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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?

Strategy

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 1 = 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} } "." } {}

Solution

Substituting known values gives

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

Discussion

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.

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.

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.
Practice Key Terms 9

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Source:  OpenStax, Concepts of physics. OpenStax CNX. Aug 25, 2015 Download for free at https://legacy.cnx.org/content/col11738/1.5
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