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Magnetic field produced by a current-carrying solenoid

A solenoid    is a long coil of wire (with many turns or loops, as opposed to a flat loop). Because of its shape, the field inside a solenoid can be very uniform, and also very strong. The field just outside the coils is nearly zero. [link] shows how the field looks and how its direction is given by RHR-2.

A diagram of a solenoid. The current runs up from the battery on the left side and spirals around with the solenoid wire such that the current runs upward in the front sections of the solenoid and then down the back. An illustration of the right hand rule 2 shows the thumb pointing up in the direction of the current and the fingers curling around in the direction of the magnetic field. A length wise cutaway of the solenoid shows magnetic field lines densely packed and running from the south pole to the north pole, through the solenoid. Lines outside the solenoid are spaced much farther apart and run from the north pole out around the solenoid to the south pole.
(a) Because of its shape, the field inside a solenoid of length l size 12{l} {} is remarkably uniform in magnitude and direction, as indicated by the straight and uniformly spaced field lines. The field outside the coils is nearly zero. (b) This cutaway shows the magnetic field generated by the current in the solenoid.

The magnetic field inside of a current-carrying solenoid is very uniform in direction and magnitude. Only near the ends does it begin to weaken and change direction. The field outside has similar complexities to flat loops and bar magnets, but the magnetic field strength inside a solenoid    is simply

B = μ 0 nI ( inside a solenoid ) , size 12{B=μ rSub { size 8{0} } ital "nI"` \( "inside a solenoid" \) ,} {}

where n size 12{n} {} is the number of loops per unit length of the solenoid ( n = N / l size 12{ \( n=N/l} {} , with N size 12{N} {} being the number of loops and l size 12{l} {} the length). Note that B size 12{B} {} is the field strength anywhere in the uniform region of the interior and not just at the center. Large uniform fields spread over a large volume are possible with solenoids, as [link] implies.

Calculating field strength inside a solenoid

What is the field inside a 2.00-m-long solenoid that has 2000 loops and carries a 1600-A current?


To find the field strength inside a solenoid, we use B = μ 0 nI size 12{B=μ rSub { size 8{0} } ital "nI"} {} . First, we note the number of loops per unit length is

n = N l = 2000 2.00 m = 1000 m 1 = 10 cm 1 . size 12{n rSup { size 8{ - 1} } = { {N} over {l} } = { {"2000"} over {2 "." "00" m} } ="1000"" m" rSup { size 8{ - 1} } ="10"" cm" rSup { size 8{ - 1} } "." } {}


Substituting known values gives

B = μ 0 nI = × 10 7 T m/A 1000 m 1 1600 A = 2 .01 T.


This is a large field strength that could be established over a large-diameter solenoid, such as in medical uses of magnetic resonance imaging (MRI). The very large current is an indication that the fields of this strength are not easily achieved, however. Such a large current through 1000 loops squeezed into a meter’s length would produce significant heating. Higher currents can be achieved by using superconducting wires, although this is expensive. There is an upper limit to the current, since the superconducting state is disrupted by very large magnetic fields.

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There are interesting variations of the flat coil and solenoid. For example, the toroidal coil used to confine the reactive particles in tokamaks is much like a solenoid bent into a circle. The field inside a toroid is very strong but circular. Charged particles travel in circles, following the field lines, and collide with one another, perhaps inducing fusion. But the charged particles do not cross field lines and escape the toroid. A whole range of coil shapes are used to produce all sorts of magnetic field shapes. Adding ferromagnetic materials produces greater field strengths and can have a significant effect on the shape of the field. Ferromagnetic materials tend to trap magnetic fields (the field lines bend into the ferromagnetic material, leaving weaker fields outside it) and are used as shields for devices that are adversely affected by magnetic fields, including the Earth’s magnetic field.

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Section summary

  • The strength of the magnetic field created by current in a long straight wire is given by
    B = μ 0 I 2 πr ( long straight wire ) ,
    where I size 12{I} {} is the current, r size 12{r} {} is the shortest distance to the wire, and the constant μ 0 = × 10 7 T m/A size 12{μ rSub { size 8{0} } =4π times "10" rSup { size 8{ - 7} } `T cdot "m/A"} {} is the permeability of free space.
  • The direction of the magnetic field created by a long straight wire is given by right hand rule 2 (RHR-2): Point the thumb of the right hand in the direction of current, and the fingers curl in the direction of the magnetic field loops created by it.
  • The magnetic field created by current following any path is the sum (or integral) of the fields due to segments along the path (magnitude and direction as for a straight wire), resulting in a general relationship between current and field known as Ampere’s law.
  • The magnetic field strength at the center of a circular loop is given by
    B = μ 0 I 2 R ( at center of loop ) , size 12{B= { {μ rSub { size 8{0} } I} over {2R} } " " \( "at center of loop" \) ,} {}
    where R size 12{R} {} is the radius of the loop. This equation becomes B = μ 0 nI / ( 2 R ) size 12{B=μ rSub { size 8{0} } ital "nI"/ \( 2R \) } {} for a flat coil of N size 12{N} {} loops. RHR-2 gives the direction of the field about the loop. A long coil is called a solenoid.
  • The magnetic field strength inside a solenoid is
    B = μ 0 nI ( inside a solenoid ) , size 12{B=μ rSub { size 8{0} } ital "nI"" " \( "inside a solenoid" \) ,} {}
    where n size 12{n} {} is the number of loops per unit length of the solenoid. The field inside is very uniform in magnitude and direction.

Conceptual questions

Make a drawing and use RHR-2 to find the direction of the magnetic field of a current loop in a motor (such as in [link] ). Then show that the direction of the torque on the loop is the same as produced by like poles repelling and unlike poles attracting.

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Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
I am now in this group
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
i dont think dere is any potential energy... by d virtue of no height present
there is compressed energy,dats only potential energy na?
yes.. but... how will u approach that question without The Height in the question?
Can you explain how you get 54J?
Because mine is 36J
got 36J too
OK the answer is 54J Babar is correct
Conservation of Momentum
woow i see.. can you give the formula for this
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
please help find dy/dx 2x-y/x+y
By using the Quotient Rule dy/dx = 3y/(x +y)²
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
i think i m correct
But how?
what's the big bang?
kwame Reply
yes what is it?
it is the explanation of how the universe began
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
thanks guys
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
pls í need help
4000 work is done
speed=distance /time distance=speed/time
now use this formula
what's the answer then
great Mudang
please Ana explain 4000 ?
hey mudang there is a product of force and acceleration not force and displacement
@Mohammed answer is 0.8hours or 48mins
its not possible
í want the working procedure
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
I think from kinematics
You can find physics books at the library or online. That's how I started.
And yes, kinematics is usually where you can begin.
study basic algebra and calculus and can start from classical mechanics
yes think so but dimension is the best starting point
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
solve the formula's please
those are the three .. what you wanna solve ?
For first equation simply integrate formula of acceleration in the limit v and u
For second itegrate velocity formula by ising first equation
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
what is the question dear
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
find the distance
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
I think the formulae used for this question is lambda=(ax)/D
Practice Key Terms 9

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