Torque equilibrium question (1 Viewer)

[Hypem]

Consider a square, single coil of wire (ABCD) of side length 0.30 m which is free to rotate about its central axis and placed in a uniform magnetic field of 81 mT.
The diagram shows this looking downwards onto the plane of the coil.

Diagram: http://puu.sh/2Cioj

When the coil is connected to an external circuit and a current is passed through it, the coil is held stationary when a 5.0 g mass is attached to point P so that its weight force is into the page.
What is the magnitude and direction of the current that must flow in the coil to just balance the mass at the instant shown in the diagram?

-- It has a sample answer, but I don't really understand how they worked it out/got the final answer.

 

[Parvee]

Alright for it to be in equilibrium the weight force of the mass must equal the torque
So ma=BIA
I= ma/BA
= 0.005x9.8/(81x10^-3 x0.3^2)
=6.72A

 

[Fizzy_Cyst]

Alright for it to be in equilibrium the weight force of the mass must equal the torque
So ma=BIA
I= ma/BA
= 0.005x9.8/(81x10^-3 x0.3^2)
=6.72A

Be careful, you have equated force with torque! They are different quantities Parv!
nBIA is total torque (torque on both sides)

You would need to equate mg = BIl

I = mg/Bl

I = (0.005*9.8)/(81x10^-3 * 0.3)

I = 2.01A

 

[Parvee]

~whoops~
Got some formulas mixed up lol

 

[Fizzy_Cyst]

Which paper is this from? I'm sure I used this question in my Half-Yearly. Can't remember where i got it from, lol.

~whoops~
Got some formulas mixed up lol

I forgive you... This time.

 

[Hypem]

The answer's 1.01 A anticlockwise through the circuit...

edit: but seeing's they were wrong in another question, they're probably wrong here lol

 

[Hypem]

Which paper is this from? I'm sure I used this question in my Half-Yearly. Can't remember where i got it from, lol.




I forgive you... This time.

2011 or 2012 CSSA exam not sure which though cbf looking.

"Sample answer:

Weight force on CD due to mass, W = mg = 0.005 * 9.8 = 0.049 N
Since length is 0.30 m, distance of P to axis is 0.15 m, hence torque due to P = length * w = 0.15 * 0.049 = 0.00735 Nm (up the page)
Since length = 0.30 m, cross-sectional area (l^2) = 0.090 m^2
Magnitude of torque on coil due to motor effect - nBIA = 1*0.081*I*0.09 = 0.00729I (down the page)

For torque equilibrium, I = 0.00735/0.00729 = 1.01 A"

So basically:

torque = torque
mgd = nBIA

I have no idea what they mean by up/down the page though. Wouldn't it be (into/out of) the page? So P would produce a torque into the page, and the motor effect a torque out of the page?

 

[Fizzy_Cyst]

OIC, so the current goes through the entire coil, not just the side opposite to where the mass is hanging?

If this is the case, then the way they have done it is indeed correct!

If the current is ONLY going through the side of the coil opposite to the mass, then the way which I have done it is correct.

Looking at the diagram, it would be in/out the page, rather than up/down.

 

[Hypem]

Ok thanks!

Also, sorry I couldn't get the CSSA trial for you

 

[Fizzy_Cyst]

No probs!

If anyone has 2012 CSSA with solns, pls hook me up

 

[anomalousdecay]

Be careful, you have equated force with torque! They are different quantities Parv!
nBIA is total torque (torque on both sides)

You would need to equate mg = BIl

I = mg/Bl

I = (0.005*9.8)/(81x10^-3 * 0.3)

I = 2.01A

This is exactly what I got, with the current flowing from D to C to B to A.

However I guess the reason why it is 1.01A is because the torque is directed from both sides of the coil.
T = Fr (where r is the radius) 2.02A is for the diameter of the coil.
(Think of the coil's turn as a point locus directed from the centre, you get a circle).
For the radius, the current is 1.01A.