# Homework on Magnetic Field

1. Oct 19, 2009

### vu95112

Please help me on this homework. Thank you.
A long piece of wire with a mass of 0.100 Kg and a total length of 4.00 m is used to make a square coil with a side of 0.100 m. The coil is hinged along a horizontal side, carries a 3.4-A current, and is placed in vertical magnetic field with a magnitude of 0.010 0T
a) Determine the angle that the plane of the coil makes with the vertical when the coil is equilibrium
b) Find the torque acting on the coil due to the magnetic force at equilibrium.

2. Oct 20, 2009

### tiny-tim

Welcome to PF!

Hi vu95112! Welcome to PF!

i] What is the torque due to gravity when the coil is at an angle θ to the vertical?

ii] What is the magnetic flux through the coil is when the coil is at an angle θ to the vertical?

Show us what you get.

3. Oct 20, 2009

### vu95112

Hello tiny-tim,

Thank you very much. Nice to meet you.

Here is my way to solve

Torque equals to gravity force

Torque = mg = NBIASinө

So ө = Arcsine ( mg/( NBIA)
= Arcsine ( 0.5Kg x 9.8 )/ ( 10 X 0.01 T x 3.4 A x 0.01 m x 0.01m)

But the answer does not come right.
Thank You

4. Oct 20, 2009

### Staff: Mentor

Your equation for the torque is not quite right. Torque is a force applied at some distance from the hinge, and at some angle. What is NBEASine?

Can you draw a sketch? That will probably help you. Also remember that the wire loop may be above or below horizontal, depending on how high the magnetic force torque is compared to the weight torque...

5. Oct 20, 2009

### vu95112

Hello Berkeman,
Thank you very much.

Torque = (L/2) BI
Where I is current, B is magnetic field , and L is the length
Torque = (0.1/2)m * 0.01T 3.4A = 0.0017 N
4 m ---------------- 0.1Kg
0.1 m ------------- X
X = (0.1*0.1) / 4= 0.0025 Kg
F = mg = 0.0025Kg * 9.8 = 0.0245 N
How do you get angle?

#### Attached Files:

File size:
107.5 KB
Views:
117
• ###### 2.bmp
File size:
195.1 KB
Views:
111
6. Oct 20, 2009

### Staff: Mentor

I think you need to be a little more detailed in your calculations. First on the torque due to gravitational forces on the wires... The total torque is the result of the torques on the 3 wires that are not the hinge wire. The two side wires will have the same torque on them, based on their weight and length. The outer wire away from the hinge will have all of its weight acting at its distance from the hinge, not half the distance. So you should probably show all 3 torques explicitly, to be sure you're calculating the gravitational torque correctly. And remember that the torque depends on the angle...

On the torque due to the magnetic force, think about the force on each of the 3 wires (the hinged wire cannot move). What can you say about the forces on the two side wires? What can you say about the force on the wire opposite to the hinge wire?

7. Oct 21, 2009

### vu95112

Hello Berkeman,
Thank you very much.

Last edited: Oct 21, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook