# accelerating magnet and electromagnetic induction

by spidey
Tags: accelerating, electromagnetic, induction, magnet
 P: 213 hi..i know the faraday's electromagnetic induction e=-dФ/dt...supposing there is an coil of wire and i am accelerating magnet through the coil of wire and this will produce emf because accelerating magnet results in change in magnetic field...my question is how to relate this acceleration a and change in magnetic field dФ/dt...i think these two should be proportional...anybody help me..
 Mentor P: 11,907 You would have to know details of the magnet's field strength as a function of location. As long as the magnet's field varies with position (i.e. is not constant), the magnet just has to be moving. It need not be accelerating.
P: 213
 Quote by Redbelly98 You would have to know details of the magnet's field strength as a function of location. As long as the magnet's field varies with position (i.e. is not constant), the magnet just has to be moving. It need not be accelerating.
Yes as long as magnet's field varies, emf is generated..but if i vary the magnet's field by acceleration then can we relate the change in the magnet's field with acceleration...i think

e = -∂Ф/∂t α ∂v/∂t
-∂Ф/∂t α ∂v/∂t

is this correct?..do we have to introduce constant in the right hand side to make the two sides equal?..

Mentor
P: 11,907

## accelerating magnet and electromagnetic induction

No, not correct. It's proportional to v, not ∂v/∂t.
P: 213
 Quote by Redbelly98 No, not correct. It's proportional to v, not ∂v/∂t.
Hi Redbelly,

Actually,
e = -∂Ф/∂t α A(Acceleration)
therefore,
-∂Ф/∂t α ∂v/∂t [ A=∂v/∂t]

How it is proportional to v?..it should be proportional to ∂v/∂t..if this is wrong,please give correct solution and also do we have to add constant to remove proportionality?
Mentor
P: 11,907
 Quote by spidey Hi Redbelly, Actually, e = -∂Ф/∂t α A(Acceleration)
I see no basis for saying -∂Ф/∂t α acceleration here.

I wonder if you are confusing something, either:
A=acceleration vs. A=area of loop?
v=velocity vs. v=voltage?

 How it is proportional to v?..it should be proportional to ∂v/∂t..if this is wrong,please give correct solution and also do we have to add constant to remove proportionality?
Let x = position of the moving loop.
The loop experiences a changing B-field, which is
dB/dt = (dB/dx) (dx/dt)

Since v = dx/dt, there it is.

Do you have a book to look at that explains induced emf? They usually have some sample problems involving moving loops or magnets. Acceleration never comes up.
P: 213
 Quote by Redbelly98 I see no basis for saying -∂Ф/∂t α acceleration here. I wonder if you are confusing something, either: A=acceleration vs. A=area of loop? v=velocity vs. v=voltage? Let x = position of the moving loop. The loop experiences a changing B-field, which is dB/dt = (dB/dx) (dx/dt) Since v = dx/dt, there it is. Do you have a book to look at that explains induced emf? They usually have some sample problems involving moving loops or magnets. Acceleration never comes up.
Thanks for clearing my doubt...can you suggest me any good book for electromagnetism? i am not physicist but interested to learn.
 Mentor P: 11,907 I like the book Physics, by Douglas C. Giancoli. It's a non-calculus based textbook. I've seen it used for both AP high school physics, and college physics for non-physics majors. This stuff is in chapter 21 of the 5th edition. For \$24, you can get a used, paperback copy from amazon.com -- but this is a special edition that begins with Chapter 16. It contains all the electricity and magnetism chapters from the full book. http://www.amazon.com/gp/offer-listi...8/ref=dp_olp_2 Or, you could search on "induction" here: http://www.amazon.com/gp/reader/0130...pt#reader-link And start reading at p. 622
P: 213
 Quote by Redbelly98 Let x = position of the moving loop. The loop experiences a changing B-field, which is dB/dt = (dB/dx) (dx/dt) Since v = dx/dt, there it is.
dx/dt = v but what dB/dx = ?
I understand this is the change in magnetic field for distance dx... is there any term for db/dx like v=dx/dt.
 Mentor P: 11,907 Nope, just leave it as dB/dx. All it really says is B should not be constant (in space) in order to generate a change in flux this way. A physics book will explain it much better, using figures, than I can.
P: 213
 Quote by Redbelly98 Nope, just leave it as dB/dx. All it really says is B should not be constant (in space) in order to generate a change in flux this way. A physics book will explain it much better, using figures, than I can.
Hi Redbelly,

I gone through "Electromagnetism and optics" An introductory course book. In that i find the motional emf which i was refering to. Below derivation was given there

l and x are dimensions. i.e. l*x is area.

Ф = Blx
dФ = Bldx
dФ = Blvdt
dФ/dt = Blv
e = Blv

but this holds for only constant velocity i think...but what happens if we accelerate i.e. if the v is increasing. Even for increasing v also the magnetic field changes. does for increasing v also the above equation holds good? Or do i have to differentiate v again like this.

d/dt (dФ/dt) = Bl dv/dt
d²Ф/dt² = Bla . here a -> acceleration.

but e = dФ/dt and e != d²Ф/dt²

what is this d²Ф/dt²? is this also changing magnetic field?

i am going somewhere.help me.i think i am confusing you.what i want is.in the book it is given for constant velocity.can we use that for increasing or decreasing v also? if so,no problem. but if it is not,how to include that increasing or decreasing v?
P: 7,403
 Quote by spidey Ф = Blx dФ = Bldx dФ = Blvdt dФ/dt = Blv e = Blv but this holds for only constant velocity i think...but what happens if we accelerate
 Quote by spidey what is this d²Ф/dt²? is this also changing magnetic field?
dФ/dt is the changing magnetic field, or more precisely, the changing magnetic flux.

Ф = Blx
dФ/dt = d(Blx)/dt = Bl(dx/dt).

In the last step, I assumed B and l were constant in this problem, and only x was changing. For a magnet moving at constant velocity or accelerating in some weird way, x will be a function of time, so we write it as x(t). You can find dx(t)/dt for any function x(t) you wish, and then find the corresponding dФ/dt.
 Mentor P: 11,907 spidey, it works fine for when v is changing too. e = Blv (in your example) still holds true.
 P: 213 Thank you Redbelly and atyy...i got it...i have another question... Emf is generated in a wire when there is change in magnetic field..supposing i dont place any wire and i am changing magnetic field i.e. am moving or shaking magnet..now this is also changing magnetic field but there is no wire nearby,so where will the emf go or where is the electric field? is it in air? I understand the first equation for total flux of electric field ∫ E dA = (4∏ r²) * (Q /4∏e0 r²) = Q / e0 But I dont understand the maxwell's second equation∫ B dA = 0. we can also do derivation like for electric field. ∫ B dA = (4∏ r²) * (μ m/4∏ r²) = μm In the book, the derivation is not given and i couldnt find any derivation in websites also..all sites directly give this result...how maxwell derived this ∫ B dA = 0?
P: 7,403
 Quote by spidey Emf is generated in a wire when there is change in magnetic field..supposing i dont place any wire and i am changing magnetic field i.e. am moving or shaking magnet..now this is also changing magnetic field but there is no wire nearby,so where will the emf go or where is the electric field? is it in air?
The emf is always generated in "empty space". If you happen to place a wire in the right position in "empty space", the free electrons in the wire will move in response to the emf.

 Quote by spidey I understand the first equation for total flux of electric field ∫ E dA = (4∏ r²) * (Q /4∏e0 r²) = Q / e0 But I dont understand the maxwell's second equation∫ B dA = 0. we can also do derivation like for electric field. ∫ B dA = (4∏ r²) * (μ m/4∏ r²) = μm In the book, the derivation is not given and i couldnt find any derivation in websites also..all sites directly give this result...how maxwell derived this ∫ B dA = 0?
∫ E dA = Q / e0

Roughly speaking, the LHS is an electric field, the RHS is charge, so it says a charge produces an electric field. Now if you place a positive and negative charge together, the RHS will be zero, and the total electric flux through any surface surrounding both of them will also be zero. So this equation requires that you can separate positive charge from negative charge.

∫ B dA = 0

Thinking in the same way, the above equation says that you cannot separate "positive magnetic charge" from "negative magnetic charge", all magnets always come with a north and south pole. If you cut the magnet into two pieces, you will generate two magnets, each with their own north and south poles. If we ever found a magnetic monopole, this equation would have to be modified.
P: 213
 Quote by atyy The emf is always generated in "empty space". Thinking in the same way, the above equation says that you cannot separate "positive magnetic charge" from "negative magnetic charge", all magnets always come with a north and south pole.
emf in empty space is weird to me. in empty space there are no charges,then how there is motion of charges for an emf.

i understand for flux of electric field. Are you saying since magnets come with north and south combined,there is no flux in closed surface? if there is no flux then it means that there is no magnetic field,isnt it?