A hanging ring carrying current in a uniform magnetic field

[palaphys]

Homework Statement

A uniform current carrying ring of mass m and radius R is connected by a massless string as shown in diagram. A uniform magnetic field Bo exists in the region to keep the ring in horizontal position, then the current in the ring is ( l length of string)

Relevant Equations

Tau= mB

this is the diagram provided for the question
I am familiar with the equation
$$\vec{\tau} = \vec{\mu} \times \vec{B}$$
where $$\tau$$ is the torque experienced by the ring. However, I think that this is the torque experienced, about an axis passing through the center of mass of the loop. If this understanding is correct, then how can we take the point of connection of the ring and string and say that the net torque about that point is zero?

The options seem to suggest that sort of a solution, but I am not getting it why it is right

 

[kuruman]

Why not consider the force on an element $d\vec l$ on the ring, find the torque element $d \vec {\tau}$ relative to the point of support and integrate to find the net torque? You should be able to trust the result using this method.

 

[palaphys]

Why not consider the force on an element $d\vec l$ on the ring, find the torque element $d \vec {\tau}$ relative to the point of support and integrate to find the net torque? You should be able to trust the result using this method.

I lack advanced calculus skills required to perform that task. So its true that the torque about any point in space is $\vec{\tau} = \vec{\mu} \times \vec{B}$ ?

 

[kuruman]

I lack advanced calculus skills required to perform that task. So its true that the torque about any point in space is $\vec{\tau} = \vec{\mu} \times \vec{B}$ ?

It is true in this case. You cannot generalize from this case to any case. Take the torque due to gravity. In this case it is $\tau_g=mgR$ about the point of support. In the case of a point at distance $2R$ from the center of the ring, it would be $\tau_g=2mgR.$

On edit: Rescinded. See posts #6 and #7.

 

[palaphys]

It is true in this case. You cannot generalize from this case to any case. Take the torque due to gravity. In this case it is $\tau_g=mgR$ about the point of support. In the case of a point at distance $2R$ from the center of the ring, it would be $\tau_g=2mgR.$

but then, my contradiction is still not resolved. if we consider the topmost contact point(where the string is fixed to the ceiling), if we assume the system to be in rotational eqb, then the torques about this point are NOT balanced, unless there is some other external torque maintaining eqb.
In addition to this, the horizontal component of tension at the support is not balanced either..

 

[TSny]

I believe the figure given in the problem statement is incorrect.

It should be as shown on the right. [Edit: Specifically, the string must be vertical.] Can you see why?

I lack advanced calculus skills required to perform that task. So its true that the torque about any point in space is $\vec{\tau} = \vec{\mu} \times \vec{B}$ ?

The magnetic torque $\vec{\mu} \times \vec{B}$ on the loop is independent of the choice of origin for calculating the torque. Each infinitesimal element of the current will experience a magnetic force. (There are two infinitesimal elements for which the force is zero. Can you identify them?) The sum of all of these forces is zero. There is a theorem that states that for any set of forces that add to zero, the net torque due to these forces is independent of the choice of origin.

 

[kuruman]

@TSny's point in post #6 is well taken. The torque $\vec \mu \times \vec B$ is indeed independent of the choice of reference for reasons explained (how could I have missed that?) One can also verify the result by doing the integral.

The correct drawing is as shown in post #6. There is a reason that, of all the points on the ring that one could pick for tying the string, the correct point is on the radius parallel to the magnetic field.

 

[palaphys]

I believe the figure given in the problem statement is incorrect.

View attachment 363499
It should be as shown on the right. [Edit: Specifically, the string must be vertical.] Can you see why?


The magnetic torque $\vec{\mu} \times \vec{B}$ on the loop is independent of the choice of origin for calculating the torque. Each infinitesimal element of the current will experience a magnetic force. (There are two infinitesimal elements for which the force is zero. Can you identify them?) The sum of all of these forces is zero. There is a theorem that states that for any set of forces that add to zero, the net torque due to these forces is independent of the choice of origin.

i got the answer to this one, also I managed to prove the theorem you are talking about.
I also understand why the string HAS to be vertical. only if that is the case, the torque about the hinge point will be zero.

Thanks a lot once again
@kuruman, @TSny

 

[TSny]

i got the answer to this one, also I managed to prove the theorem you are talking about.

I also understand why the string HAS to be vertical. only if that is the case, the torque about the hinge point will be zero.

If the string is not vertical, then the string will exert a force on the ring that has a nonzero horizontal component. But there are no other horizontal forces acting on the ring. So, if the string is not vertical, there will be a nonzero net horizontal force on the ring and the ring will not stay at rest.

 

[palaphys]

the string will exert a force on the ring that has a nonzero horizontal component. But there are no other horizontal forces acting on the ring. So, if the string is not vertical, there will be a nonzero net horizontal force on the ring and the ring will not stay at rest.

 

[Herman Trivilino]

I also understand why the string HAS to be vertical. only if that is the case, the torque about the hinge point will be zero.

Only if the tension in the string is nonzero, and I see no reason it can't be.