A constant current I is maintained in a solenoid. Which of the following quantities will not increase if an iron rod is inserted in the solenoid along its axis?
When an iron rod is inserted inside a solenoid the magnetic field inside the solenoid increases due to the high magnetic permeability of iron. Due to the increased magnetic field, the magnetic flux also increases. The self-inductance of the solenoid also increases because self-inductance (L) of the coil is directly proportional to the permeability of the material inside the solenoid. The rate of joule heating doesn’t change.
Whenever the magnetic flux () linked with a coil changes the emf is induced in the circuit, This emf lasts
The expression for the magnetic flux
A magnet is moved towards the coil (a) quickly and (b) slowly, and then the work done is
When a magnet is moved towards the coil quickly, the rate of change of flux is larger than that if the magnetic field is moved slowly, thus larger emf is induced due to quick movement of the coil.
A cylindrical bar magnet is held along the axis of a circular conducting loop. Which one of the following movements of the bar magnet will not induce a current in the loop?
The magnetic flux () linked with a coil is related to the number (N) of turns of the coil as
The total number of magnetic lines of force crossing a surface normally is termed as
The measurement of the total magnetic field that passes through a given area is known as magnetic flux. It is helpful in describing the effects of the magnetic force on something occupying a given area.
If we consider a simple flat area A as our example and angle θ as the angle between the normal to the surface and a magnetic field vector, then the magnetic flux is given by the equation:
In the given circuit the maximum deflection in the galvanometer occurs when
A coil of area 10 cm² of 20 turns is placed in uniform magnetic field of 104 gauss. The normal to the plane of the coil makes an angle of 60° with the magnetic field. The flux in maxwell through the coil is
1 gauss=10-4 Tesla
Find the flux contained by the material when the flux density is 11.7 Tesla and the area is 2 units.
The total flux is given by φ = ∫ B.ds, where ∫ds is the area. Thus φ = BA. We get φ = 11.7 x 2 = 23.4 units.