CBSE Class 12 · 2017 · Delhi · Electromagnetic InductionShort Answer

(a) Define the term 'self-inductance' and write its S.I. unit.
(b) Obtain the expression for the mutual inductance of two long co-axial solenoids

S_{1}

and

S_{2}

wound one over the other, each of length

L

and radii

r_{1}

and

r_{2}

, with

n_{1}

and

n_{2}

turns per unit length, when a current

I

is set up in the outer solenoid

S_{2}

Answer: Self-inductance is flux linkage per unit current; unit henry (H). $M = μ_{0} n_{1} n_{2} π r_{1}^{2} L$ .

Solution

(a) The self-inductance of a coil is the magnetic flux linked with it per unit current; its S.I. unit is the henry (H).
(b) Let a current

I

flow in the outer solenoid

S_{2}

. The uniform field it produces inside is

B_{2} = μ_{0} n_{2} I

This field threads the inner solenoid

S_{1}

(cross-section

π r_{1}^{2}

, total turns

n_{1} L

). The flux linkage with

S_{1}

N_{1} ϕ_{1} = (n_{1} L) (π r_{1}^{2}) (μ_{0} n_{2} I) = μ_{0} n_{1} n_{2} π r_{1}^{2} L I

Since

N_{1} ϕ_{1} = M I

M = μ_{0} n_{1} n_{2} π r_{1}^{2} L

The inner radius

r_{1}

appears because the linking flux is confined to the inner solenoid's cross-section.

CBSE Class 12 Board · 2017 · Delhi · Subjective