29.12:
Magnetic Field of a Solenoid
Consider a solenoid with closely packed turns of wire so that its total length is greater than its radius.
A steady current flowing through this solenoid generates a magnetic field, which can be estimated by considering a rectangular Amperian loop through it.
Applying Ampere's Law, the sum of the line integral of the magnetic field along each loop path equals the permeability multiplied by the net current enclosed by the loop.
Now, the magnetic field integral along path one is the product of the magnetic field and the loop length. It is zero along paths two and four since the magnetic field is perpendicular to the path. As the magnetic field outside the solenoid is zero, the integral along path three is also null.
The net enclosed current is equal to the total turns inside the Amperian loop, multiplied by the current flowing through the solenoid.
Thus, the magnetic field inside a solenoid parallel to the solenoid axis is directly proportional to the number of turns per unit length and the current.
29.12:
Magnetic Field of a Solenoid
A solenoid is a conducting wire coated with an insulating material, wound tightly in the form of a helical coil. The magnetic field due to a solenoid is the vector sum of the magnetic fields due to its individual turns. Therefore, for an ideal solenoid, the magnetic field within the solenoid is directly proportional to the number of turns per unit length and the current. Conversely, the magnetic field outside the solenoid is zero.
Consider a solenoid with 100 turns wrapped around a cylinder of diameter 1 cm and length 10 cm. What is the magnitude of the magnetic field inside and outside the solenoid when a current of 0.5 A is passed through it?
Here, since the number of turns, solenoid length, and current are known quantities, the magnetic field inside and outside the solenoid can be estimated using Ampere's Law.
Since the length is much greater than the diameter, it can be considered an ideal solenoid. Hence, the magnetic field outside it is zero.
The magnetic field inside the solenoid is expressed as the product of vacuum permeability, the number of turns per unit length, and the current. The number of turns per unit length is given by the total number of turns divided by the solenoid length and is calculated to be 1,000 turns/m. Using the values of number of turns per unit length and current, the magnetic field inside the solenoid is estimated as 6.29 x 10-4 T.
Suggested Reading
- Young, H. D., and Freedman, R.A. (2012). University Physics with Modern Physics. San Francisco, CA: Pearson. pp. 939-940.
- OpenStax. (2019). University Physics Vol. 2. [Web version]. pp.555 Retrieved from https://openstax.org/books/college-physics/pages/22-9-magnetic-fields-produced-by-currents-amperes-law