# Magnetic force and field

## 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?
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

## The Attempt at a Solution

[/B]

Last edited by a moderator:

Related Introductory Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
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.

• SHOORY
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

TSny
Homework Helper
Gold Member
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?

• SHOORY
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

TSny
Homework Helper
Gold Member
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.

• SHOORY
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

TSny
Homework Helper
Gold Member
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.

• SHOORY
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

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

TSny
Homework Helper
Gold Member
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.

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

TSny
Homework Helper
Gold Member
• SHOORY
TSny
Homework Helper
Gold Member
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.

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

if it was md^2/2 it is still wrong

ohhhhh okay i got it
it is 1.07 my mistake
thanksssssssssssssssss thanks thanks

TSny
Homework Helper
Gold Member
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.

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

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

TSny
Homework Helper
Gold Member
ohhhhh okay i got it
it is 1.07 my mistake
thanksssssssssssssssss thanks thanks
OK, good.

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