Ruby Briggs
2022-07-23
Answered

(a) A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 5.00 by 8.00 cm rectangular coil rotates in an adjustable magnetic field so that it can produce sufficient voltage even at low rpms. What is the field strength needed to produce a 24.0 V peak emf? (b) Discuss how this required field strength compares to those available in permanent and electromagnets.

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nuramaaji2000fh

Answered 2022-07-24
Author has **18** answers

a The emf produced is given by

$\u03f5=NBA\omega $

For a given emf, it is clear that the needed field will have to be

$B=\frac{\u03f5}{NA\omega}$

Let us now remember that since the area is a rectangle, we have $A=ab$. Also, to convert from rotations per minute to radians per second, we need to multiply by $2\pi $ and divide by $60$ - that is, multiplying by $\frac{\pi}{30}$. This means that the final expression will be

Substituting numerically, we have

$B=\frac{30\u03f5}{abN\cdot \pi \cdot rpm}$

b.This required field is quite high to be achieved by permanent magnets if the distance between them is considerable, and not as easy to be achieved by electromagnets either. Let us mention that the strongest electromagnets - those used in magnetic resonance imaging - usually produce fields no greater than $5T$.

$\u03f5=NBA\omega $

For a given emf, it is clear that the needed field will have to be

$B=\frac{\u03f5}{NA\omega}$

Let us now remember that since the area is a rectangle, we have $A=ab$. Also, to convert from rotations per minute to radians per second, we need to multiply by $2\pi $ and divide by $60$ - that is, multiplying by $\frac{\pi}{30}$. This means that the final expression will be

Substituting numerically, we have

$B=\frac{30\u03f5}{abN\cdot \pi \cdot rpm}$

b.This required field is quite high to be achieved by permanent magnets if the distance between them is considerable, and not as easy to be achieved by electromagnets either. Let us mention that the strongest electromagnets - those used in magnetic resonance imaging - usually produce fields no greater than $5T$.

asked 2022-05-17

Consider one-dimensional ferromagnet namely N spin-1/2 objects placed around a circle with the Hamiltonian

$\mathcal{H}=-\mathcal{J}\sum _{n=1}^{N}{\overrightarrow{\mathcal{S}}}_{n}\cdot {\overrightarrow{\mathcal{S}}}_{n+1}$

where we assume the periodic boundary condition ${\overrightarrow{\mathcal{S}}}_{N+1}\equiv {\overrightarrow{\mathcal{S}}}_{1}$ and $\mathcal{J}>0$

I'm trying to show that total spin ket is a good quantum number that is they commute with $\mathcal{H}$ and finding out the energy corresponding to

$|{\psi}_{0}\u27e9=|\uparrow {\u27e9}_{1}\otimes |\uparrow {\u27e9}_{2}\otimes \cdots \otimes |\uparrow {\u27e9}_{N}$

By definition:

${\mathcal{S}}^{2}={\left(\sum _{n}{\overrightarrow{\mathcal{S}}}_{n}\right)}^{2}=\sum _{n}{\mathcal{S}}_{n}^{2}+\sum _{i,j}{\overrightarrow{\mathcal{S}}}_{i}\cdot {\overrightarrow{\mathcal{S}}}_{j}$

The second term has our Hamiltonian but there are other terms also. I don't understand, How do I proceed from here?

$\mathcal{H}=-\mathcal{J}\sum _{n=1}^{N}{\overrightarrow{\mathcal{S}}}_{n}\cdot {\overrightarrow{\mathcal{S}}}_{n+1}$

where we assume the periodic boundary condition ${\overrightarrow{\mathcal{S}}}_{N+1}\equiv {\overrightarrow{\mathcal{S}}}_{1}$ and $\mathcal{J}>0$

I'm trying to show that total spin ket is a good quantum number that is they commute with $\mathcal{H}$ and finding out the energy corresponding to

$|{\psi}_{0}\u27e9=|\uparrow {\u27e9}_{1}\otimes |\uparrow {\u27e9}_{2}\otimes \cdots \otimes |\uparrow {\u27e9}_{N}$

By definition:

${\mathcal{S}}^{2}={\left(\sum _{n}{\overrightarrow{\mathcal{S}}}_{n}\right)}^{2}=\sum _{n}{\mathcal{S}}_{n}^{2}+\sum _{i,j}{\overrightarrow{\mathcal{S}}}_{i}\cdot {\overrightarrow{\mathcal{S}}}_{j}$

The second term has our Hamiltonian but there are other terms also. I don't understand, How do I proceed from here?

asked 2022-04-06

I'm trying to learn the connection between special relativity and magnetism. I know that if I place a positive charge, at rest, next to wire with current, I should not observe any force on it because there is no electric field and there is no magnetic force as my charge is at rest.

But here is what confuses me - the wire contains moving electrons and according to what I learned, the stationary charge should observe a length contraction of those electrons and so the density of them will increase and a negative electric field should be observed.

This is definitely not the case and I wonder if someone can explain to me what is wrong in my analysts.

But here is what confuses me - the wire contains moving electrons and according to what I learned, the stationary charge should observe a length contraction of those electrons and so the density of them will increase and a negative electric field should be observed.

This is definitely not the case and I wonder if someone can explain to me what is wrong in my analysts.

asked 2022-05-07

Lets say you have an uniform prism magnet of Iron for example. How would you calculate the demagnetization field H which the bar magnet produces? As I understand, first you need the magnetization M which is the magnetic moments per volumen. But then what ? Or would you apply and external field to the bar magnet and see how it reacts, if this is the case, how does this work ?

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asked 2021-03-21

In the figure below, the rolling axle, 1.43 m long, is pushed along horizontal rails at a constant speed $v=3.36m/s$.

A resistor $R=0.325$ ohm is connected to the rails at points a and b, directly opposite each other. (The wheels make good electrical contact with the rails, and so the axle, rails, and R form a closed-loop circuit. The only significant resistance in the circuit is R.) There is a uniform magnetic field $B=0.0850T$vertically downward. Calculate the induced current I in the resistor and what horizontal force F is required to keep the axle rolling at constant speed?

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asked 2022-05-18

Why is magnetic field vector perpendicular to magnetic force vector?

So recently in physics class, we learned about the magnetism right hand rules.

One of them states that the index finger points in the direction of the velocity of a particle, the middle finger points in the direction of the magnetic field, and the thumb points in the direction of the magnetic force.

I'm curious why the magnetic field vector is perpendicular to the magnetic force vector.

So recently in physics class, we learned about the magnetism right hand rules.

One of them states that the index finger points in the direction of the velocity of a particle, the middle finger points in the direction of the magnetic field, and the thumb points in the direction of the magnetic force.

I'm curious why the magnetic field vector is perpendicular to the magnetic force vector.