# Homework Help: Conducting loop placed in magnetic field

1. Apr 11, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
A uniformly conducting wire is bent to form a ring of mass m and radius r, and the ring is placed on a rough horizontal surface with its plane horizontal. There exists a uniform and constant horizontal magnetic field of induction B. Now a charge q is passed through the ring in a very small time interval Δt. As a result the ring ultimately just becomes vertical. Find the value of g. Assume that friction is sufficient to prevent slipping and ignore any loss in energy.

2. Relevant equations

3. The attempt at a solution

Let the initial angular velocity of the loop be ω. Since energy is conserved,
$\dfrac{I \omega ^2}{2} = mgr$

I know the moment of inertia but I'm still perplexed as why does the loop becomes vertical when a charge is passed through it? I have spent almost a day thinking about this. I really need some help.

2. Apr 11, 2014

### TSny

When it says a charge q is passed through the ring in time Δt, I guess it means that there is something that induces a brief current I in the ring that lasts a time Δt such that q = IΔt.

If so, would the magnetic field exert a torque on the ring while it is carrying current?

You should eventually be able to use your energy equation.

Are you asked to find a value of g or a value of q?

Last edited: Apr 11, 2014
3. Apr 11, 2014

### utkarshakash

The force on the loop will be Bilsinθ. But what should I substitute for length? Should it be 2πr?....or something else? What is the point of application of force? Is it a point on the loop situated at a diametrically opposite end about the axis of rotation? I've made no progress so far. ::Sigh::.

Don't confuse g with q. The question asks to find the value of g.

4. Apr 11, 2014

### TSny

There is a well-known expression for the torque on a current loop in a uniform magnetic field. For example see here

OK. I was just checking. It seemed more natural to me to ask for q rather than g, since g is often assumed to be known.

5. Apr 11, 2014

### utkarshakash

Thanks a lot

6. Apr 12, 2014

### Vibhor

Hello utkarshakash

Thanks

7. Apr 12, 2014

### utkarshakash

$\left( \dfrac{q \pi B}{m} \right) ^2 \dfrac{r}{3}$

8. Apr 13, 2014

### Vibhor

Work done by torque due to magnetic field is $dw = \tau dθ$

$\tau = i(\pi r^2)Bsinθ$

$W = \int_{\frac{\pi}{2}}^0 i(\pi r^2)Bsinθdθ = -\frac{q}{Δt}\pi r^2B$

Why is work coming out to be negative ?

Work done by magnetic field = Increase in GPE = mgr

This doesn't give right answer . What is the mistake ?

Thanks

9. Apr 13, 2014

### TSny

The time interval Δt that the current lasts in the ring is assumed to be very small. So the torque due to the magnetic field is assumed to be "impulsive" and you can consider the ring to still be lying horizontally when the current stops. But the ring will now have an angular velocity that you can take to be the initial angular velocity for rotating up to the vertical position.

10. Apr 13, 2014

### Vibhor

Thanks TSny ! That gives correct answer .

But what is wrong with my work in post#8 ? Please explain flaw in my reasoning .

11. Apr 13, 2014

### TSny

It looks to me that you worked a different problem where there is a steady current in the ring the entire time that the ring rotates from horizontal to vertical and that you assumed the ring doesn't change kinetic energy as it rotates.

You should get a positive work done by the magnetic torque. I think the problem there is associated with the way you chose your positive direction for θ. You have the ring starting horizontally at θ = π/2 and ending vertically at θ = 0. So, the ring rotates in the "negative θ" direction. Then the magnetic torque must be in the "negative direction" for your convention. Therefore, you would need to write $\small \tau = -i(\pi r^2)B \sin\theta$. Then $\small dW =\tau d\theta = -i(\pi r^2)B \sin\theta d\theta$ will be positive since $\small d\theta$ is negative.

12. Apr 13, 2014

### Vibhor

No . I took initial and final kinetic energy as zero . It starts from rest and comes to rest on being vertical . Is it wrong ?

Suppose the problem had a steady current ,then would we approach in the manner I had done in post#8 ?

Would then it be right to say : Work done by torque = increase in GPE ?

13. Apr 13, 2014

### TSny

When the ring is inclined at angle θ to the vertical, the magnitude of the magnetic torque is $\small i(\pi r^2)B\sin \theta$. The opposing gravitational torque will have a magnitude $\small mgr\sin\theta$. Assuming the current is time independent while the loop is rotating, then both torques are equal to constants multiplied by $\small \sin \theta$. If the loop is not to end up with KE in the vertical position then the two torques must essentially add to zero throughout the rotation. Then the KE of the loop will remain constant during the rotation of the loop and the KE can be taken to be small enough to be neglected. Of course the time Δt that the current needs to exist is not going to be "very small" in this case.

But you are correct. In your rendition of the problem, the work done by the magnetic torque would equal the change in GPE.

14. Apr 13, 2014

### Vibhor

Thanks a lot :)