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The magnetic flux through one turn of a solenoid of self-inductance $8.0 \times 10^{- 5} H$ and cross sectional area A ,when a current of $3.0 A$ flows through it is $ϕ$ . Assume that the solenoid has 1000 turns and is wound from wire of diameter $1.0 m m$ .

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$ϕ =$ $4.4 \times 10^{- 7} W b$

$ϕ = 2.4 \times 10^{- 7} W b$

$A = 4.37 \times 10^{- 5} m^{2}$

$A = 6.37 \times 10^{- 5} m^{2}$

answer is B.

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Detailed Solution

From the relation, $L = \frac{N ϕ}{i}$

The flux linked with one turn,

$ϕ = \frac{L i}{N} = \frac{(8.0 \times 10^{- 5}) (3.0)}{1000} = 2.4 \times 10^{- 7} W b$

Now , $ϕ = B S = (μ_{0} n i) (S)$

a. What is the magnetic flux through one turn of a solenoid of self inductance 8.0xx10^-5H when a current of 3.0 A flows throgh it? Assume that the solemoid has 1000 turns

Here, n = number of turns per unit length

$= \frac{N}{l} = \frac{N}{N d} = \frac{1}{d} ϕ = \frac{μ_{0} i S}{d}$

$S = \frac{ϕ d}{μ_{0} i} = \frac{(2.4 \times 10^{- 7}) (1.0 \times 10^{- 3})}{(4 π \times 10^{- 7}) (3.0)} = 6.37 \times 10^{- 5} m^{2}$

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