Find the magnitude and direction of the Magnetic field required

  • #1

Homework Statement:

The rectangular loop of wire shown in Fig.has a mass of 0.15 g per centimeter of length and is pivoted about side ab on a friction less axis. The current in the wire is 8.2 A in the direction shown.Find the magnitude and direction of the magnetic field parallel to the y-axis that will cause the loop to swing up until its plane makes an angle of 30Deg with the yz-plane.

Relevant Equations:

F = IlBsin(Θ); F = mg
1597498624365.png


i tried to draw the directions of the parameters
1597507895813.png

The direction of B is clear since then the Force will be in the positive X direction. I am bit confused with the direction of Force, how would i draw it and the components. Is the gravitational force i have drawn is correct? Do we have better tools to draw these diagrams in 3D which will make life easy. Can i neglect the Force on ab? Please advise.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,796
3,153
What conditions must be satisfied in order for the loop to be in static equilibrium?

It looks like you have the correct directions for the magnetic forces on the top and bottom segments of the loop. Are there magnetic forces on the other two segments? I'm not sure what your notation "0 F" indicates in the drawing.

Your force of gravity has the correct direction. At what point should this force be considered to act?

Your direction for ##\vec B## also looks correct.
 
Last edited:
  • Like
Likes Physicslearner500039
  • #3
I'm not sure what your notation "0 F" indicates in the drawing.
The 0F indicates 0 force acting on the side segments when they are in the initial position, since the B field is parallel and antiparallel to the side segements. But when they have moved to 30 Deg, there is a force acting on the side segments as well.

The diagram showing the direction of force on the lower segment. Is it correct?

1597559094889.png


At what point should this force be considered to act?
I consider this applying on the lower segment. Is it correct? I am mixing everything, very confused. Please advise.
 

Attachments

  • #4
TSny
Homework Helper
Gold Member
12,796
3,153
I'm having a little trouble interpreting your last figure. In your first post, I think you have the correct directions for the forces on the top and bottom segments of the rectangle:
1597592767657.png


The force, ##F_1## on the lower segment ##cd## is parallel to the x-axis and points towards positive x.
The force, ##F_2## on the upper segment ##ab## is parallel to the x-axis and points towards negative x.

Similarly, can you describe the directions of the forces on sides ##bc## and ##ad##?

It looks like you are drawing the gravitational force ##mg## as acting on the lower segment ##cd##. But, ##mg## acts at the average location of all of the mass. That is, it acts at the center-of-mass of the loop. This is important if you are going to calculate torques.

The side ##ab## is mounted on the z-axis so that the loop can rotate about the z-axis. The z-axis will provide a supporting force ##\mathbf R## acting on the side ##ab##. Without this force, the whole loop would fall. However, if you take the z-axis to be your axis for calculating torques, you will not need to worry about this force since ##\mathbf R## produces zero torque about the z-axis.
 
  • Like
Likes Physicslearner500039 and etotheipi
  • #5
Similarly, can you describe the directions of the forces on sides ##bc## and ##ad##?
Yes i messed up everything in the 2nd post, now i think i got it back again. For the side segments i can directly say
bc -> +Z direction
ad -> -Z direction
It looks like you are drawing the gravitational force ##mg## as acting on the lower segment ##cd##. But, ##mg## acts at the average location of all of the mass. That is, it acts at the center-of-mass of the loop. This is important if you are going to calculate torques.
I need more time to think and work out, I will try and update.
 
  • #6
It looks like you are drawing the gravitational force ##mg## as acting on the lower segment ##cd##. But, ##mg## acts at the average location of all of the mass. That is, it acts at the center-of-mass of the loop. This is important if you are going to calculate torques.
It means the total mass of rectangular loop is ##2*(8+6)*0.15 = 4.2 g##. Other thing is I found that the gravitational force having a component in the X direction and Force F1 acting along the same X direction, there is one more component of gravitational force acting in the +Z direction. I do not see any component of force acting in the negative X direction. Then how it will balance? My other question Do I need to equate the Forces or torques? What is the difference? Please advise.
 
  • #7
TSny
Homework Helper
Gold Member
12,796
3,153
For the side segments i can directly say
bc -> +Z direction
ad -> -Z direction
Yes, these are the correct directions.
 
  • #8
TSny
Homework Helper
Gold Member
12,796
3,153
It means the total mass of rectangular loop is ##2*(8+6)*0.15 = 4.2 g##.
Yes, this is the correct mass of the loop. Keep in mind that the gram is not the SI unit for mass.

Other thing is I found that the gravitational force having a component in the X direction and Force F1 acting along the same X direction, there is one more component of gravitational force acting in the +Z direction.
I do not see any component of force acting in the negative X direction. Then how it will balance?
The gravitational force acts vertically downwards, as you indicated in your first post. So, there is no x or z component of the force of gravity.

To see how the forces can add to zero, you must include all of the forces. There is a magnetic force on each of the four sides of the loop. There is the gravitational force. And then there is the force of support at the mounting along the z-axis.

My other question Do I need to equate the Forces or torques? What is the difference? Please advise.
For static equilibrium, the forces must add to zero and the torques must add to zero. To solve the problem, you might be able to use just the torque condition.
 
  • Like
Likes Physicslearner500039
  • #9
I am not sure if i am correct
a. On the sides of the rectangle the forces are equal and act along the same line of reference hence they cancel each other.
b. The force on the lower side is ##F1=8.2*0.06*B = 0.492B## and the perpendicular distance is ##d = 8 sin(30)=4##. Hence the torque on the lower side is ##\tau = 0.492B * 6.92=3.404B ##
c. The gravitational force is ##F = mg = 0.0042*9.8=0.04116## the perpendicular distance is ## \frac 8 2 * sin(30) =2##. Hence the torque is ##\tau = 0.04116*2 = 0.08232##. I am assuming this force is acting at the center of the loop.
Equating both of them
##3.404B = 0.0823 => 0.024 T##
 
  • #10
TSny
Homework Helper
Gold Member
12,796
3,153
I am not sure if i am correct
a. On the sides of the rectangle the forces are equal and act along the same line of reference hence they cancel each other.
Yes

b. The force on the lower side is ##F1=8.2*0.06*B = 0.492B## and the perpendicular distance is ##d = 8 sin(30)=4##.
No. Using sin(30) will not give you the perpendicular distance for this force. Also, it's always a good idea to denote the units.

Hence the torque on the lower side is ##\tau = 0.492B * 6.92=3.404B ##
So, you decided to use 6.92 cm instead of 4 cm for the perpendicular distance. Good.


c. The gravitational force is ##F = mg = 0.0042*9.8=0.04116##
Ok. Always include the units.

the perpendicular distance is ## \frac 8 2 * sin(30) =2##.
Yes. Include the units.

Hence the torque is ##\tau = 0.04116*2 = 0.08232##. I am assuming this force is acting at the center of the loop.
Equating both of them
##3.404B = 0.0823 => 0.024 T##
Yes, I think this is the right answer. Looking back, note that you used units of cm for the perpendicular distances in calculating the torques. This means that your torques were in units of N⋅cm. This is OK, even though the SI unit for torque is N⋅m. All the cm units cancelled out in the calculation.
 
  • Like
Likes Physicslearner500039

Related Threads on Find the magnitude and direction of the Magnetic field required

Replies
2
Views
5K
Replies
5
Views
497
Replies
4
Views
597
Replies
3
Views
6K
Replies
6
Views
5K
Replies
3
Views
2K
Replies
7
Views
813
Top