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B · d l = 1 B · d l = B l .
Figure shows the closed rectangular path and the infinite solenoid. Segment 1 is inside the solenoid and is parallel to the path. Segments 2 and 4 are perpendicular to the path. Segment 3 is outside the solenoid.
The path of integration used in Ampère’s law to evaluate the magnetic field of an infinite solenoid.

The solenoid has n turns per unit length, so the current that passes through the surface enclosed by the path is nlI . Therefore, from Ampère’s law,

B l = μ 0 n l I

and

B = μ 0 n I

within the solenoid. This agrees with what we found earlier for B on the central axis of the solenoid. Here, however, the location of segment 1 is arbitrary, so we have found that this equation gives the magnetic field everywhere inside the infinite solenoid.

Outside the solenoid, one can draw an Ampère’s law loop around the entire solenoid. This would enclose current flowing in both directions. Therefore, the net current inside the loop is zero. According to Ampère’s law, if the net current is zero, the magnetic field must be zero. Therefore, for locations outside of the solenoid’s radius, the magnetic field is zero.

When a patient undergoes a magnetic resonance imaging (MRI) scan, the person lies down on a table that is moved into the center of a large solenoid that can generate very large magnetic fields. The solenoid is capable of these high fields from high currents flowing through superconducting wires. The large magnetic field is used to change the spin of protons in the patient’s body. The time it takes for the spins to align or relax (return to original orientation) is a signature of different tissues that can be analyzed to see if the structures of the tissues is normal ( [link] ).

Photo shows an MRI system. It consists of the cylindrical solenoid that is used to generate a large magnetic field.
In an MRI machine, a large magnetic field is generated by the cylindrical solenoid surrounding the patient. (credit: Liz West)

Magnetic field inside a solenoid

A solenoid has 300 turns wound around a cylinder of diameter 1.20 cm and length 14.0 cm. If the current through the coils is 0.410 A, what is the magnitude of the magnetic field inside and near the middle of the solenoid?

Strategy

We are given the number of turns and the length of the solenoid so we can find the number of turns per unit length. Therefore, the magnetic field inside and near the middle of the solenoid is given by [link] . Outside the solenoid, the magnetic field is zero.

Solution

The number of turns per unit length is

n = 300 turns 0.140 m = 2.14 × 10 3 turns/m .

The magnetic field produced inside the solenoid is

B = μ 0 n I = ( 4 π × 10 −7 T m/A ) ( 2.14 × 10 3 turns/m ) ( 0.410 A ) B = 1.10 × 10 −3 T .

Significance

This solution is valid only if the length of the solenoid is reasonably large compared with its diameter. This example is a case where this is valid.

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Check Your Understanding What is the ratio of the magnetic field produced from using a finite formula over the infinite approximation for an angle θ of (a) 85 ° ? (b) 89 ° ? The solenoid has 1000 turns in 50 cm with a current of 1.0 A flowing through the coils

a. 1.00382; b. 1.00015

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Toroids

A toroid is a donut-shaped coil closely wound with one continuous wire, as illustrated in part (a) of [link] . If the toroid has N windings and the current in the wire is I , what is the magnetic field both inside and outside the toroid?

Figure A shows a toroid that is a coil wound into a donut-shaped object. Figure B shows a loosely wound toroid that does not have cylindrical symmetry. Figure C shows a tightly wound toroid with the symmetry that is very close to the cylindrical. Figure D shows several paths. Paths D1 and D3 are external to the toroid. Path D2 lies within the toroid.
(a) A toroid is a coil wound into a donut-shaped object. (b) A loosely wound toroid does not have cylindrical symmetry. (c) In a tightly wound toroid, cylindrical symmetry is a very good approximation. (d) Several paths of integration for Ampère’s law.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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