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Magnetic force and field

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Homework Statement


A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails that are d=12.0 cm apart and L=45.0 cm long. The rod carries a current of I=48.0 A in a direction that makes it rolls along the rails without slipping. A uniform magnetic field of magnitude 0.240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?
the answer is 1.07m/s
but I keep on getting wrong answers and I don't know what is the mistake
I used both
Energy and work equation
equations of moving in one dimension

but it didn't work
so how to get the right answer
and what is the mistake in using energy and equations of moving in one dimension

Homework Equations




The Attempt at a Solution


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Answers and Replies

  • #2
TSny
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We ask that you please fill out all three sections of the template in the future.

Using work and energy is a good way to approach the problem. Without seeing more details of your attempt, we cannot tell where you made a mistake.
 
  • #3
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E
We ask that you please fill out all three sections of the template in the future.

Using work and energy is a good way to approach the problem. Without seeing more details of your attempt, we cannot tell where you made a mistake.
Ei=Ef
there is no magnetic fied in the end and its at rest in the beggining so
I A B = 0.5 m v^2
v=(2IAB/m)^0.5
then the answer is wrong
 
  • #4
TSny
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Ei=Ef
there is no magnetic fied in the end and its at rest in the beggining so
I A B = 0.5 m v^2
What does "A" represent? What type of energy is represented by I A B?

Also, does 0.5 m v^2 include the rotational energy associated with rolling?
 
  • #5
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What does "A" represent? What type of energy is represented by I A B?

Also, does 0.5 m v^2 include the rotational energy associated with rolling?
ENERGY OF MAGNETIC FIELD IS
I A B
WHERE I is the current
A is the cross-sectional area
B is the magnetic field
 
  • #6
TSny
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ENERGY OF MAGNETIC FIELD IS
I A B
WHERE I is the current
A is the cross-sectional area
B is the magnetic field
I've never seen the formula I A B for "magnetic field energy". However, it is possible to show that I A B is the work done by the magnetic force on the rod if A denotes the distance between the rails multiplied by the distance the rod travels. I would think that you would need to show how you arrive at the expression I A B for magnetic field energy, unless this was already derived in your class or textbook.
 
  • #7
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I've never seen the formula I A B for "magnetic field energy". However, it is possible to show that I A B is the work done by the magnetic force on the rod if A denotes the distance between the rails multiplied by the distance the rod travels. I would think that you would need to show how you arrive at the expression I A B for magnetic field energy, unless this was already derived in your class or textbook.
A is pi r^2 of the rod
 
  • #8
TSny
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A is pi r^2 of the rod
Hmm. If A is ##\pi r^2## of the rod, then I don't think I A B can be interpreted as the initial energy Ei of the system. Instead, you can use the work-energy theorem: the net work done on the rod equals the change in KE of the rod. For the KE, you will need to take into account that the rod rolls without slipping.
 
  • #9
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Hmm. If A is ##\pi r^2## of the rod, then I don't think I A B can be interpreted as the initial energy Ei of the system. Instead, you can use the work-energy theorem: the net work done on the rod equals the change in KE of the rod. For the KE, you will need to take into account that the rod rolls without slipping.
what made you think we can't use it if we can't use it now when can we use it is there any conditions of using it
so if we used the work equation it will be
I L B d = 0.5mv^2
 
  • #10
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Hmm. If A is ##\pi r^2## of the rod, then I don't think I A B can be interpreted as the initial energy Ei of the system. Instead, you can use the work-energy theorem: the net work done on the rod equals the change in KE of the rod. For the KE, you will need to take into account that the rod rolls without slipping.
when we use the work equation we get v=1.314m/s and the answer is 1.07 so i really need to know the conditions of using work and energy
 
  • #11
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what made you think we can't use it if we can't use it now when can we use it is there any conditions of using it
Sorry, I'm not following what you said here. Where have you seen the formula I A B for magnetic field energy?
so if we used the work equation it will be
I L B d = 0.5mv^2
The left side is good for the work done by the magnetic force. The right side does not represent the total KE of the rod. The rod rotates about its central axis as it rolls without slipping. So, there is an additional KE associated with this rotation.
 
  • #12
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Sorry, I'm not following what you said here. Where have you seen the formula I A B for magnetic field energy?
The left side is good for the work done by the magnetic force. The right side does not represent the total KE of the rod. The rod rotates about its central axis as it rolls without slipping. So, there is an additional KE associated with this rotation.
and what is that KE equal to
do you mean angular velocity or something
sorry but its the first time I hear about an additional kinetic energy for rolling
 
  • #13
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  • #17
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I=md^2/12
This is not the correct expression for the moment of inertia of a cylinder that rotates about its central axis. You’ve used the formula for the moment of inertia about a different axis and for a thin rod.
 
  • #18
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w
This is not the correct expression for the moment of inertia of a cylinder that rotates about its central axis. You’ve used the formula for the moment of inertia about a different axis.
what is it
if it is md^2/3 it is still a wrong answer
 
  • #19
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if it was md^2/2 it is still wrong
 
  • #20
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ohhhhh okay i got it
it is 1.07 my mistake
thanksssssssssssssssss thanks thanks
 
  • #21
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w

what is it
if it is md^2/3 it is still a wrong answer
That’s not the correct formula either. Be sure you are clear on the orientation of the axis of rotation.
 
  • #22
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That’s not the correct formula either. Be sure you are clear on the orientation of the axis of rotation.
is it F=IqI v B
or
F=q v B
 
  • #23
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That’s not the correct formula either. Be sure you are clear on the orientation of the axis of rotation.
there are questions when you solve them you get a positive magnetic field but when you check your answer it is negative why
does it have anything to do with either its a proton or electron
 
  • #24
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ohhhhh okay i got it
it is 1.07 my mistake
thanksssssssssssssssss thanks thanks
OK, good.
 
  • #25
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That’s not the correct formula either. Be sure you are clear on the orientation of the axis of rotation.
can you solve this plz
An electron has an initial velocity of (12 j + 15 k) km/s and a constant acceleration of 2*10^12 i m/s^2 in a region in which uniform electric and magnetic fields are present. If B= (400*10^-6 i) T, find the electric field.
I SOLVED IT AND FOUND Ex Ey Ez but all of them was positive when i checked just Ez was positive
 

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