# Calculating Area & Direction of Magnetic Field

• JoelKTH
In summary, the problem involves a homogeneous magnetic field in the ex and ez direction, acting on a pendulum that is swinging in the xy-plane. The magnetic flux is defined as Phi=integral(B dS)=(Area)exB=(Area_triangle + (L^2/2)*(β + α(t)))ezB. The direction of the magnetic field is given in the problem statement, and the time derivative of the area is needed to calculate the emf. The component of the magnetic field in the x-direction does not contribute, so there is no need to calculate the cross product. It is not necessary to determine the area of the pendulum, as it is not needed for the exercise. f

#### JoelKTH

Homework Statement
A homogenous B-field B= B e_x + B e_z is acting in space. A metalbar is attached in a horisontal axis, mounted on the height y=L on the y-axis. The metal bar is swinging freely in the xy-plane around the axis. The metal bars pendulummovement is described as α(t) = β cos(ωt) which is time dependent on t. β > 0 och ω > 0 are constants. The metal bar lower end is in electrical contact with a metal rail with the geometry of a circular arc with radius L. The metal bar is through the resistance R coupled to the axis and the bar as the figure shows. The the self-inductance of the circuit is being neglected as the circuits other resitive losses.

Calculate the current i(t) and use Lenz law to validate that the currents sign is correct.
Relevant Equations
emk= - d(phi)/dt, i(t)= emk/R, phi=surface integral(B dS)
Hi, the problem statement is above. I have some questions about how to calculate the area and the direction of the magnetic field of this problem.
As the magnetic flux, my professor have defined it as Phi= integral(B dS)=(Area)e_x B= (Area_triangle + (L^2/2) *(β + α(t)))*B e_z.
How can one know that the magnetic field is in the e_z direction? As far as I know the right hand rule makes the direction of the magnetic field in e_phi direction. However to convert this e_phi=-e_x sin(phi) + e_y cos(phi) does not help me.

Another question is how the Area is calculated. There is a figure attached. As the figure is time dependent, I am aware that the Area_triangle will change.

Kind regards,

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• Skärmbild_20221228_181728.png
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Homework Statement:: A homogenous B-field B= B e_x + B e_z is acting in space. A metalbar is attached in a

How can one know that the magnetic field is in the e_z direction?
It's given in the problem statement!

As the magnetic flux, my professor have defined it as Phi= integral(B dS)=(Area)e_x B= (Area_triangle + (L^2/2) *(β + α(t)))*B e_z.
I don't undestand this. The pendulum is swinging in the ##xy## plane, so the ##x## component of the magnetic field is of no interest.

Note that with emk = ##\varepsilon = {d\over dt} \Phi \ , \ ## you only need the time derivative of the area

##\ ##

It's given in the problem statement!

I don't undestand this. The pendulum is swinging in the ##xy## plane, so the ##x## component of the magnetic field is of no interest.

Note that with emk = ##\varepsilon = {d\over dt} \Phi \ , \ ## you only need the time derivative of the area

##\ ##

Yes, its given in the problem statement, but also e_x. Even if the pendulum is swinging in the xy-plane, why wouldn't the magnetic field also work in the e_x direction? if according to the Right Hand Rule the magnetic field is in e_phi direction, which is defined as e_phi=-e_x sin(phi) + e_y cos(phi)?

Also you are correct that one only needs the time derivative of the area, but if it were an exam question I believe my professor unfortenately would deduct points if the Area is not properly defined :/

Yes, its given in the problem statement, but also e_x. Even if the pendulum is swinging in the xy-plane, why wouldn't the magnetic field also work in the e_x direction? if according to the Right Hand Rule the magnetic field is in e_phi direction, which is defined as e_phi=-e_x sin(phi) + e_y cos(phi)?

Also you are correct that one only needs the time derivative of the area, but if it were an exam question I believe my professor unfortenately would deduct points if the Area is not properly defined :/
It is confusing to introduce a second ##\phi## (admittedly lower case, but still...) and an e_phi direction. What is ##\phi## ? Where did you get this definition ? How does it help you ?

The emf that causes a current to flow in the circuit is in a direction along the metal bar. The component of the magnetic field in the x-direction does not contribute: the vector cross product ##\vec v \times \vec B## is in the ##z## direction (see Lorentz force).

I would consider to deduct points for wasting time determining an area that isn't needed for the exercise .

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• berkeman
It is confusing to introduce a second ##\phi## (admittedly lower case, but still...) and an e_phi direction. What is ##\phi## ? Where did you get this definition ? How does it help you ?

The emf that causes a current to flow in the circuit is in a direction along the metal bar. The component of the magnetic field in the x-direction does not contribute: the vector cross product ##\vec v \times \vec B## is in the ##z## direction (see Lorentz force).

I would consider to deduct points for wasting time determining an area that isn't needed for the exercise .

##\ ##

The reason why I introduce ##\phi## is because the current is in the y-direction and when determining the B-field the direction of the current is "rotating" around the axis. Just as the B-field is calculated for a long straight conductor. I think I am trying to apply that same logic here. The ##\phi## is basis vector relationship from transforming e_##\phi##=-e_x sin(theta) + e_y cos(theta) for cylindrical coordinates.

Is it necessary to calculate the cross product? According to my professors solution its defined like this: ##\phi##= integral(B dS)= Area(dot)e_z(dot)B. How would you recommend me knowing how to define dS?

The reason why I introduce ##\phi## is because the current is in the y-direction and when determining the B-field the direction of the current is "rotating" around the axis. Just as the B-field is calculated for a long straight conductor. I think I am trying to apply that same logic here. The ##\phi## is basis vector relationship from transforming e_##\phi##=-e_x sin(theta) + e_y cos(theta) for cylindrical coordinates.

Is it necessary to calculate the cross product? According to my professors solution its defined like this: ##\phi##= integral(B dS)= Area(dot)e_z(dot)B. How would you recommend me knowing how to define dS?
You really want to separate this exercise into two parts: 1. Calculate the induced current and 2. Look at the magnetic field this current causes around the rod. Your ##\phi## may come in in part 2, but has no role in part 1.

For part 1 you indeed need ##{d\Phi\over dt} ##. With ##\vec B_z## constant (the given B) the flux is something like ##\ \ ## B times Area ##\ \ ##and ##\ \ {d\Phi\over dt} \ \ ## is ##\ \ ## B times ##\displaystyle{d\;{\text{Area}}\over dt}##.
With the given ##\alpha(t)## you should be able to write down an expression for this. Right ?

Is it necessary to calculate the cross product?
Where do you need it ?

How would you recommend me knowing how to define dS?
The area of a circle sector, e.g. the sector between the bar and the y-axis in the figure in post #1 is ...
(expressed in terms of ##\alpha## and ##L##)

The time derivative is ...

##\ ##

You really want to separate this exercise into two parts: 1. Calculate the induced current and 2. Look at the magnetic field this current causes around the rod. Your ##\phi## may come in in part 2, but has no role in part 1.

For part 1 you indeed need ##{d\Phi\over dt} ##. With ##\vec B_z## constant (the given B) the flux is something like ##\ \ ## B times Area ##\ \ ##and ##\ \ {d\Phi\over dt} \ \ ## is ##\ \ ## B times ##\displaystyle{d\;{\text{Area}}\over dt}##.
With the given ##\alpha(t)## you should be able to write down an expression for this. Right ?

Yes, from this I must conclude that the ##\vec B##= ##\vec B_z## and not in##\vec B_x##. This I can conclude from that its given and that ##\vec B_x##=0 because the pendulum is swinging in the xy plane, so the component of the magnetic field is of no interest. Not sure why its of no interest, you provided me a link to Lorentz law where v##\vec v##x ##\vec B##= only in z-direction.
Where do you need it ?

The area of a circle sector, e.g. the sector between the bar and the y-axis in the figure in post #1 is ...
(expressed in terms of ##\alpha## and ##L##)
So the A_sector= ((Beta + alpha(t))*L^2 )/2. From what part is the total Area then A_tot= A_triangle + A_sector? When taking the derivative the A_triangle part is 0 anyways as you said.
The time derivative is ...

##\ ##
The time derivative is quite easy to make. Answer is emf=((L^2 * B_z)/2) *Beta*w*sin(wt)

Also the time derivative to get emf is logical because at t=0 (when the bar is at its highest peak x>0 and the angle is positive because sin(wt)>0. When the time increases(bar going down), the Area will also decrease and then the current will decrease because of i(t)=(w*Beta*B_z*L^2 sin(wt)/2R.

Then Lenz law wants to counteract this by increasing the B-field which then decreases the time varying angle alpha(t)?

Not sure why its of no interest, you provided me a link to Lorentz law
The Lorentz law will help you to double check the direction of the emf that is generated.
(grant you that the Lorentz force is my personal favorite since it's at the base of so much of all the other stuff )

The given ##\vec B_x \ne 0##, but it doesn't affect the induced emf.

Good. So now you have the time derivative, and thereby the induced emf, and the current.
Part 1 is almost done: a final expression with attention to the sign and a check.
Part 2 is a breeze: Lenz is only qualitative.

Then Lenz law wants to counteract this by increasing the B-field which then decreases the time varying angle alpha(t)?
You are correct ##\dagger## : the problem statement does not specify that the pendulum is forced (it says: 'freely' ). So (implicitly, from the ' α(t) = β cos(ωt) ' ) we have a damped harmonic oscillation where the kinetic energy from the pendulum is transferred to the resistor and disspated there. I.e. ##\beta## will decrease (unless R is very small, the pendulum will be underdamped and α oscillates around zero as the ' α(t) = β cos(ωt) ' suggests) .

By now I feel I've painted us into a corner, because a rigorous treatment would modify α(t) = β cos(ωt) to α(t) = β(t) cos(ωt) with a more complicated equation of motion. I don't think that's intended* by prof, but perhaps I'm too optimistic.

*I.e. the intention is that the time derivative of β can be neglected

##\dagger## well, (nitpicking): 'decrease' in one direction of the swing, increasing in the other !

Pinging @haruspex for a second opinion (congrats with the umpteenth reward! Well earned; hats off !).

##\ ##

The area of a circle sector, e.g. the sector between the bar and the y-axis in the figure
Not sure that's quite the right area to use. Doesn't the current only arise from the p.d. between the axle and the point of contact with the rail?
Not sure why its of no interest
Motion in the x direction does not interact with Bx.
Motion in the y direction interacts with Bx to produce a p.d. in the z direction, but the bar has no thickness in that direction, so does not produce a current.