
#1
Jul2604, 05:46 AM

P: 160

a solenoid having 200 turns per metre and carrying a current of 0.050A lies with its axis east west. Well inside the solenoid is a small compass whose needle points 37 west of north. calculate earth's horizontal magnetic field component B
i already find the magnetic flux density of solenoid which is 1.26 X 10^5 T. what should i do next? and is there any site that have a good explanation on earth's horizontal magnetic field. it is not covered in my syllabus, but i just want to know. 



#2
Jul2604, 06:22 AM

P: 1,772

You seem to be having some problem relating the field of the solenoid and the earth's field with the deflection. The magnetic field intensity (sometimes called magnetic induction) vectors for the solenoid and the earth get superposed at every point in space when both these fields are active. So you need to figure out (a) how to vectorially add these fields and (b) the deflection of the compass needle due to this net field (because after all, the needle is subjected to this net field and this field is what causes it to deflect from its initial angle).
Its funny they should teach all this stuff when its been replaced by more modern theories of magnetism. We were taught all this at school too (in class 12). Nevertheless, here are some sites that might help you: http://www.earth.nwu.edu/people/seth/demos/DIP/dip.html http://www.fas.harvard.edu/~scdiroff...eticField.html http://www.physics.ucla.edu/demoweb/...tic_field.html Hope that helps... Cheers Vivek 



#3
Jul2604, 11:04 AM

P: 160

im still confused. i have a few questions which is of the same type.
mavick, do u mind if you give me an example how to work on the question above? thank you. 



#4
Jul2604, 11:57 AM

P: 1,772

magnetic field calculation
Denian, you need to figure out which component of the earth's magnetic field, say H, adds to the magnetic field of the solenoid (given by [tex]B_{solenoid} = \mu_{0}nI[/tex] where n is the no of turns per unit length and I is the current carried by it). The component would be of the form [tex]H cos\theta[/tex] or [tex]H sin\theta[/tex] depending on how you measure theta.
The best way to go about it is to draw a diagram showing the two vectors B and H due to the solenoid and the earth at a point (remember that B is axial and will be strongest inside the solenoidthe above expression is that approximated for a closed pack coil). Then, you should realize that if the needle comes into equilibrium 37 west of north (or whatever you're given in a general case), the net force on the needle is zero (for if it were not zero, there could have been a torque that could cause further deflection..but its static so reversing this logic, I can say that the NET force on it at this point in time is zero). Use the equation for the force/torque. BUT: You need to understand that this whole solenoid+earth system can be replaced (in effect) by a net field [tex]B_{net}[/tex] which is the vector sum of the fields of the solenoid (axial, direction given by Right Hand Rule or Fleming rule whatever you wish to use) and that of the earth (I call the earth's field H). The earth's field has a horizontal component and a vertical component. Resolve it using the angle of dip/declination. If you are having trouble doing this, the sites I have given in my previous post should help. Finally if you just can't get things to work, send your complete solution and I'll try to see what the problem is. There is no big deal here, its just the directions you and I have to take care of and the rest is mincemeat math. Cheers Vivek EDIT: Give me a few hours denian, while you follow the advice given in this post and I'll figure out the equations myself and give you the answer. 



#5
Jul2704, 04:38 AM

P: 160

i try first.




#6
Jul2804, 01:32 AM

P: 160

this is what i did. please help to check. thx
flux density due to solenoid, B(solenoid) = 1.26 X 10^5 from the vector diagram, use sine rule, i will get B' = 1.67 X 10^5 T the answer is correct, but i doubt if my working in not correct. 



#7
Jul2904, 08:03 PM

P: 1,772

Hi denian
The vector diagram is okay. If you used it to write, [tex] B_{resultant} = \sqrt{B_{earth}^{2} + B_{solenoid}^2} [/tex] then all your subsequent computions should be okay too (as they are indeed as you get the correct answer). Do you still have doubts about your working? Cheers Vivek 



#8
Jul3004, 08:23 AM

P: 160

No. Thanks maverick.



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