Using Ampere's circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis
Answers
Answer:
Ampere's circuital law states that line integral of magnetic field around any closed loop is equal to μ
o
times the electric current flowing through the cross-section area enclosed by that loop.
Mathematically, ∮B.dl=μI
Let the current flowing in the solenoid having number of turns per unit length n be I.
Magnitude of magnetic field inside the solenoid is B while at outside is zero.
Now ∮
loop
B.dl=∫B
ab
.L+∫B
bc
.L
′
+∫B
cd
.L+∫B
da
.L
′
The value of first term ∫B
ab
.L=BL
The second and fourth term are zero because angle between magnetic field and the length loop is 90
o
.
The third term is also zero as the value of magnetic field outside the solenoid is zero.
Total current flowing through the loop I
total
=(nL)I
From Ampere's circuital law, we get BL=μ
o
(nLI)
Answer:
(a) Magnetic Field Due to a Current Carrying Long Solenoid: A solenoid is a long wire wound in the form of a close-packed helix, carrying current. To construct a solenoid a large number of closely packed turns of insulated copper wire are wound on a cylindrical tube of card-board or china clay. When an electric current is passed through the solenoid, a magnetic field is produced within the solenoid. If the solenoid is long and the successive insulated copper turns have no gaps, then the magnetic field within the solenoid is uniform; with practically no magnetic field outside it. The reason is that the solenoid may be supposed to be formed of a large number of circular current elements. The magnetic field due to a circular loop is along its axis and the current in upper and lower straight parts of solenoid is equal and opposite. Due to this the magnetic field in a direction perpendicular to the axis of solenoid is zero and so the resultant magnetic field is along the axis of the solenoid. If there are ‘n’ number of turns per metre length of solenoid and I amperes is the current flowing, then magnetic field at axis of long solenoid B = μ0nI If there are N turns in length l of wire, then n = N/l or B = μ0NI/l Derivation: Consider a symmetrical long solenoid having number of turns per unit length equal to n. Let I be the current flowing in the solenoid, then by right hand rule, the magnetic field is parallel to the axis of the solenoid. Field outside the solenoid: Consider a closed path abcd. Applying Ampere’s law to this path (since net current enclosed by path is zero) As dl ≠ 0 ∴ B = 0 This means that the magnetic field outside the solenoid is zero. Field Inside the solenoid: Consider a closed path pqrs. The line integral of magnetic field vector B along path pqrs is
(b) In a toroid, magnetic lines do not exist outside the body./Toroid is closed whereas solenoid is open on both sides./Magnetic field is uniform inside a toroid whereas for a solenoid, it is different at the two ends and cenre
(c) The magnetic field lines of toroid are circular having common centre. Inside a given solenoid, the magnetic field may be made strong by (i) passing large current and (ii) using laminated coil of soft .