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Emf

a coil is being rotated at a frequency of 50Hz in a magnetic field of field density B, what is the average emf induced and maximum emf induce? Area = a and number of turns = N.

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

You can also find the Exam Thread list for A-levels here and GCSE here. :dumbells:

Just quoting in Puddles the Monkey so she can move the thread if needed :h:

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Reply 2

hezzlington

Original post by jonnypdot

a coil is being rotated at a frequency of 50Hz in a magnetic field of field density B, what is the average emf induced and maximum emf induce? Area = a and number of turns = N.

What issues do you have with this problem?

Are you able to show me an attempt at the question?

Reply 3

jonnypdot

Original post by hezzlington

What issues do you have with this problem?

Are you able to show me an attempt at the question?

i just thought of it because emf graphs are usually sine or cosine waves so how can i determine average emf.

Reply 4

hezzlington

Original post by jonnypdot

i just thought of it because emf graphs are usually sine or cosine waves so how can i determine average emf.

Well, what is the average for a periodic waveform? The positive values cancel out the negative values, giving you an average of zero.

Okay, for some reason latex is really messing up for me.

Maximum emf is w*BA*sinw*t

w = omega

(edited 7 years ago)

Reply 5

jonnypdot

Original post by hezzlington

but i thought the current coming from mains supply is created from alternating pd

Reply 6

hezzlington

Original post by jonnypdot

but i thought the current coming from mains supply is created from alternating pd

The problem you're describing sounds very similar to an alternator (used in cars). The emf varies sinusoidally and you get an alternating current that also varies sinusoidally in magnitude and direction.

Reply 7

jonnypdot

Original post by hezzlington

so i was describing a different concept? If so lets forget about it. Also can i ask you a question regarding electric fields?

Reply 8

hezzlington

Original post by jonnypdot

so i was describing a different concept? If so lets forget about it. Also can i ask you a question regarding electric fields?

It sounds like you were possibly trying to describe a direct-current generator that produces emf of the same sign (e.g. no negative values). The DC generators used commercially have lots of coils and commutator segments which smooth out the bumps in emf so that your terminal voltage is one directional and pretty much constant.

For a problem involving a dc generator, you can calculage average back emf.

Sure, ask away :smile:

(edited 7 years ago)

Reply 9

jonnypdot

Original post by hezzlington

It sounds like you were trying to describe a direct-current generator that produces emf of the same sign (e.g. no negative values). The DC generators used commercially have lots of coils and commutator segments which smooth out the bumps in emf so that your terminal voltage is one directional and pretty much constant.

For a problem involving a dc generator, you can calculage average back emf.

Sure, ask away :smile:

so i have 2 positive charges if they are separated by a distance d what will the electric field strength at the mid point be? Would i just add the electric field strength for each charge at a distance d/2?

If i had unlike charges would i still have to add the electric field strength? a
if i had 2 negative charges i would have to find the net electric field strength right?

Reply 10

hezzlington

Original post by jonnypdot

Yes. This is known as the superposition of Electric Fields and you use it a lot when dealing with distributions of point charges.

Original post by jonnypdot

If i had unlike charges would i still have to add the electric field strength? a

Two unlike charges is also known as an electric dipole.

Total field is denoted as a. So you need to calculate both the field strengths of each test charge, then add them together, yes.

Original post by jonnypdot

if i had 2 negative charges i would have to find the net electric field strength right?

Calculate the E of each, taking into account direction, and then add them. Same process.

Reply 11

jonnypdot

Original post by hezzlington

Yes. This is known as the superposition of Electric Fields and you use it a lot when dealing with distributions of point charges.

Two unlike charges is also known as an electric dipole.

Total field is denoted as a. So you need to calculate both the field strengths of each test charge, then add them together, yes.

Calculate the E of each, taking into account direction, and then add them. Same process.