Find magnetic dipole moment of coiled wire

[Jimbob999]

Homework Statement

A 45-m long wire is coiled so that it makes a coil containing 100 circular loops, one on top of the other. If the wire carries a current of 13 A, what is the magnetic dipole moment of the coil?

21 A·m2

6.7 A·m2

3.3 A·m2

2.6 A·m2

1.2 A·m2

Homework Equations

B(z) = u0/2pi x u/z^3
Where u0 is the permeability constant and u is the magnetic dipole moment

The Attempt at a Solution

[/B]
I can get as far as B(z) = (2 x 10^7 u)/z^3
I know I need to find u.
It doesn't seem to mention where exactly z is, thus can't work out magnetic field. Do you think there is enough information provided in this question to solve it?

Thanks.

 

[Jimbob999]

Would this be a solenoid and thus use equation B = u0 x i x n ?

 

[Qwertywerty]

I'm curious - how would finding B help ?

Use simply the formula - M = n*i*A , where n is no of loops , and i the current passing through the loops , and A the area of each loop .

Also , here's a hint : M due to n circular loops would be the same as due to one large , single loop of current - carrying wire . This might help .

 

[Jimbob999]

Oh I couldn't find that equation.

But solving for this equation gives me 2.07 x 10^6.
I guess I need to convert this to A/m^2, but I am not sure the conversion and couldn't find it on Google anywhere?

 

[Jimbob999]

Anyone got any idea?

 

[Qwertywerty]

Anyone got any idea?

Firstly ,

Also , here's a hint : M due to n circular loops would be the same as due to one large , single loop of current - carrying wire . This might help .

Unfortunately , I have made a mistake here . Discard this .
Secondly ,

But solving for this equation gives me 2.07 x 10^6.
I guess I need to convert this to A/m^2, but I am not sure the conversion and couldn't find it on Google anywhere?

Try the working again now . You will get the answer . Also , I believe you mean Am² .

Hope this helps .

 

[Jimbob999]

Ok, let me show you my working.

n is 100
I is 13
A = Pi R^2, which is 506.3. However, I am inclined to take this as one big loop. Thus 45m divided 100 loops.
A = 0.159?

M = 206.7
or M = 2.07 if take n as 1.

So don't think any of my answers are right?

 

[Qwertywerty]

However, I am inclined to take this as one big loop.

I said I made a mistake - you shouldn't do this .

Find area of each individual loop , and use the formula M = n*i*A .

A = Pi R^2, which is 506.3. However, I am inclined to take this as one big loop. Thus 45m divided 100 loops.
A = 0.159?

Check your area .

Hope this helps .

 

[Jimbob999]

Thanks qwerty.
Was stupidly using circumference instead of radius. Got it now.

 

[Almamoun]

Homework Statement

A 45-m long wire is coiled so that it makes a coil containing 100 circular loops, one on top of the other. If the wire carries a current of 13 A, what is the magnetic dipole moment of the coil?

21 A·m2

6.7 A·m2

3.3 A·m2

2.6 A·m2

1.2 A·m2

i = 13 , L = 45 , N = 100

100 circular loop

the circumference of the all loops L = 2*pi*r *100
then the r of 100 circular loop equal r = L / ( 2*pi*100 )

magnetic diploe = N i A = 100 * pi * r^2 * 13 = 21 A·m^2

solved the Q - Almamoun and Feras Alhazmi from King Fahad of petroleum and minerals university

 

[kuruman]

Good solution, @Almamoun, but please note that the question is almost 7 years old so it is unlikely that it will help the person who posted it.