Work done on a magnetic dipole (compass needle rotating)

[Ugnius]

Homework Statement

Compass needle magnetic dipole moment is 4mA*m^2 , in the location of the compass Earth's magnetic field is 55 microteslas , and pointing north and downwards at 48 degree angle with compass needle. What amount of work is needed to turn compass needle from pointing North to South?

Relevant Equations

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So this is a sketch I made of the situation

and this is my approach

my approach is incorrect , and Idon't seem to find the mistake , maybe B*p isn't correct. Any ideas?

 

[Charles Link]

On this one, you need to work in the x-y plane: The $B$ that generates the torque in the direction of interest lies in the x-y plane. (Calling north as the positive x-axis and and west as the positive y-axis, the $B$ of interest is $B_x$). The torque is proportional to $\sin{\theta}$, and the work is $\int \tau_o \sin{\theta} \, d \theta$, where $\tau_o$ is the value for $\theta=90$ degrees. (Note: The axis of rotation is the z-axis. The $B_z$ that you computed will not cause any torque in the z-direction=that is why your calculation doesn't work).

Edit: Note: The problem can also be worked as a rotation about the y-axis, but you will find that the $B_z$ has a null effect, with the y component of the torque being in opposite directions as the dipole moment passes through the z-axis. Once again it is $B_x$ that determines how much work is needed.

 

[Ugnius]

On this one, you need to work in the x-y plane: The $B$ that generates the torque in the direction of interest lies in the x-y plane. (Calling north as the positive x-axis and and west as the positive y-axis, the $B$ of interest is $B_x$). The torque is proportional to $\sin{\theta}$, and the work is $\int \tau_o \sin{\theta} \, d \theta$, where $\tau_o$ is the value for $\theta=90$ degrees. (Note: The axis of rotation is the z-axis. The $B_z$ that you computed will not cause any torque in the z-direction=that is why your calculation doesn't work).

Edit: Note: The problem can also be worked as a rotation about the y-axis, but you will find that the $B_z$ has a null effect, with the y component of the torque being in opposite directions as the dipole moment passes through the z-axis. Once again it is $B_x$ that determines how much work is needed.

Sorry , but I didn't managed to understand. Can you maybe explain a bit simpler?

 

[kuruman]

The problem is telling you that the Earth's magnetic field vector at the location of the needle has a magnetic dip angle of 48^° below the xy-plane. The needle is constrained to move in the xy-plane. Initially, it is aligned with the horizontal component of the magnetic field, B_h which is less than the magnitude of 55 μT. (What is the value of B_h?) You are asked to find how much work you need to do on the needle in order to rotate it by 180^° in the xy-plane. It's easier to use energy considerations instead of torques. What is the potential energy of a magnetic moment in a magnetic field?

 

[Ugnius]

Yes , I have calculated x projection of Magnetic field, and calculated the torque , but I don't know what method to use to calculate the work.

 

[kuruman]

Yes , I have calculated x projection of Magnetic field, and calculated the torque , but I don't know what method to use to calculate the work.

I pointed that out in post #4.

It's easier to use energy considerations instead of torques. What is the potential energy of a magnetic moment in a magnetic field?

 

[Ugnius]

isn't E = τ*θ , θ being the dip angle?

 

[kuruman]

isn't E = τ*θ , θ being the dip angle?

Nope. Read about magnetic potential energy here.

 

[Ugnius]

Nope. Read about magnetic potential energy here.

I see , and should I use the full magnetic field , or just the x projection ?

 

[kuruman]

I see , and should I use the full magnetic field , or just the x projection ?

Do it both ways and see if it matters. Be careful with the "before" and "after" angles in each case.

 

[Charles Link]

The energy $E=-m \cdot B$ is the easier way to do this, but it can be useful to know where this formula comes from, and $W=\int \tau \, d \theta$ is basically the same result as the energy formula in what may be a not-so-obvious form.

In post 2 above, I pointed out that you can consider the rotation to be about the z-axis or the y-axis, and you get the same answer. It's much easier to just consider energy though, as others have pointed out.

Note: You can rotate the compass needle in the plane of the compass or rotate out of that plane=the needle still winds up in the same final position. The energy difference is the same regardless of the path you take.

 

[Ugnius]

Thank you guys , I solved it , and answer seems to be correct. Couldn't have done it without your help!

 

[Pipiras]

Please show how u did it, because i have the same task