Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area $\mathrm{A}= 10 \mathrm{~cm}^{2}$ and length $=20 \mathrm{~cm}$. If one of the solenoids has 300 turns and the other 400 turns , their mutual inductance is ( $\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$ )

(D) $2.4 \pi \times 10^{-4} \mathrm{H}$

(A) $2.4 \pi \times 10^{-5} \mathrm{H}$

(B) $4.8 \pi \times 10^{-4} \mathrm{H}$

(C) $4.8 \pi \times 10^{-5} \mathrm{H}$

JEE Main · 2008 · 27 Apr 2008 · Offline · Electromagnetic Induction

Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross sectional area

A = 10 cm^{2}

and length

= 20 cm

. If one of the solenoids has 300 turns and the other 400 turns , their mutual inductance is (

μ_{0} = 4 π \times 1 0^{- 7} Tm A^{- 1}

)

Answer: (D)

Two coaxial solenoids wound on the same pipe share essentially all their flux, so their mutual inductance is

M = \frac{μ _{0} N _{1} N _{2} A}{ℓ}

. Inserting

N_{1} = 300

N_{2} = 400

, cross-section

A = 10 c m^{2} = 1 0^{- 3} m^{2}

, length

ℓ = 0.20 m

and

μ_{0} = 4 π \times 1 0^{- 7}

, the product

N_{1} N_{2} A = 300 \times 400 \times 1 0^{- 3} = 120

, so

M = \frac{4 π \times 1 0 ^{- 7} \times 120}{0.20} = 2.4 π \times 1 0^{- 4} H

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option