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Two coaxial solenoids A and B spaced close to each other and have 400 and 700 turns, respectively. A current of $3.50 A$ in coil A produced an average flux of $300 µ T - m ²$ through each turn of A and a flux of $900 µ T - m ²$ through each turn of B.
(a) Calculate the mutual inductance of the two solenoids.
(b) What is the self-inductance of A?
(c) What emf is induced in B when the current in A increases at the rate of $0.5 A / s$ ?

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$5.8 \times 10^{- 2} H, 3.43 \times 10^{- 2} H, 7.0 \times 10^{- 3} V$

$1.8 \times 10^{- 2} H, 7.43 \times 10^{- 2} H, 7.0 \times 10^{- 3} V$

$7.8 \times 10^{- 2} H, 5.43 \times 10^{- 2} H, 9.0 \times 10^{- 3} V$

$1.8 \times 10^{- 2} H, 3.43 \times 10^{- 2} H, 9.0 \times 10^{- 3} V$

answer is D.

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Detailed Solution

(a)

$M = \frac{N_{B} ϕ_{B}}{i_{A}} = \frac{(700) (90 \times 10^{- 6})}{3.5} = 1.8 \times 10^{- 2} H$

(b) $L_{A} = \frac{N_{A} ϕ_{A}}{i_{A}}$

$= \frac{(400) (300 \times 10^{- 6})}{3.5} = 3.43 \times 10^{- 2} H$

(c)

$e_{B} = M (\frac{d i_{A}}{d t}) = (1.8 \times 10^{- 2}) (0.5) = 9.0 \times 10^{- 3} V$

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