# Torque on current carrying loop/Diploe moment of loop

1. Jul 9, 2017

### Tanishq Nandan

1. The problem statement, all variables and given/known data
The current loop of the radius b as shown in figure is mounted rigidly on the axle,midway between the two supporting cords.In the absence of the external magnetic field,the tension in the two cords are equal and are T.What will be the tension when a vertical magnetic field B is present??

2. Relevant equations
Torque on a current carrying loop= (cross product of dipole moment vector and magnetic field vector)
Magnetic moment of loop=I × A,
I is current in the loop,A is it's area
And of course,
Torque at a point=r×F
3. The attempt at a solution
I found out the dipole moment of the whole disc by integration,through the following method.

Now,as I stated in the first line of section 2,I found out the torque on the disc as well as it's direction,
which came out to be,say in +x direction.Then,the torque due to the left string (T1) and the right string (T2) are coming out to be in -x and +x directions respectively.Now,since the disc is in rotational equilibrium,all the torques along the x direction should cancel out.
Which means,
(Torque due to B)+ (Torque due to T2)=(Torque due to T1)

Last line of section 2 is the formula used for finding direction of torques..
Also,the torque without any field is T,therefore,we get two equations,which can be sloved to obtain T1 and T2.

HOWEVER,the answer given varies SLIGHTLY from what I have got.

Where am I going wrong??I doubt it's the integration part..
One thing it MAY be is...the disc is NOT IN ROTATIONAL EQUILIBRIUM,dunno...maybe
Help appreciated..

2. Jul 9, 2017

### TSny

From the wording of the problem, I would interpret the current $I$ as running around the edge of the disk. That is, all of the current $I$ is at a radius $b$ so that you have a current "loop" of radius $b$.

However, if your interpretation is correct, it looks like you miss-copied the expression for $\mu$ when going to the second page of your notes. The exponent of 2 has moved from $b$ to $I$.

Did you drop a factor of 2 in the denominator when going from the next-to-last line to the last line?

Also, why doesn't the magnetic field B enter into your result?

3. Jul 10, 2017

### Tanishq Nandan

Idiotic mistake...integrating where there was no need to.
Yup,the answer is coming out correctly taking the current in a loop.
Thanks
And sorry for so many errors.. :p
I just remembered the given answer and mine varied on a coefficient of 2..so I wrote that down..should have checked.

4. Jul 10, 2017

### TSny

OK. I realize now that what you wrote as "Ans" (at the end) is the answer given to you rather than your answer.
Glad it all works out now.