

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I just tried doing this again and my work is below...

V=IR

I=V/R

F=BIL

Fnet=Fg-F

ma=mg-BIL

ma=mg-BVL/r

a=(mg-BVL/r)/m

Not sure if this is right but its not as a function of time!

- #5

ehild

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ehild

- #6

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If the magnetic flux changes in time, that means the area would change or the magnetic field would change.

The magnetic flux would change when the crossbar falls because there will be a greater area.

So to get the equation, would I have to start off with F=BIl and since Fnet=ma, ma=BIl and then incorporate time in it somehow, or would I have to start with ε=-NΔΦ/Δt?

- #7

ehild

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So to get the equation, would I have to start off with F=BIl and since Fnet=ma, ma=BIl and then incorporate time in it somehow, or would I have to start with ε=-NΔΦ/Δt?

You need both equations. Faraday's law of induction gives you the unknown emf in the circuit from what you find the current as I=ε/R. From the current, you get the force of the magnetic field on the crossbar. From Lenz's law you know that the induced emf works against the effect that produced it. The resultant of the magnetic force and gravity determines the acceleration of the bar.

ehild

- #8

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Then get ε on one side which gives

ε=FR/BL

and then substitute Faradays-Lenz law for ε which gives

-NΔΦ/Δt=FR/BL

Then make another equation for just forces.

Fnet= F+Fg

ma=F+mg

F=ma-mg

Sub it in for F and then do the math and get acceration on one side by itself which gives

mghΔt-NΔΦBL/ΔtmR=a

Is this right?

- #9

ehild

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mghΔt-NΔΦBL/ΔtmR=a

Is this right?

It is almost right, but you have not finished yet.

What is h? Why is it there in your formula?

What is N? Is it needed in this case?

The magnetic flux is related to B. How?

The time derivative of the magnetic flux is related to the velocity of the bar. How?

ehild

- #10

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Sorry the h was a typo, it was supposed to be R. I guess you can just remove the N since there aren't coils wrapped around and we can consider it as one coil. Which gives

(mgRΔt-ΔΦBL)/ΔtmR=a

Magnetic Flux is related to magnetic field by Flux=FieldxArea, and since the area of a rail gun would change as the bar moves down, ΔΦ=BΔA so that gives

(mgRΔt-(B^2)AL)/ΔtmR=a

Would this be right?

And I don't get the last question, but I do not think I need calculus for my physics course so maybe my teacher just wanted us to include time in our equation.

(mgRΔt-ΔΦBL)/ΔtmR=a

Magnetic Flux is related to magnetic field by Flux=FieldxArea, and since the area of a rail gun would change as the bar moves down, ΔΦ=BΔA so that gives

(mgRΔt-(B^2)AL)/ΔtmR=a

Would this be right?

And I don't get the last question, but I do not think I need calculus for my physics course so maybe my teacher just wanted us to include time in our equation.

Last edited:

- #11

ehild

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You left out Δ in front of A. So a=(mgRΔt-(B^2)ΔAL)/ΔtmR wich equals(mgRΔt-ΔΦBL)/ΔtmR=a

Magnetic Flux is related to magnetic field by Flux=FieldxArea, and since the area of a rail gun would change as the bar moves down, ΔΦ=BΔA so that gives

(mgRΔt-(B^2)ΔAL)/ΔtmR=a

Would this be right?

a=(mgR-(B^2)L (ΔA/Δt))/mR

The area of the loop is A*y . y is the vertical length. As the bar falls, y increases. ΔA = LΔy. The velocity of the bar is v=Δy/Δt, so ΔA/Δt=Lv. Substitute for ΔA/Δt into the formula for the acceleration.

As the bar falls, its speed increases, but the magnetic force also increases, so the acceleration decreases with time. Can you find the velocity when the acceleration is zero?

ehild

- #12

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I got mgR/B^2L^2=v when I did all the math. So would that be the answer for the second question?

- #13

ehild

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If you meant v=mgR/(B^2L^2) then it looks correct. Use parentheses.

ehild

ehild

- #14

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Wait, for (mgRΔt-(B^2)ΔAL)/ΔtmR=a , would that work? If the change in time is 0, then the acceleration is undefined.

Would this work?

Can you apply the kinematics equation into this since it is for constant acceleration?

The question says that the object falls from rest and with this solution, at t=0, a would =g and that would work since at t=0, the magnetic force would be 0 by F=qvb.

Would this work?

Can you apply the kinematics equation into this since it is for constant acceleration?

The question says that the object falls from rest and with this solution, at t=0, a would =g and that would work since at t=0, the magnetic force would be 0 by F=qvb.

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- #15

ehild

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Wait, for (mgRΔt-(B^2)ΔAL)/ΔtmR=a , would that work? If the change in time is 0, then the acceleration is undefined.

Can you apply the kinematics equation into this since it is for constant acceleration?

The question says that the object falls from rest and with this solution, at t=0, a would =g and that would work since at t=0, the magnetic force would be 0 by F=qvb.

I do not understand why the acceleration should be undefined at t=0.

I explained that ΔA=LvΔt. I do not know why you replaced ΔA by Lat. It is wrong. When you take into account that ΔA=LvΔt in the formula for acceleration you can simplify with Δt. Δt means a short time interval, it is not zero! There is no "t" in the formula.

You get

At t=0, the acceleration is g as v=0. As the rod falls, its velocity increases, but the acceleration decreases. The velocity increases with smaller and smaller rate. The acceleration tends to zero with time and the velocity tends to the value v(max)=mgR/(B^2L^2) as you have figured out.

Using that a=dv/dt, the formula in blue means a differential equation for v. I quess you have not studied differential equations yet. The solution is of the form

v=v(max)(1-e

You can not apply the constant-acceleration formula v=at!

ehild

- #16

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- #17

ehild

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I am glad to be of help.

ehild

ehild

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