Magnetic Field In A Solenoid Formula

Magnetic Field in a Solenoid Formula

A magnetic field is created when a current flows through a conductor. A solenoid experiences the same thing when an electrical current flows through it. A solenoid turns into an electromagnet when it receives current. B=0nI is the Magnetic Field in a Solenoid Formula

Additionally, the loop of the current-carrying solenoid generates a magnetic field, which is why solenoids are frequently used in electronics and electromagnets to produce a uniform magnetic field. Ampere’s Circuital Law is used in magnetics to determine the magnetic field of a highly symmetric configuration carrying a constant current. The Magnetic Field in a Solenoid Formula, the formula for a solenoid, the magnetic field caused by a current in a solenoid, and the magnetic field of a solenoid are all covered in this article.

A solenoid is a long wire that has been wound in a helix shape. Using this arrangement, we can create a magnetic field that is fairly uniform. When the turns of the solenoid are closely spaced, we can think of each turn as a circular loop, and the net magnetic field is the vector sum of the fields produced by all the turns. With increasing solenoid length, the interior field becomes more uniform.

We can think of it as the ideal solenoid when the turns are closely spaced and the length is significantly greater than the turn radius. In this case, the interior field is uniform over a sizable volume, and the external field is zero. The magnetic field in an ideal solenoid can be expressed using Ampere’s law.

Introduction to Solenoid

Take into account the solenoid, which is where the imaginary loop ‘c’ in any figure is situated. According to Ampère’s law, the magnetic flux density vector B’s line integral around this loop is zero. It occurs because it contains no electrical currents. The horizontal portions of loop c do not contribute anything to the integral because, as we have already demonstrated, the field is pointing upwards inside the solenoid. As a result, the integral of the figure’s upper side, which stands for number one, equals an integral portion of side two, which is moving downward. The integrands are equal because the dimensions of the loop can be changed arbitrarily and still produce the same outcome.

Magnetic Field in a Solenoid Formula is as follows:

B=μ0 nl

The magnetic flux density is represented here by B.

The value of the magnetic constant, 0 is 4 x 10-7 Hm.

or 12.57 x 107 Hm, where l is the length of the solenoid, I is the current flowing through it, and N is the number of turns.

Magnetic Field Inside a Solenoid Formula

A solenoid’s magnetic field can be calculated using the Magnetic Field in a Solenoid Formula B = 0 nl. The magnetic constant, 0 in the Magnetic Field in a Solenoid Formula, has a value of 4 x 10-7 Hm-1 or 12.57 x 10-7 Hm-1. N in the Magnetic Field in a Solenoid Formula stands for the number of turns, and I in the Magnetic Field in a Solenoid Formula stands for the current flowing through the solenoid. The strength of the magnetic field in a long solenoid is homogeneous and independent of the solenoid’s cross-sectional area or its distance from the axis.

Magnetic Field Outside a Solenoid Formula

The field outside the solenoid can be inferred to be radially uniform or constant by using a similar argument for the loop. The absence of the flux density outside the solenoid can also be demonstrated intuitively. Magnetic field lines can only be found as loops; unlike electric field lines, they cannot diverge or converge to a point. Since the longitudinal path is followed by the magnetic field lines inside the solenoid, these magnetic field lines outside the solenoid must move the other way. The lines creating a loop are what causes this to occur. Nevertheless, the volume outside the solenoid is significantly larger than the volume inside.

• A bar magnet’s magnetic field is 1000 times weaker than Earth’s magnetic field.
• Our solar system contains planets with magnetic fields. Earth, Saturn, Jupiter, Neptune, and Uranus are among these planets.
• Because of the presence of a substantial iron core at its core, Earth has a magnetic field.
• When operating, maglev trains make use of magnetic field-related concepts.
• Many roller coasters push their cars onto the track using an electromagnet.
• When the earth’s magnetic field lines and the sun’s incoming solar winds interact, the aurora phenomenon is visible close to the poles.

Solved Examples

Determine the magnetic field produced by the solenoid of length 80 cm under the number of turns of the coil is 360 and the current passing through is 15 A.

Solution:

Given:

Number of turns N = 360

Current I = 15 A

Permeability μo = 1.26 × 10−6 T/m

Length L = 0.8 m

The Magnetic Field In A Solenoid Formula is given by,

B = μoIN / L

B = (1.26×10−6 × 15 × 360) / 0.8

B = 8.505 × 10−3 N/Amps m

The magnetic field generated by the solenoid is 8.505 × 10−4 N/Amps m.

Conclusion

The Magnetic Field in a Solenoid Formula is a key component of the Class 12 Physics curriculum as well as numerous entrance exams at the federal, state, and local levels.

Theoretical and mathematical formulas relating to magnetic fields in a solenoid have been covered by Extramarks. At the end of the article, additional interesting information about magnetic fields is added to keep students interested in the topic.

Extramarks is dedicated to giving its students the best service possible in every way. As a result, it has provided you with all of these study materials at no cost. Students who have read Magnetic Field in a Solenoid Formula from Extramarks will be able to answer all the questions that might come up in this subject on any given exam.