Two long co-axial solenoid of same length are shown below: b Derive an expression for mutual inductance of two co-axial solenoids.

Let r 1 and r 2 be the radii of inner (let it be primary P ) and outer (let it be secondary S ) co-axial solenoids respectively, and n 1 and n 2 be the number of turns per unit length of the two solenoids. Let N 1 and N 2 be the total number of turns in two solenoids and each of length L . Let the secondary solenoid carry current I 2 . This current sets up magnetic flux ϕ 1 through the primary (inner) solenoid. The total flux linkages with the primary solenoid are given by N 1 ϕ 1 = M 12 I 2 where , M 12 is the mutual inductance of the primary (inner) solenoid with respect to the secondary (outer) solenoid. Magnetic field at the centre of the secondary solenoid due to a current I 2 is given by B 2 = μ 0 n 2 I 2 . The total flux linkages with the primary solenoid are given by N 1 ϕ 1 = ( n 1 L ) B 2 I 1 N 1 ϕ 1 = ( μ 0 π n 1 n 2 L r 1 2 ) I 2 M 12 = N 1 ϕ 1 I 2 = μ 0 π n 1 n 2 L r 1 2 Similarly M 21 = N 2 ϕ 2 I 1 = μ 0 π n 1 n 2 L r 1 2

Two long co-axial solenoid of same length are shown below: Write the dimension of mutual inductance.

We know, The emf induced in the secondary coils is directly proportional to the rate of change of current in primary coil. i.e e 2 ∝ di dt . ⇒ e 2 = M di dt Where M is the coefficient of matual inductance. Now, The dimension of mutual inductance M = Dimension of emf induced Dimension of current × Dimension of time = ML 2 T - 3 A - 1 A T = ML 2 T - 2 A - 2 .

Two long co-axial solenoid of same length are shown below: Write the dimension of mutual inductance.

We know, The emf induced in the secondary coils is directly proportional to the rate of change of current in primary coil. i.e e 2 ∝ di dt . ⇒ e 2 = M di dt Where M is the coefficient of matual inductance. Now, The dimension of mutual inductance M = Dimension of emf induced Dimension of current × Dimension of time = ML 2 T - 3 A - 1 A T = ML 2 T - 2 A - 2 .

Two long co-axial solenoid of same length are shown below: b Derive an expression for mutual inductance of two co-axial solenoids.

Let r 1 and r 2 be the radii of inner (let it be primary P ) and outer (let it be secondary S ) co-axial solenoids respectively, and n 1 and n 2 be the number of turns per unit length of the two solenoids. Let N 1 and N 2 be the total number of turns in two solenoids and each of length L . Let the secondary solenoid carry current I 2 . This current sets up magnetic flux ϕ 1 through the primary (inner) solenoid. The total flux linkages with the primary solenoid are given by N 1 ϕ 1 = M 12 I 2 where , M 12 is the mutual inductance of the primary (inner) solenoid with respect to the secondary (outer) solenoid. Magnetic field at the centre of the secondary solenoid due to a current I 2 is given by B 2 = μ 0 n 2 I 2 . The total flux linkages with the primary solenoid are given by N 1 ϕ 1 = ( n 1 L ) B 2 I 1 N 1 ϕ 1 = ( μ 0 π n 1 n 2 L r 1 2 ) I 2 M 12 = N 1 ϕ 1 I 2 = μ 0 π n 1 n 2 L r 1 2 Similarly M 21 = N 2 ϕ 2 I 1 = μ 0 π n 1 n 2 L r 1 2

What do you mean by eddy current? Write any two applications of it.

When magnetic flux in a metal block changes, induced currents are produced. This current is called eddy current. Applications: Magnetic breaking in trains, Damping, Induction furnace Hence, it is the required answer.

Derive an expression for self inductance of a solenoid.

We are considering a solenoid with nn turns with length l . The area of cross-section is A . The solenoid carriers current I and B is the magnetic field inside the solenoid. The magnetic field B is given as, B = μ ∘ nI l Where, μ ∘ is the permeability of free space, n is the number of turns and I is the current in the solenoid and l is the length. The magnetic flux is the product of the magnetic field and area of the cross-section. Here the magnetic flux per turn is given as, ϕ = B × A Substituting the values in the above expression, ϕ = μ ∘ nI l × A Hence, there is n number of turns, the total magnetic flux is given as, ϕ = μ ∘ nI l × A × n ϕ = μ ∘ n 2 IA l . . . . . 1 If L is the coefficient of self-inductance of the solenoid, then ϕ = LI . . . . . 2 Comparing the two equations we get, LI = μ ∘ n 2 IA l L = μ ∘ n 2 A l So the expression for the coefficient of self-inductance is μ ∘ n 2 A l .