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R = L 2 π

The magnetic field due to current I in the circular wire is :

B C = μ 0 I 2 R = 2 π μ 0 I 2 L = π μ 0 I L = 3.14 μ 0 I L

In the case of straight wire, let us consider that wire is long enough for a point around middle of the wire. For comparison purpose, we assume that perpendicular linear distance used for calculating magnetic field due to current in straight wire is equal to the radius of circle. The magnetic field at a perpendicular distance “R” due to current in long straight wire is given as :

B L = μ 0 I 4 π R = μ 0 I X 2 π 4 π L = μ 0 I 2 L = 0.5 μ 0 I L

Clearly, the magnetic field due to current in circular wire is 6.28 times greater than that due to current in straight wire at comparable points of observations. Note that this is so even though we have given advantage to straight wire configuration by assuming it to be long wire. In a nutshell, a circular configuration tends to concentrate magnetic field along axial direction which is otherwise spread over the whole length of wire.

Problem : A current 10 A flowing through a straight wire is split at point A in two semicircular wires of radius 0.1 m. The resistances of upper and lower semicircular wires are 10 Ω and 20 Ω respectively. The currents rejoin to flow in the straight wire again as shown in the figure. Determine the magnetic field at the center “O”.

Magnetic field due to current in wire

Magnetic field due to current in wire

Solution : The straight wire sections on extension pass through the center. Hence, magnetic field due to straight wires is zero. Here, the incoming current at A is distributed in the inverse proportion of resistances. Let the subscripts “1” and “2’ denote upper and lower semicircular sections respectively. The two sections are equivalent to two resistances in parallel combination as shown in the figure. Here, potential difference between “A” and “B” is :

Currents in semicircular segments

Currents in semicircular segments

V A B = I X R 1 X R 2 R 1 + R 2 = I 1 R 1 = I 2 R 2 I 1 = I X R 2 R 1 + R 2 = 10 X 20 30 = 20 3 A I 2 = I X R 1 R 1 + R 2 = 10 X 10 30 = 10 3 A

We see that current in the upper section is twice that in the lower section i.e. I 1 = 2 I 2 . Also, the magnetic field is perpendicular to the plane of semicircular section (plane of drawing). The current in the upper semicircular wire is clockwise. Thus, the magnetic field due to upper section is into the plane of drawing. However, the current in the lower semicircular is anticlockwise. Thus, the magnetic field due to lower section is out of the plane of drawing. Putting θ = π for each semicircular section, the net magnetic field due to semicircular sections at “O” is:

B = μ 0 I 1 π 4 π R μ 0 I 2 π 4 π R B = μ 0 I 1 π 4 π R μ 0 I 1 π 8 π R = μ 0 I 1 π 8 π R B = 10 - 7 X 20 3 X 8 X 0.1 = 8.3 X 10 - 7 T

The net magnetic field is into the plane of drawing.

Exercises

An electron circles a single proton nucleus of radius 3.2 X 10 - 11 m with a frequency of 10 16 Hz. The charge on the electron is 1.6 X 10 - 19 Coulomb. What is the magnitude of magnetic field due to orbiting electron at the nucleus?

The equivalent current is given by :

I = q T = q ν

where ν and T are frequency and time period of revolutions respectively. The magnitude of magnetic field due to circular wire is given by :

B = μ 0 I 2 R

Substituting for I, we have :

B = μ 0 I 2 R = μ 0 q ν 2 R

Putting values,

B = 4 π 10 - 7 X 1.6 X 10 - 19 X 10 16 2 X 3.2 X 10 - 11 B = 31.4 T

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
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Sherica
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right! what he said ⤴⤴⤴
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
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Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Electricity and magnetism. OpenStax CNX. Oct 20, 2009 Download for free at http://cnx.org/content/col10909/1.13
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