Whenever electric and magnetic fields act on a moving charge, the charge feels a force known as Lorentz Force, a scalar quantity of forces owing to electromagnetic fields. When the electromagnetic field, magnetic Force, and charge motion are mutually perpendicular to one another, they are crossed fields, and forces owing to electric and magnetic fields operate in opposing directions.
As a result, the Lorentz force F will be:
F = qEî+ (qvî x Bk^) = qE – qVBî = q(E – vB) When the intensity of the electric and magnetic fields is modified to equalise the forces due to the electric and magnetic fields (FE = FB), the charge can move freely in the field. ∴ v = E/B
This exceptional situation is employed when charged particles with a specific velocity selector (of value E/B) must pass through the crossing fields undeflected, and this phenomenon is known as a frequency selector. In 1897, JJ. Thomson used it to calculate the charge-to-mass ratio.
This velocity selector principle is also utilised in mass spectrometers to identify charged particles based on their charge to mass ratio. Velocity selector is the region to find magnetic and uniform fields.
A solenoid magnetic field is a wire coil that is intended to create a strong magnetic field inside it. When an electric current is fed via a wire that has been wrapped several times around a cylinder, it produces a powerful magnetic field. The magnetic field becomes stronger as the number of loops increases. The number N specifies the number of revolutions of the solenoid.
A solenoid magnetic field is a type of electromagnet designed to generate a regulated magnetic field. If the objective of a solenoid is to obstruct changes in electric current, it is characterised as an inductor.
The magnetic field of a solenoid may be calculated using the formula,
Please keep in mind that the magnetic field in the coil is equal to μo times the supplied current as well as the inductance per unit length.
Solved Examples 1
Determine the magnetic field created by an 80-cm-long solenoid with 360 coil turns and a current of 15 amps flowing through it.
Solution:
Given:
μo= 1.26 107 T/m permeability
In a solenoid formula, the magnetic field is given by,
Example 2 A 40 cm diameter solenoid has a magnetic field of 2.9105 N/Amps m. Calculate the current that passes through it if it has 300 turns.
The magnetic field formula is as follows:
The magnetic field inside a solenoid is a tightly coiled helical coil of wire with a tiny diameter relative to its length. The magnetic field created in the core of a current-carrying solenoid is virtually homogenous and oriented along the solenoid’s axis.
The magnetic field is much weaker outside of the solenoid. The solenoid is made of a single helical wire that carries an electric current I. The wrapping is sufficiently tight that each turn of the solenoid may be approximated as a circular wire loop located in the plane perpendicular to the solenoid’s axis and carrying a current I.
A current-carrying conductor in a magnetic field feels a force. If the ground and current directions are perpendicular to one another, then the resultant force on the conductor will be perpendicular to both using Fleming’s rule.
When current flows through a conductor, it is displaced, indicating the presence of a force on the conductor.
Fleming’s Left-Hand Rule: Make a straight angle with your thumb and the first two fingers on your left hand. The thumb will point in the direction of Force if the forefinger points in the direction of the field and the second finger in the direction of the Current.
Charged particles have long been known to travel in circular orbits under a magnetic field. Magnetic fields are also utilised in accelerators for both scientific and medicinal applications to direct the travel of charged particles. Velocity selector is the region to find magnetic and uniform fields. The circling motion of charges in a magnetic field is used to calculate an atom’s mass.