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22.9 Magnetic fields produced by currents: ampere’s law  (Page 3/13)

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Magnetic field produced by a current-carrying circular loop

The magnetic field near a current-carrying loop of wire is shown in [link] . Both the direction and the magnitude of the magnetic field produced by a current-carrying loop are complex. RHR-2 can be used to give the direction of the field near the loop, but mapping with compasses and the rules about field lines given in Magnetic Fields and Magnetic Field Lines are needed for more detail. There is a simple formula for the magnetic field strength at the center of a circular loop    . It is

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 is very similar to that for a straight wire, but it is valid only at the center of a circular loop of wire. The similarity of the equations does indicate that similar field strength can be obtained at the center of a loop. One way to get a larger field is to have N size 12{N} {} loops; then, the field is B = Nμ 0 I / ( 2 R ) . Note that the larger the loop, the smaller the field at its center, because the current is farther away.

Figure a illustrates use of the right hand rule 2 to determine the direction of the magnetic field around a current-carrying loop. The right hand thumb points in the direction of I while the fingers curl around in the direction of B. Figure b shows the magnetic field lines circling the wire, as viewed from the side.
(a) RHR-2 gives the direction of the magnetic field inside and outside a current-carrying loop. (b) More detailed mapping with compasses or with a Hall probe completes the picture. The field is similar to that of a bar magnet.

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?

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 = 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 = 4π × 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.

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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